Thermodynamics
January 7, 2017 | Author: Priyansh Mishra | Category: N/A
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THERMODYNAMICS & THERMOCHEMISTRY
Contents Topic
Page No.
Theory
01 - 29
Exercise - 1
30 - 38
Exercise - 2
39 - 44
Exercise - 3
45 - 49
Exercise - 4
50 - 55
Exercise - 5
56 - 57
Answer Key
58 - 59
Syllabus THERMODYNAMICS & THERMOCHEMISTRY First law of thermodynamics; Internal energy, work and heat, pressure-volume work; Enthalpy, Hess's law; Heat of reaction, fusion and vapourization; Second law of thermodynamics; Entropy; Free energy; Criterion of spontaneity.
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THERMODYNAMICS & THERMOCHEMISTRY Introduction : Thermodynamics : The branch of science which deals with different forms of energy & their interconversion. THERMODYNAMICS
First law of Thermodynamics
Thermochemistry (application of I law in chemical reactions)
Second law of Thermodynamics
Application of thermodynamics : In chemistry using thermodynamics
We can predict feasibility of the reaction that is if two substances are mixed then the reaction between them will takes place or not. If reaction does take place then what are the energy changes involved during the reaction. If in a chemical reaction, equilibrium is going to get attained then what will be the equilibrium concentrations of different reactants & products, can be calculated with thermodynamics.
Limitations of Thermodynamics : Laws of thermodynamics are applicable to matter in bulk or on system as a whole, these can not be applied on individual particles(temperature, pressure, enthalpy etc have meanings only for system as a whole). Using thermodynamics we cannot calculate the time taken for completion of a reaction or for attainment of chemical equilibrium.
Terms to be used in Thermodynamics :
System : Part of the universe which is under study for energy changes.
Ex.
Air in a room, water in a bottle, any living body.
Surrounding : Rest of the universe.
Universe : Universe = System + Surroundings
Boundary : Anything which separates system & surroundings is called boundary. Boundary can be real or imaginary. Boundary can be flexible or rigid e.g. - air in a flexible balloon (flexible boundary) while air in a room (fixed boundary). Boundary can be adiabatic(non-conducting) or diathermic(conducting).
Types of system :
Open system : System which can exchange energy & matter both with the surroundings. e.g. : Living systems( any living organism) are open systems, air in an open room
Closed system : System which can exchange only energy but cannot exchange matter with the surroundings is called closed system. e.g. : any matter in a closed container.
Isolated system : System which cannot exchange energy and matter both with the surroundings. e.g. : Water in thermos flask.(Though not a perfectly isolated system but can be taken as, for small interval of time as the energy exchanges are negligible).
Whole of universe is a perfect isolated system.
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THERMODYNAMICS_JEE MAINS # 1
.
Open System
Close System
Isolated system
State of a system :
It means the condition in which the system is present. It can be specified/defined by measuring/ specifying some observable/measurable properties of the system like pressure, volume, temperature, amount of substance, elasticity, heat capacity etc. e.g. For an ideal gaseous system state of the system can be defined by specifying volume, temperature and pressure. We may have to specify more properties of the system depending on the complexity of the system.
State function : Property of a system which is dependent only on the state of the system i.e. it is a point function It is independent of the path adopted to attain a particular state. e.g. In Mechanics, Displacement of any object will a state function but distance travelled by the object will be a path function. For any thermodynamic system, Temperature, Pressure, Volume, Total internal energy (E or U), Enthapy(H), Gibbs free energy (G), Entropy (S) are all state functions. e.g.
In the above example the final temperature, pressure, and the volume will be same in both the above ways but the work involved and the heat exchanged during the processes will be different.
Path function :
Quantities which are dependent on the path/way the system has achieved a particular state. e.g. Heat, work, Heat capacities(Molar heat capacities, specific heat capacities etc.). These quantities are define when there is a process going on. These can not have any definite (particular) value in any particular state of the system.
Types of preperties
Extensive properties : Functions or properties of the system which are dependent on mass or on size of the system are called Extensive Properties .
Extensive functions are additive in nature( The addition of the volumes of the two parts equals the volume of the whole of the room.) e.g. Volume, Mass, Total heat capacity, Total internal energy (E), Enthalpy(H), Gibbs Free Energy(G), Entropy(S).
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THERMODYNAMICS_JEE MAINS # 2
Intensive properties : Functions or properties which are not mass dependent or size dependent are called intensive function. Intensive properties are not additive in nature. eg. Temperature, pressure, molar heat capacity, specific heat capacity, density,concentration, vapour pressure. How to identify extensive or intensive properties If a system in a particular state is divided into two equal or unequal parts, the properties which have value equal to the original value of that property for the whole of the system is called an Intensive property. While the properties which have values different from the values for whole of the system are called Extensive Properties.
For example consider air in a room at temp of 300K, 1 atm pressure. Now, if the room is divided by some boundary( imaginary or real) into two parts( equal or unequal) then in these two parts : The temperature, pressure, density of the gas, concentration of gaseous molecules etc. will have the same value as that of for whole of the system. (intensive) While the volume of two parts, mass of gas in two parts, total energy of the gaseous molecules in the two parts, entropy the two parts etc. will be different from the values of these properties as for the whole of the system initially. (extensive)
Thermodynamic equilibrium :
When there is no change in any observable or measurable property of a system with time then the system is said to be in thermodynamic equilibrium. Thermodynamic equilibrium consist of three types of equilibrium. (a) Mechanical equilibrium (b) Thermal equilibrium (c) Chemical equilirbrium
Mechanical equilibrium :
There should not be any pressure gradient with time or with space (for any ideal gaseous system, for a liquid system there can be pressure gradient with space as pressure at the bottom of the container in which a liquid is filled will be greater than the pressure at the surface of the liquid.) in the system.
Thermal equilibrium : There should not be any temperature gradient (difference). Temperature may have different values at different places/locations in a system but it should remain constant with time.
Chemical equilibrium : There should not be any concentration gradient of any of the species in the system.
Types of thermodynamic process on basis of state/conditions Thermodynamic process : Any method/process by which system can change its state from one state of thermodynamic equilibrium to another state of thermodynamic equilibrium. There can be infinite type of thermodynamic processes, out of these the following are important ones: ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 3
1.
Isothermal process :
2.
Isochoric process :
3.
Isobaric process :
4.
Adiabatic process :
T = constant dT = 0 T = 0 V = constant dV = 0 V = 0 P = constant dP = 0 P = 0 q = constant or heat exchange with the surrounding = 0(zero)
Types of thermodynamics processes on basis of the way the processes are carried out : Reversible process :
The process that can be reversed by a very small change is known as reversible process. If a process is carried out in such a manner so that the system is always in thermodynamic equilibrium at every stage of the process. If the process is carried out such that the difference in driving force and opposing force is infinitesimally small so that process takes place at infinitesimally slow rate. Fdriving – Fopposing = dF and dF 0
An ideal reversible process will take infinite time to get completed.
It is carried out infinitesimally slowly.
Strictly speaking there is no ideal reversible process in universe.
To get an idea of a reversible process we can consider the following system. An ideal gas is enclosed in a container and a massless piston is put on the gas on which a pile of sand is placed having particles of negligible mass. To carry out a reversible expansion we will slowing decrease the mass of the sand lets say by removing the particles one by one, so the expansion of the gas will take place at infinitesimally small rate and we can always assume the system to in thermodynamic equilibrium. So, the expansion will be of reversible type. FBD of piston For piston to be in equilibrium : Pgas = Patm + Mg/A
Irreversible process : The process can not be reversed by a small change is known as irreversible. If a process is carried out in such a manner so that the system is in thermodynamic equilibrium (I) Only at initial & final state of the process but not at the intermediate stages. (II) System may be in thermodynamic equilibrium state at some finite number of intermediate stages only - for example - n step irreversible expansion of a gas If during the process there is a finite difference in driving force and opposing force so that process takes place with a finite rate. Irreversible processes will get completed in finite time. At intermediate stages of the irreversible process, different state function such as Pressure, temperature etc. are not defined. All real process are irreversible.
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THERMODYNAMICS_JEE MAINS # 4
Consider the above system. If the stopper placed over the piston is removed, then the piston will move with almost infinite acceleration and will keep moving to a position where the pressure of the gas becomes equal to the external pressure. Since the process will get completed in finite time and there was a finite difference between the driving force and the opposing force so, process is irreversible.During the process, the pressure of the gas can not be defined as it will be having different values at different locations.
Modes of energy exchange : These are two ways by which a system can interact or can exchange energy with its surroundings. (i) Heat & (ii) Work Heat & Work both are forms of energy. Heat : When the energy transfer takes place because of temperature difference between system & surroundings. It is known as heat. Work : Energy transfer which is not heat or which is not because of temperature difference is called work. Work can be of many types : Mechanical work, Electrical work, Magnetic work, Gravitational work etc. The same energy transfer can be called work or can also be called heat depending on choice of the system. To understand this, consider a system shown below in which water is taken in a closed container at 25ºC, the surroundings is also at temperature of 25ºC and there is a heater coil in the dipped in the water which is connected to a battery through a switch S.Heater coil is also at 25ºC initially.
Now, there are two ways in which system can be chosen –System : All contents of the container (water + Heater coil). When switch is turned on there will be increment in the temperature of the system. Since the temperature of the surroundings was equal to temperature of the system so, heat can not flow but still there is increment in the energy of the system and hence, there is temperature increment. This must be because of electrical work done by the battery on the system not because of the heat transfer as initially temperatures were equal. –System : Water only is our system. Heater coil will be part of the surroundings. In this case when switch is turned on the temperature of the heater coil will increase first so there will be a temperature difference between system & surroundings. Hence, this energy transfer will be called heat.
IUPAC Sign convention about Heat and Work Any energy given to system is taken positive so heat given to system = positive heat given out from system / taken out from system = Negative Work done on the system = Positive Work done by the system = Negative The sign convention is different from physics, but the meaning always comes out to be same only in equation we have to use a different sign convention for work. So if in any problem, w = – 10 J It means system has done work of 10 Joule on surroundings. According to Chemistry :
Q
Q In Physics
w U = Q + W
w U = Q – W
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THERMODYNAMICS_JEE MAINS # 5
IUPAC convention of Heat 1. Heat given to the system = +ve 2. Heat coming out of the system = –ve 3. Work done on the system = +ve 4. Work done by the system = –ve
First Law of Thermodynamics (FLOT) : Law of energy conservation : Energy of total universe is always conserved. or Total energy of an isolated system is always conserved. or Hence absolute value of E can never be calculated only change in value of E can be calculated for a particular process. Mathematical form of First Law of thermodynamics. If a system is intially in a particular state in which its total internal energy is E1. Now q amount of heat is given to it and w amount of work is done on it so that in new state its total internal energy becomes E2. Then according to 1st Law of thermodynamics. E2 = E1 + q + w so E = (E2 – E1) = q + w Application of First Law U = Q + W U = Q – PV
Since
W = – P V
Calculation of different quantities in First Law of Thermodynamics (FLOT) : Calculation of E : Thermodynamic definition of an ideal gas : If for a gas the internal energy is directly proportional to its absolute temperature then the gas is termed as an ideal gas. E E =0, =0 so V T P T
Internal Energy (E, also denoted by U) :
Every system having some quantity of matter is associated with a definite amount of energy, called internal energy.
It is the sum of all forms of energies present in the system. E = ETranslational + ERotational + EVibrational + Ebonding + ..... E = EFinal EInitial .
E = qv, heat supplied to a gas at constant volume, since all the heat supplied goes to increase the internal energy of the gas .
It is an extensive property & a state function . It is exclusively a function of temperature. If
T = 0 ;
E = 0 as well.
With change in temperature only kinetic energy changes.
Degree of freedom The total no of modes on which a molecule of an ideal gas can exchange energy
during collisons is known as its degrees of freedom. Translational degree of freedom = 3 - for all type of gases. Rotational degree of freedom =0 – monoatomic gases =2 – diatomic or linear polyatomic gases =3 – non-linear polyatomic gases. If “f” is initial degree of freedom for that gas. f=3 for monoatomic =5 for diatomic or linear polyatomic =6 for non - linear polyatomic ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 6
Law of equipartion of energy : 1 kT is associated with each degree of freedom per ideal gas molecule 2 Where k is Boltzmann constant Energy equal to
U/molecule = f x U/mole =
1 kT 2
f RT (k × NA = R) 2
For n moles, f U = nRT only for ideal gas. 2 f U = nRT 2
Calculation of Heat (q)
Heat is a path function and is generally calculated indirectly using Ist Law of thermodynamics First calculate E and W & then q or heat can be calculated if heat capacity of any process is given to us. Heat capacity is a path function and different type of heat capacities are defined
Remember heat capacity of a substance is not fixed it is dependent on type of process which is being performed on that substance
Total Heat Capacity (CT) Heat required to raise the temperature of system by 1ºC under the given process is known as total heat capacity. Mathematically, CT =
q dq J/ºC T dT
It is extensive properties and path function. So, dq = CTdT on integrating
q = CTdT
Molar heat capacity (C) Heat required to raise temperature of 1 mole of a subtance by 1ºC q dq J mole–1 K–1 nT ndT dq = nCdT
Mathematically, C = So,
q=
nCdT nCT
C is intensive path function. Cp is molar heat capacity at constant pressure CV is molar heat capacity at constant volume Cp and CV are intensive but not a path function
Specific heat capacity (s) : Heat required to raise temperature of unit mass (generally 1 g) of a substance by 1ºC. q dq Jg–1 K–1 m T mdT dq = msdT
S= So,
q=
dq ms dT msT
S is intensive path function SP is specific heat capacity at constant pressure SV is specific heat capacity at constant volume SP & SV are intensive but not a path function ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 7
Total heat capacity, molar heat capacity & specific heat capacity of a process on a substance are related as CT = nC = ms & C = Ms Where m – weight of substance M – molar mass of substance n – no. of moles of the substance For For
isothermal process isochoric process
C =± C = Cv
For For
isobaric process adiabatic process
C = Cp C=0
Heat capacity can have value from – to + depending on the process.
WORK DONE (w) : Energy that is transmitted from one system to another in such a way that difference of temperature is not directly involved is known as work. It is a path function. This definition is consistent with our understanding of work as dw = Fdx. The force F can arise from electrical, magnetic, gravitational & other sources. Units : Heat & work both are forms of energy . Hence, their units are units of energy. i.e.S system: Joules (J). Much data is available in the old units of calories (cal) as well. P × V = (litre. atmosphere) term which has unit of energy . It is useful to remember the conversion 1 litre. atm = 101.3 Joules = 24.206 cal
For irreversible processes, state parameters such as P,T etc cannot be defined. Hence, work cannot be estimated using Pgas. But by the work energy theorem W gas = – W ext + Kpisston
When the piston comes to rest again Kpiston = 0
W gas = – W ext = –
P
ext dv
as the external pressure is always defined hence, for all processes work can be calculated using
W ext = – Pext .dv
Work : dW = Fext dx P-V Work : Assuming that under an external force Fx, the piston moves down by a distance ‘dx’.
Fext dx
Fext ( A dx ) dW = Pext (dV) W ext = Pext dV & W gas = – W ext = – Pext dV A As the work done in the above case by the external agent is +ve and as the expression conveys otherwise, hence a -ve sign is introduced. dW
dW = – Pext. dV
Calculation of work for different type of process on an ideal gas.
(A) Isothermal expansion : There are many ways in which a gas can be expanded isothermally. (a)
Isothermal reversible expansion :
mg = Pgas = P (always) A In reversible process, Pext = P0 +
Pext= Pgas (thermodynamic equilibrium always) Since process is isothermal ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 8
nRT V
P=
Vf
W= –
Vf
Vf
Pext dV = –
Vi
P dV
W= –
Vi
Vi
Vf W = –nRT ln V i
nRT dV V
PV diagram Representation In expansion work is done by system on the surroundings and Vf > Vi W = –ve Work = Area under the P–V diagram (b)
Irreversible isothermal expansion :
(i) Single step isothermal expansion We are assuming expansion against atmospheric pressure which need not be the case in a given problem. A mass equal to m0 is placed on piston initially to maintain equilibrium. Initially, Pi = Pgas = Patm + m0 g/A
PV diagram Representation For expansion to take place, m0 mass is suddenly removed so gas expands against constant external pressure of Patm In this case, the pressure of the gas will not be defined as the sudden expansion has taken place so all the molecules of sample will not get the information of expansion simultaneously, there will be a time gap and hence, there will be a state of turbulence. From some intermediate state of volume ‘V’, the work done is slight expansion from V (V + dV) dw = – Pext . dV
(IUPAC sign convention) Vf
So,
W=
dw Pext .dv
Vi
W = – Pext (Vf – Vi) ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 9
Only initial and final states can be located (as at intermediate stages pressure of the gas is not defined) (ii) Two step isothermal expansion : Mass M0 is divided into two mass (may be equal or unequal)
M0 (m1 m 2 ) Now, if m1 only is removed, then the expansion of gas will take place against constant external pressure
Pext1 = (Patm + m2g/A) and this expansion will take place only upto volume V1 such that
Pext1 . V1 = Pi Vi (isothermal) Now, if second mass m2 is also removed then expansion V1 Vf will take place against constant pressure
Pext2 = Patm
So,
work done is expansion Vi V1 W 1 = – (Patm +M2g/A) (Vi – V1) & work done is expansion V1 Vf W 2 = – Patm (Vf – Vi) Total work = W 1 + W 2
Patm M1
Vf
M0
V Vi
PV Diagram representation
Work done in this irreversible expansion is greater than work done by gas during the single stage expansion of gas and so on for three step expansion we divide the mass m0 into three masses m1, m2 and m3 and remove these step by step and so on. (iii) For n step expansion and n Irreversible process becomes
PV Diagram representation
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THERMODYNAMICS_JEE MAINS # 10
(B) isothermal compression of an ideal gas : (a)
Reversible isothermal compression of an ideal gas
This can be achieved by placing particles of sand one by one at a very slow take in the assembly which keeps the temperature of gas constant in this case the expression of work done will be exactly similar to as obtained in case of reversible expansion of gas W = – nRT ln(Vf/Vi) This will automatically come out to be +ve as Vf < Vi
(Patm + M0g/A) = Pf Patm
Vi Vi
Patm Vf
(b)
Irreversible isothermal compression of an ideal gas
(i)
Single step compression : To compress gas a mass m0 is suddenly placed on massless piston dw = – Pext. dv = – (Patm + m0g/A) dv so. to calculate total work done on the gas Vf
W=
dw V
i
(ii)
Pext . dv ;
Vi
W = – Pext (Vf – Vi)
Two step compression : Place mass m0 in two fragments (m1 + m2) the graphical representation will make the calculation of work done.
If m1 is placed first, then the first compression has taken place aganist external pressure of (Patm + m1g/A) So, W 1 = – (Patm + m1 g/A) (V1 – Vi) Simlarly, W 2 = – (Patm + m0 g/A) (Vf – Vi) Note : If process takes place in n steps and n then process will be like reversible compression ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 11
Conclusion : Whenever work is done on the gas then it will be minimum in case of reversible process. Thats why different machines/engines are designd to work reversibly so maximum output can be obtained but minimum input is given to it output – work done by engine/machine/system input – work done by system us a surrounding If expansion/compression takes place against constant external pressure then it is irreversible. If there is sudden change then the process is irreversible. A Reversible and irreversible isothermal process.
Except the infinite stage compression/expansion, all are irreversible. We can redefine reversible and irreversible as follows :
B
Reverssible process : If a process operates is such a fashion that when it is reversed back both the system as well as surroundings are restored to their initial position w.r.t. both work and heat, is known as reversible process. If for the process A B work = w, heat = Q then if for the process B A, work = – w, heat = –Q then the process is reversible. If the external pressure is constant in isothermally process, process is irreversible
Calculation of Cp and CV (a)
Constant volume process (Isochoric) dU = dq + dw dU = dq – pdV dU = (dq)v Heat given at constant volume = change in internal energy du = (dq) dU = (nCdT)v dU = nCvdT Cv is Specific molar heat capacity at constant volume. 1 dU 1 d ( fnRT ) / 2 fR = n dT n dT 2
Cv = . (b)
Constant pressure process (Isobaric) : dU = dQ + dW dU = dQ – PdV dQ = dU + PdV ............... (i)
Defining a new thermodynamic function H Enthalpy It is a state function and extensive property It is mathematically defined as : H = U + PV as dH = dU + d (Pv) as P = constant dH = dU + PdV .............. (ii) from equation (i) & (ii) dH = (dq)p only at constant pressure. Heat given at constant pressure = Change in enthalpy dH = (nCdT)p dH = nCp dT
Relation between Cp ad Cv for an ideal gas H = U + PV dH = dU + d(PV) for an ideal gas PV = nRT d(PV) = d(nRT) = nRdT nCp dT = nCvdT + nRdT Cp – Cv = R only for ideal gas Mayer’s Relationship ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 12
ADIABATIC PROCESS :
dQ = 0 (no heat changed b/w system and surrounding dU = dQ + dW
nCV dT = – PdV T2
T1
V2
C v .dT T =–
T2 ln T = ln 1
nc
T2 V2 Cv ln T = – P ln V 1 1
T2 T1
R
V . dT
V1
V2 V1
R / c v
T2 V2 –1 = T1 V1 –1
or
V dT
nRT . dv V
V1 V2
1
T V–1 = constant
PV = constant
–1
This is only valid when the quantity PV or TV is constant only for a quasi-static or reversible process. For irreversible adiabatic process these equations are not applicable.
Operation of adiabatic process (a)
Reversible Adiabatic
Operation wise adiabatic process and isothermal process are similar hence all the criteria that is used for judging an isothermal irreversible processes are applicable to adiabatic process. Also, volume in case of isothermal volume is more than that of adiabatic at constant pressure and no of moles, V T
w=–
Pext. dv ,
w=–
K
V
but
. dv ,
work done =
Pext = Pint =
V
1 V1 1 2
V W = – K
P2 V2 P1V1 1
K
1
=
P2 V2 .V2
1
P1V1 .V11 1
(as K = P2V2 = P1V1 )
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THERMODYNAMICS_JEE MAINS # 13
(b)
Irreversible Adiabatic
Adiabatic irreversible expansion –
W=
P
ext.dv
= – Pext (V2 – V1) ,
and
du dw
W = u
W = nCv (T2 – T1)
=
nR(T2 T1 ) P2 V2 P1V1 ) = 1 1
Note: If two states A and B are connected by a reversible path then they can never be connected by an irreversible path.
If the two states are linked by an adiabatic reversible and irreversible path then wrev. = urev. But as u is a state function urev. = uirrev. wirrev. = wrev. as work is a path function. If we assume that wirrev. = wrev. It implies that urev. uirrev.. which again is a contradiction as U is a state function.
Two states A and B can never lie both on a reversible as will as irreversible adiabatic path.
There lies only one unique adiabatic path linkage between two states A and B.
Comparison of Adiabatic Expansion (single stage Vs Infinite stage) Single stage means irreversible process Infinite stage means reversible process In adiabatic expansion process, (W gas)rev > (W gas)irrev urev. > uirrev. (T2)rev. < (T2) irrev (P2)rev.< (P2) irrev (If volume change are same) (V2)rev.< (V2) irrev (If pressure change are same)
Comparison of Adiabatic Expansion (single stage Vs Infinite stage) Single stage means irreversible process Infinite stage means reversible process In adiabatic compression process, (W gas)rev < (W gas)irrev urev. < uirrev. (T2)rev. < (T2) irrev (P2)rev.< (P2) irrev (If volume change are same) (V2)rev.< (V2) irrev (If pressure change are same)
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Comparison of single stage Vs two stage expansion(adiabatic). If the expansion is carried out in two stages then work done in two stage by the gas > work done in one stage by the gas. So, Utwo stage > Usingle stage Tf in two stage < Tf in single stage because decrease in internal energy in two stage is > decrease in internal energy in one stage.
Adiabatic Irreversible process (calculation of state parameters) irrev State A State B P1,V1,T1 P2,V2,T2
W=
P1V1 P2 V2 nR(T2 T1 ) P2 V2 P1V1 = – Pext. (V2 – V1) , T = T 1 1 1 2
Free expansion –
Always going to be irrerversible and since Pext = 0 So, dW = – Pext . dV =0 If no heat is supplied q = 0 then E = 0 So, T = 0.
Calculation of H, U, work, heat etc. Case - I
For an ideal gas undergoing a process. the formula to be used are dU = nCvdT =
f nRdT 2 f
dH = nCpdT = 2 1 nRdT W=–
P
ext . dV
dH = dU + d (PV) H = U + nR T du = dQ + dw
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Case - II
For solids and liquid system : dU = nCvdT dH = W=–
f nRdT 2
(as it is not an ideal gas)
f nCp dT 1 nRT (as it is not an ideal gas) 2
P
ext . dV
H = U + (P2V2 – P1V1) dU = dq + dw Case - III
For chemical reactions : aA (s) + bB() + cC(g) a’A’(s) + b’B’ () + c’ C’ (g) W =– =
Pext. dV
Pext. Vf Vi
= – Pext. [(VA’(s) + (VB’() + (VC’(g)– (VA(s) + (VB() + (VC(g)] = W = – Pext. [VC’(g) – VC(g)]
n'c RT nc RT Pext. (n' c nc )RT =– = – Pext. P Pext. Pext ext W = – (n’c – nc)RT W = – ng RT dU = dq + dw if at constant pressure dq = dH dU = dH – pdV dU = dH – ng RT dH = dU + ng RT Case - IV During phase transformation. Phase transitions generally take place at constant pressure (unless specified) and at constant temperature. So, T = 0 But still H and E are non zero (different from process on an ideal gas) Because during phase transitions though the kinetic energy of molecules of substance remains same but the potential energy gets modified or changed and since E is summation of all type of energies. So, E 0. Also, during this phase transition (PV) 0 Hence, H 0. Now, to calculate heat – generally latent heats of transitions are given and since process is taking place at constant pressure. So, q = H To calculate W, use dW = – Pext dV and then calculate E using Ist law of thermodynamics
IInd Law of thermodynamics : Requirement : Ist law insufficient to Explain Neutralization, Melting of ice > 0° Combustion, Freezing of water < 0° Fall of water , Boiling of water > 100° C Cooling of a hot body, Free expansion of a gas, mixing of gases. Spontaneous /Natural /Irreversible Processes : If a system moves along in a direaction on its own and system, surrounding interactions without the help of any external agency, then that direction is known as spontaneous direction and the process is knwon as a spontaneous process.
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Limitation : First law cannot predict the direction of any process. But in nature a process is found to have a specific direction at certain conditions. 1. Energy can be the factor : To achieve stability * All exothermic processes are spontaneous. * All endothermic processes are non spontaneous. Exceptions * melting of ice (above 0°C) although H > 0 , E > 0 * Vaporization of water (above 100°C) H > 0 , E > 0 * Spreading of ink in water H = 0 , E = 0 * Mixing of two gases * Endothermic solution of salt H > 0 , E > 0 2.
It is observed that nature tries to distribute energy and matter uniformaly (randomization)
Second law of thermodynamics : Entropy of the universe is constanlty inereasing. S universe = S system + S surrounding > 0
for a spontaneous process.
Where (S) = entropy is the measure of randomness or disorder. It is a state function and extensive property. Mathematically ds =
dq rev or S = T
dq rev T
dq is state function only for reversible process T 1 T At Higher T - entropy is already high. Therefore, heat addition will not introduce much change in the entropy. dS dqrev , dS
Entropy Calculation :
Calculation of Ssystem
As ‘S’ is a state function ‘S’ will not depend on path. S irrev = S rev B
S =
A
dqrev T reversible path
Entropy calculation for an ideal gas undergo in a process. irr
State A
State B
Sirr
P1, V1, T1
P2, V2, T2
As ‘S’ is a state function.So, S will not depend upon path B
Sirrev = Srev, system =
A
dq rev T
from the first law du = dQ + dw dQ = du – dw for reversible path dq = dqrev dqrev = du + Pgas dV Srev, system =
rev path.
{Pext = Pgas reversible}
B
du Pgas dV
T
A
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for an ideal gas Pgas
T
nR & du = ncV dT V T2
S sys =
T1
nc v dT T
V2
V1
nRdV V
T2 V2 Ssystem = ncv n T + nR n V 1 1
Calculation of Ssurrounding : As the surrounding is an infinite heat reservoir hence, no. amount of heat given
to the surrounding can cause turbulance in it, hence all heat exhanges with surroundings are considered to be reversible. dq surroundin g
Ssurrounding = For
T surrounding T is constant
q surroundin g 1 dq surr = T T But according to the law of conservation of energy qsurr = – qsystem
Ssurr =
Ssurr =
q system T
Isothermal process : (a)
Reversible Re v State A State B P1,V1,T P2,V2,T
S system = nCVn
V T2 + nRn 2 V1 T1
Since T1 = T2 S system = nRn
Ssurrounding =
V2 V1
q system
T du = dq + dw (since T1 = T2 . So, du = 0) dq = – dw
V2 q = –w as w = – nRTn V 1 V2 q = nRTn V 1
nRTn Ssys. =
V2 V1
T
–nRn
V2 V1
Suniverse= Ssystem + Ssurr = nRn (b)
V2 V2 –nRn =0 V1 V1
rreversible irrev State A State B P2,V2,T P1,V1,T
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Ssystem = nRn
SSurrounding =
V2 V1
qsystem
T qsystem is calculated using FLOT qsystem = Pext(V2– V1)
Ssurrounding =
p ext ( V2 V1 ) T
Suniverse = nRn
p ext ( V2 V1 ) V2 – V1 T
=
1 T
V2 ) Pext ( V2 V1 ) (nRT n V1
1 [W irrev – W rev ] > 0 T As irreversible work of gas is more than reversible work. =
Adiabatic Process (Isoentropic process) (a)
Reversible Adiabatic rev State A State B P1,V1, T1 P2,V2, T2
T2 V2 Ssystem = nCV n T + nR n V 1 1 TV–1 = constant
T2 V1 = T1 V2
1
V1 (S)system = nCVn V2 S surr = –
1
V2 + nRn V = 0 1
qsys
= 0 (qsys= 0) T S universe = 0 + 0 = 0 (b)
Irreversible Adiabatic irre State A State B P1,V1,T1 P2,V2,T2
Using irreversible adiabatic process it is not possible to reach same state B which was reached by reversible adiabatic.
T2 V2 S system = nCV n T + nr n V > 0 1 1 S surr = –
[(T2)irrev > (T2)rev]
q sys
=0 as q = 0 T For irreversible expansion the decrease in temprature will be lesser as work done is lesser hence, decrease in entropy due to fall in temprature will be lesser in case of irrversible expansion. Hence, net entropy would increase.
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Isobaric process : (a)
Reversible Isobaric rev State A State B P1,V1, T1 P2,V2, T2
Ssys =
dqrev = T
nC p dT
T T2
Ssurrounding
dqrev = = T T1
– dqsys
T2 = nCpn T 1
T
T2 = – nCpn T 1
Suniverse = Ssys + Ssurrounding = 0 (b)
Irreversible Isobaric T2
Ssys =
dqrev = T
nC p dT
T2 = nCpn T 1
T
T1
– nC p T2 – T1
Ssurrounding =
T2
Suniverse = Ssys + Ssurrounding > 0
Isochoric Process (a)
Reversible Isochoric rev State A State B P1,V1, T1 P2,V2, T2 T2
Ssys =
dqrev = T
T1
nC v dT T2 = nCvn T T 1 T2
dqrev Ssurrounding = = T T1
– dqsys
T
T2 = – nCvn T 1
Suniverse = Ssys + Ssurrounding = 0 (b)
Irrversible Isochoric T2
Ssys =
dqrev = T
T1
nC v dT T2 = nCpn T T 1
– nC v T2 – T1 T2
Ssurrounding =
Suniverse = Ssys + Ssurrounding > 0
Entropy Calculation for phase transformations Sfusion=
dq
T
1 T
Q
dq T
Hfus T Ufusion for constant volume Sfus. = T for constant pressure Sfus. =
Svap =
dQ
T
1 Q dQ T T
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for constant Pressure, Svap = for constant Volume, Svap =
H vap T
U vap T
Entropy Calculation solid or liquid systems: (a)
‘Cu’ block kept in open atmosphere T2
Ssystem =
T1
msdT T2 T = msln T1
q surr dqsurr T T 2 but q surr = – qsystem = – ms (T2 – T1) = ms (T1 – T2) Ssurr =
Suniverse = Ssystem + Ssurr (b)
T2 ms (T1 T2 ) Suniverse = ms n T + >0 T2 1 Two copper block kept in contact in thermostat Two blocks are of same mass
//////////////////////
//////////////////////
///////////////////////////////////// Cu (M)
T2 T1 /////////////////////////////////////
Thermostart
Ssurr = 0 (since no heat is lost to surrounding) S system = SA + SB System will reach a common temperature (Tf ) and T1 > Tf > T2 Tf
S system =
T1
Tf
=
T1
dq A + T
Tf
T2
dqB T
Tf msdT msdT = ms T + T
T2
Tf Tf n n T T 1 2
2
Ssystem = msn Since Tf =
Tf T1T2
T1 T2 2
Ssystem = msn
T1 T2 2 4T1T2
Third Law Of Thermodynamics : The entropy of perfect crystals of all pure elements & compounds is zero at the absolute zero of temperature. Thus, absolute value of entropy unlike the absolute value of enthalpy for any pure substance can be calculated at any given temperature . In standard state (298 K, 1 atm), it is standard absolute entropy Sº . T
S = ST(K)– S0(K) =
0
nCdT T
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For chemical reaction aA + bB cC + dH 0 0 S0system = ( ni Sm )product – ( ni Sm )reactant 0 where Sm = standard molar entropy. It can calculated using third law of thermodynamics.
Ssurr =
Q system
=–
Hsystem
T T * For a perfactly crystline substance at 0 K, entropy = 0
Gibb’s free energy (G) : A system parameter to predict the spontaneity of chemical reaction was introduced by Gibb’s so that entropy calculation for the surrounding need not be carried out. It is a free energy at constant pressure. G = H – TS dG = dH –TdS – SdT & H = E + PV dH = dE + PdV + VdP dG = dE + PdV + VdP –TdS –SdT From Ist law of thermodynamics dq = dE + PdV & from 2nd law , dq = TdS dG = dq + VdP –TdS – SdT dG = – SdT + VdP At constant temperature, dG = VdP For ideal gas , V = dGm = VmdP = P2
2
dG
m
= RT
1
P1
nRT P
RT dp P
dP P
Gm2 – Gm1 = RTln
P2 P1
Where Gm is free energy for 1 mole If Gm1 is free energy for 1 mole in standard state and Gm2 is free energy for one mole in any other state. Gm – G°m = RTln
P 1
& G – G° = nRTln
For the reaction, a A (g) + b B (g) (G)reaction= GC + GD – GA – GB
P 1
C (g) + d D (g)
PC GC = G°C + cRTln PC 1 GD= G°D + dRTln PD
GC – G°C = cRTln Similarly,
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GA= G°A + aRTln PA ,
GB= G°B + bRTln PB
(G)reaction= (G°C + cRTln PC) + (G°D + dRTln PD ) – (G°A + aRTln PA ) – (G°B + bRTln PB) = (G)°reaction + RTln
P c C .P dD
P a A .P b B Greaction= G°reaction + RTlnQ
Gibb’s Helmholtz Equation : dG = VdP – SdT At constant Pressure dG G =–S& = – S dG = – SdT dT P T P
& G = H – TS
G–H =–S T
G G–H = T P T G G = H + T T P
G T P
G = H + T
G = H –T S Sunivense = Ssystem + Ssurrounding Ssurr =
Q system
=–
Hsystem
T T TS univ = TS system – Hsystem
Introducing a new thermodyamic function G = Gibb’s free energy (State function and an extensive property) Gsystem = H system – TS system G = H – TS Gsystem = H system – TS system .................. (ii) Comparing equation (i) and (ii) Gsystem = – TS universe
New criteria of spontaneity : (i) If G system is (–ve) < 0 (ii) If G system is > 0 (iii) If G system = 0
process is spontaneous process is non spontaneous system is at equilibrium
So, at every temparature G 0 G system = H system – TS system
H system
S system
+ Ve
– Ve
+
– Ve
+ Ve
–
+ Ve
+ Ve
– Ve at high temprature
– Ve
– Ve
+ Ve at high temprature
Gº = standrad free energy change : When the reactants under standrad conditions gets converted into products which is also under standered condition, then the free energy change is known as Gº (it is a constant) for 1 mole at 1 bar.
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At standared conditions : G° = H° – TS° = constant for a GIVEN REACTION AB C + D G° = (Gbb’s energy of product) – (Gibb’s energy of reactant) 0 0 0 0 0 0 G° = Gm ,C + Gm,D – Gm,A – G m,B , where Gm can not be calculated as Hm cannot be
calculated. hence, we can convert this relation in to Gibb’s energy of formation of substance. G° = G0f ,C + G0f ,D – G0f ,A – G 0f ,B
G0f : standered Gibb’s energy of formation. G0f (elements in their standard states) = 0 G° = G0f ,product – G0f ,reac tan ts For a reaction in progress G = Gº + RT n Q Q is reaction quotient , G > 0 backward is feasible , G < 0 forward is feasible At equilibrium G = 0 Q=K Gº = – RT n K at equilibrium
THERMOCHEMISTRY
Enthalpy of a substance :
Every substance has a fixed value of enthalpy under any particular state. Though, its exact value cannot be
calculated but it has some finite fixed value. The enthalpy of one mole of a substance called molar enthalpy in any particular state can be assigned symbol Hm(substance) : For example molar enthalpy of water vapours at 398 K and 1 atm pressure may be represented
as Hm (H2O, g, 398 K, 1 atm). In very simple works, enthalpy can be considered as heat constant (amount) of substance, and during reaction this heat is being released or absorbed. Molar enthalpy of substance under standard conditions is called standard molar enthalpy of a substance. Standard state of any substance means. For a GAS standard state means ideal gas at 1 bar partial pressure at any give temperature. For a LIQUID – pure liquid at one bar pressure at 1 bar pressure at any given temperature. For a PURE CRYSTALLINE SOLID – pure crystalline solid at 1 bar pressure and at any given temperature For any SUBSTANCE or ION IN SOLUTION – the species should be in unit molality (can also be taken as 1M concentration), at one bar pressure and at any given temperature. Molar standard enthalpy of water vapours at 398 K will be represented as H°(H2O, g , 398 K) and molar standard enthalpy of liquid water at 398 K will be represented as Hm° (H2O, l, 398 K) (It is hypothetical but can be calculated). We cannot exactly calculate enthalpy content of a substance only the change in enthalpy can be calculated when substance is taken from one state to other. For example :
0 0 Let enthalpy contant initially be Hm ,1 & finally enthalpy content be Hm , 2 0 0 Then, H° = Hm , 2 – Hm ,1
= heat added at constant pressure to change temperature from 25°C to 50°C. = CPT = (18 cal/mole °C) (25°C) = 450 cal
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Enthalpy of formation : It is not possible to determine the absolute value of the molar enthalpy of a substance. However, based on the following convention, the relative values of standard molar enthalpies of formation of various substances can be built. “The standard enthalpy of formation of every element in its stable state of aggregation at one bar pressure and at specified temperature is assigned a zero value”.The specified temperature is usually taken as 25 °C. A few exmaples are Hf° (O2 , g) = 0 Hf° (C, graphite) = 0 Hf° (C, diamond) 0 Hf° (Br2 , ) = 0 Hf° (S, rhombic) = 0 Hf° (S, monoclinic) 0 Hf° (P, white) = 0 Hf° (P, black) 0 “The standard enthalpy of formation of a compound is the change in the standard enthalpy when one mole of the compound is formed starting from the requisite amounts of elements in their stable state of aggregation”. The chemical equations corresponding to enthalpy of formation of few substances are given below. Enthalpy of formation of HBr(g) :
1 1 H (g) + Br2 (l) HBr(g) 2 2 2
0 0 = B Hm (B) = Hm (HBr, g) –
Hf°(HBr, g)
Enthalpy of formation of SO2 (g) : Hf° (SO2, g) =
0 Hm
(SO2, g) –
1 0 1 0 Hm (H2, g) – H (Br2, 1) 2 2 m
...(1)
S (rhombic) + O2 (g) SO2 (g) 0 (S, Hm
0 rhombic) – Hm (O2, g)
...(2)
But above equations cannot be for calculation of enthalpy of reaction as the molar enthalpies of different species can not be exactly known.
Enthalpy of Reaction from Enthalpies of Formation: The enthalpy of reaction can be calculated by Hr° = B Hf°,products – B Hf°,reactants B is the stoichiometric coefficient above equation holds good for any reaction as the same reference state is used for reactants and products (shown in figure).
Hess’s Law of constant heat summation : The heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one
step or several steps. The chemical equation can be treated as ordinary algebraic expressions and can be added or subtracted to
yield the required equation. The corresponding enthalpies of reactions are also manipulated in the same way so as to give the enthalpy of reaction for the desired chemical equation. Since rH stands for the change of enthalpy when reactants (substances on the left hand side of the arrow) are converted into products (substances on the right hand side of the arrow) at the same temperature and pressure, if the reaction is reversed (i.e., products are written on the left hand side and reactants on the right hand side), then the numerical value of rH remains the same, but its sign changes. The utility of Hess’s law is considerable. In almost all the thermochemical numericals, Hess’s law is used. One of the important applications of Hess’s law is to determine enthalpy of reaction which is difficult to
determine experimentally. For example, the value rH for the reaction C(graphite) +
1 O (g) CO(g) 2 2
which is difficult to determine experimentally, can be estimated from the following two reactions for which rH can be determined experimentally. C(graphite) + O2(g) CO2(g) CO(g) +
1 O (g) CO2(g) 2 2
rH1 rH2
Substracting the latter from the former, we get C(graphite) + Consequently,
1 O (g) CO(g) 2 2
rH = rH 1 – rH 2
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Enthalpy of Combustion : It is the enthalpy change when one mole of a compound combines with the requisite amount of oxygen to
give products in their stable forms. For example, the standard enthalpy of combustion of methane at 298 K is – 890 kJ mol–1. This implies the following reaction : CH4 (g) + 2O2(g) CO2(g) + 2H2O () H° = – 890 kJ mol–1 The standard enthalpy of combustion of methane at 298 K may be written as Hc° (CH4, g, 298 K) = – 890 kJ mol–1 The data on the enthalpy of combustion can be determined experimentally. With the help of such data, we can determine the enthalpy of formation of a compound, which otherwise is difficult or impossible to determine experimentally. Consider for example, the enthalpy of formation of CH4(g): C(graphite) + 2H2(g) CH4(g) First of all, the combination of carbon and hydrogen does not occur readily. Secondly, if the reaction is even completed, the end product would not be pure methane. Therefore, the enthalpy of formation of methane can be determined indirectly through the enthalpy of combustion of methane : CH4(g) + 2O2(g) CO2(g) + 2H2O() HC°(CH4, g) = Hf°(CO2, g) + 2Hf°(H2O , ) – Hf° (CH4, g) therefore Hf°(CH4, g) = Hf°(CO2, g) + 2Hf°(H2O , ) – HC° (CH4, g) The enthalpies of formation of CO2 and H2O can be determined experimentally by the combustion of carbon
(graphite) and hydrogen. Thus, knowing the mesured value of HC°(CH4, g), the enthalpy of formation of CH4 can be calculated. The value is Hf°(CH4, g) = Hf°(CO2, g) + 2 Hf°(H2O , ) – HC° (CH4, g) = [– 393 + 2 (–285) – (–890)] kJ mol–1 = – 73 kJ mol–1 or, equivalently, we may add the following three chemical equations. C(graphite) + O2 (g) CO2 (g) 2 [H2(g) +
1 O (g) H2O()] 2 2
CH° = – 393 kJ mol–1 rH° = 2(– 285) kJ mol–1
– [CH4(g) + 2O2 (g) CO2 (g) + 2H2O()] –––––––––––––––––––––––––––––––––––––––
rH° = –(– 890) kJ mol–1 ––––––––––––––––––––––
C(graphite) + 2H2(g) CH4 (g)
fH° = – 73 kJ mol–1
Measurement of Enthalpy of Combustion : Enthalpy of combustion are usually measured by placing a known mass of the compound in a closed steel container (known as bomb calorimeter) which is filled with oxygen at about 30 bar pressure. The calorimeter is surrounded by a known mass of water. The entire apparatus is kept in an insulated jacket to prevent heat entering into or leaving from the container, as shown in figure. The sample is ignited electrically to bring about the combustion reaction. The heat evolved is used in raising the temperature of water and the calorimeter. If total heat capacity of calorimeter and all of its contents = C, rise in temperature = T then heat released = q = CT of this heat is because of mass m of substance then M q = EC° m
due to 1 mole, heat released =
(constant volume reaction). Now, HC° can be calculated by using HC° = EC° + ng RT. Where ng is the change in stoichiometric number of gaseous species in the balanced chemical equation representing the combustion process.
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Integral enthalpy of solution : The integral enthalpy of solution at the given concentration is the enthalpy change when one mole of the solute is dissolved in a definite quantity of solvent to produce a solution of a desired concentration. While recording integral enthalpies of solution it is a general practice to state the amount of the solvent in which 1 mole of solute is dissolved ; Thus HCl(g) + 10H2O() HCl (10H2O) H1 = –69.5 kJ mol–1 indicates that when 1 mole of hydrogen chloride gas is dissolved in 10 mol of water, there is an evolution of 69.5 kJ of heat. Other values are (i) HCl(g) + 25 H2O() HCl (25H2O) H2 = –72.3 kJ mol–1 (ii) HCl(g) + 40 H2O() HCl (40H2O) H3 = –73.0 kJ mol–1 (iii) HCl(g) + 200 H2O() HCl (200H2O) H4 = –74.2 kJ mol–1 (iv) HCl(g) + aq HCl (aq) H5 = –75.0 kJ mol–1 Whenever amount of solvent is not specified then take its amount to be very large just like in equation no. (iv).
Enthalpy of Hydration :
Enthalpy of hydration is used in following to ways. Enthalpy of hydration of anhydrous or partially hydrated salts : Enthalpy of hydration of a given anhydrous or partially hydrated salt is the enthalpy change when it combines with the requisite amount of water to form a new hydrated stable salt. For example, the hydration of anhydrous cupric sulphate is represented by CuSO4 (s) + 5H2O () CuSO4 . 5 H2O(s) There is a almost invariably a liberation of heat in such reactions, i.e. the value of H is negative. CuSO4(s) + 800 H2O () CuSO4 (800 H2O) Hr° = – 68 kJ mol–1 CuSO4 . 5H2O (s) + 795 H2O () CuSO4 (800 H2O) Hr° = + 10 kJ mol–1 by subtraction, we get CuSO4(s) + 5H2O () CuSO4 . 5 H2O(s)Hr° = – 78 kJ mol–1 Enthalpy of hydration of gaseous ions. Enthalpy of hydration of any gaseous ion is the enthalpy change when 1 mole of the gaseous ion is hydrated in large amount of water to form aqeous ion. By convention, the standard enthalpy of formation of H+(aq) is taken to be zero. Enthalpy of hydration of Cl¯ gaseous ions will be represented by : Cl¯(g) + aq. Cl¯(aq) Hr° = Hf° (Cl–, aq)
Enthalpy of Neutralization : The amount of heat released when one gram equivalent of an acid is neutralised by one gram equivalent of a
base. or The amount of heat released in formation of one mole of water when an acid is neutralised by a base.
or Enthalpy of neutralization is defined as the enthalpy change when one mole of H+ in dilute solution combines with one mole of OH¯ to give rise to undissociated water, i.e. H+(aq) + OH¯(aq) H2O() H= –57.1 kJ/mole = –13.7 kcal/mol Remember : For Strong Acid + Strong Base, heat of neutralisation is always equal to –13.7 kcal/mole or –57.1 kJ/mole. For any other combination of acid and base this heat is less than –13.7 kcal/mole or –57.1 kJ/mole.
Enthalpy of Ionization : Whenever a weak acid (or base) reacts with a strong base (or acid), the release of heat is less than 57.1 kJ
mol–1. It is because of the fact that these acids or bases are not completely ionized in solution. Some of the heat is consumed in ionizing there acids and bases this heat is known as enthalpy of ionization. Examples are : HCN + Na+ OH– Na+ + CN– + H2O rH° = – 12 kJ mol–1 + – + – CH3COOH + Na OH Na + CH3COO + H2O rH° = – 49 kJ mol–1 The enthalpy of ionization can be calculated as follows. The neutralization of a weak acid, say HCN, may be represented in two steps, namely, Ionization HCN H+ + CN¯ H°1 = x + Neutralization H + OH¯ H2O H°2 = – 57.1 kJ/mole ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
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The complete reaction is obtained by adding the above two steps. Thus HCN + OH¯ H2O + CN¯ Obviously,
H° = – 12 kJ/mole
H° = H°1 + H°2 H°1 = H°H°2 = [–12 – (–57.1)] = 45.1 kJ/mole
Greater the enthalpy of ionization of any weak acid or weak base, weaker will be the acid or base.
Enthalpy of Transition : Enthalpy of transition is the enthalpy change when one mole of one allotropic form changes to another. For example : C(graphite) C(diamond) Htrs0 = 1.90 kJ mol–1 so if C(graphite) + O2(g) CO2(g) HC0 = – 393.51 kJ mol-1 and C(diamond) + O2(g) CO2(g) HC0 = – 395.41 kJ mol–1 Subtracting, we have, C(graphite) C(diamond) Htrs0 = 1.90 kJ mol–1
Enthalpy of Precipitation : Enthalpy of precipitation is the enthalpy change when one mole of a precipitate is formed. For example : BaCl2(aq.) + Na2SO4(aq) BaSO4(s) + 2NaCl(aq) rH0=–24.27 kJ mol–1
Enthalpy of Formation of ions : The enthalpy change when one mole of hydrated ions is obtained from element in its standard state as. ½ Cl2 (g) + aq Cl– (aq) Hr° = Hf° (Cl–, aq) By convention, the standard enthlpy of formation of H+(aq) is taken to be zero. We have seen that H+ (aq) + OH–(aq) H2O(l ) rH0 = – 57.1 kJ mol–1 0 0 0 + 0 For this reaction, Hr = Hf (H2O,l ) – {Hf (H , aq) + Hf (OH–,aq)} Hence, at 25ºC, we get Hf0 (H+, aq) + Hf0 (OH– ,aq) = Hf0 (H2O, l ) – Hr0 so Hf0 (OH– ,aq) = {– 286.1 – (– 57.1)} kJ mol–1 = – 229.00 kJ mol–1 With the enthalpies of formation of these two ions, the enthalpy of formation of any other ion can be found from the enthalpies of formation and solution of its pure compound with H+ or OH¯. for example, the enthalpy of formation of Na+ can be calculated from the enthalpy of formation and enthalpy of infinite dilute solution of NaOH. The two values are : The chemical equation for the formation of infinite dilute solution of NaOH(s) is
NaOH(s) + nH2O() Na+(aq) + OH¯(aq) aqH°(NaOH, s) = –44.50 kJ mol–1 Since there are equal amounts of water on both sides of the above equation, the two enthalpies give no net effect and thus aqH°(NaOH, s) = fH°(Na+, aq) + fH°(OH¯, aq) – fH°(NaOH, s) or fH° (Na+, aq) = aqH° (NaOH, s) – fH°(OH°, aq) + fH°(NaOH, s) = [–44.50 – (–229.99) + (–425.61)] kJ mol–1 = –240.12 kJ mol–1 Similarly, from NaCl(aq) or HCl(aq), the enthalpy of formation of Cl¯(aq) can be determined, and so on. The
changes in enthalpy of any ionic reaction can then be found from these ionic enthalpies of formation and the usual enthalpies of formation of compounds.
Bond Enthalpies : The bond enthalpy is the average of enthalpies required to dissociate the said bond present in different gaseous compounds into free atoms or radicals in the gaseous state. While bond dissociation enthalpy is the enthalpy required to dissociate a given bond of some specific compound.for example the enthalpy of dissociation of the O–H bond depends on the nature of molecular species from which the H atom is being separated. For example, in the water molecule. H2O(g) H(g) + OH(g) Hr0 = 501.87 kJ mol–1 However, to break the O–H bond in the hydroxyl radical required a different quantity of heat : OH(g) O(g) + H(g) Hr0 = 423.38 kJ mol–1 The bond enthalpy, OH, is defined as the average of these two values, that is : OH =
501.87mol 1 423.38kJmol1 = 462.625 kJ mol–1 2
In the case of diatomic molecules, such as H2, the bond enthalpy and bond dissociation enthalpy are identical because each refers to the reaction. H2(g) 2H(g) H – H = Hr0 = 435.93 kJ mol–1 Thus, the bond enthalpy given for any particular pair of atoms is the average value of the dissociation enthaplies of the bond for a number of molecules in which the pair of atoms appears. ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
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Estimation of Enthalpy of a reaction from bond Enthalpies : Let the enthalpy change for the gaseous reaction C2H4(g) + HCl(g) C2H5Cl(g) (g) be required from the bond enthalpy data. This may be calculated as follows : Enthalpy required to break reactants Enthalpy released to form products from the gasesous atoms
H = into gasesous atoms = =
[4C–H + C=C + H–Cl] + [–5C–H – C–C – C–Cl] (C=C + H–Cl) – (C–H + C–C + C–Cl)
Resonance Energy :
Difference between energy of resonance hybrid and resonating structure in which
resonance
hybrid have lower energy because stabilised by resonance. H°resonance = H°f, experimental – H°f, calclulated = H°combustion, calclulated– H°combustion, experimental
Relation between energy and enthalpy of a reaction : rH = rU + (vg)RT where vg is the change in the stoichiocmetric number of gaseous species in going from reactants to products. It should be noted that while computing vg of a reaction, only the stoichiometric numbers of gaseous is counted and those of liquids and solids are completely ignored.
Kirchoff’s equation (Variation of Hº with temprature) Since the enthalpy (or standard enthalpy) of a substance is dependent on state of the substance, value of enthalpy of a substance changes with temperature and hence the enthalpy change of reaction is also dependent on temperaturte at which the reaction is being carried out. This change is enthalpy change (or E, of reaction is carried out at constant volume) is represented by Kirchoff’s Equations. P + Q R + S at temperature T1 let the standard enthalpy of reaction be H°1 , then H1° = H°m (R,T1) + H°m (S, T1) – H°m (P, T1) – H°m (Q, T1) If the same reaction is carried out at temperature T2, then H2° = H°m (R,T2) + H°m (S, T2) – H°m (P, T2) – H°m (Q, T2) Then, the change in enthalpy (or difference in enthalpy at these two temperatures) H° = H2° – H1° = {Hm° (R,T2) – Hm° (R,T1)} + {Hm° (,T2) – Hm° (,T1)} – {Hm° (P,T2) – Hm° (P,T1)} - {Hm° (Q,T2) – Hm° (Q,T1)} Hm° (R, T2) – Hm° (R,T1) = CP, R (T2 – T1) = Heat required at constant pressure to increase temperature of one mole of R from T1 to T2 Similarly Hm° (S,T2) – Hm° (S,T1) = CP,S (T2 – T1) Hm° (P,T2) – Hm° (P,T1) = CP,P (T2 – T1) and Hm° (Q,T2) – Hm° (Q,T1) = CP,Q (T2 – T1) so H° = H2° – H1° = CP,R (T2 – T1) + CP,S(T2 – T1) – CP,P (T2 – T1) CP,Q(T2 – T1) = [ CP,R + CP,S – CP,P – CP,Q] (T2 – T1) = CP (T2 – T1) CP = CP, R + CP,S – CP,P – CP,Q = Difference in molar heat capacities of products and reactants. so
H2° = H1° +
C
P .dT
for example for the reaction N2 (g) + 3H2 (g) 2NH3 (g) H2° = H1° + CP (T2 – T1) where
CP = 2CP,NH3 CP,N2 3CP,H2
* for a constant volume reaction
E02 E10 C V .dT
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*Marked Questions are having more than one correct option.
1. Basic Terminology 1.1
Warming ammonium chloride with sodium hydroxide in a test tube is an example of : (A) Closed system (B) Isolated system (C) Open system (D) None of these
1.2
Out of boiling point (), entropy (), pH () and e.m.f. of a cell (V), intensive properties are : (A) , (B) , , (C) , , V (D) All of the above
2. First law of thermodynamics, calculations of E, w & q. 2.1
What is U for the process described by figure. Heat supplied during the process q = 100 kJ.
(A) + 50 kJ
(B) – 50 kJ
(C) –150 kJ
(D) + 150 kJ
2.2
Two mole of an ideal gas is heated at constant pressure of one atomosphere from 27ºC to 127ºC. If Cv,m = 20 + 10–2 T JK–1 mol–1, then q and U for the process are respectively : (A) 6362.8 J, 4700 J (B) 3037.2 J, 4700 J (C) 7062.8, 5400 J (D) 3181.4 J, 2350 J
2.3
One mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. If the work done by the gas in the process is 3 kJ, the final temperature will be equal to (CV=20 J/K mol) (A) 100 K (B) 450 K (C) 150 K (D) 400 K
2.4
What is the change in internal energy when a gas contracts from 377 ml to 177 ml under a constant pressure of 1520 torr, while at the same time being cooled by removing 124 J heat? [Take : (1 L atm) = 100 J ] (A) –24 J (B) – 84 J (C) – 164 J (D) –248 J
2.5
A sample of liquid in a thermally insulated container (a calorimeter) is stirred for 2 hr. by a mechanical linkage to a motor in the surrounding, for this process : (A) w < 0; q = 0; U = 0 (B) w > 0; q > 0; U > 0 (C) w < 0; q > 0; U = 0 (D) w > 0; q = 0; U > 0
2.6
An ideal gas is taken around the cycle ABCDA as shown in figure. The net work done during the cycle is equal to : D P B 2
P P1
C V1
(A) zero 2.7
(B) positive
V
V2
(C) negative
(D) we cannot predict
In the cyclic process shown in P-V diagram, the magnitude of the work done is : P P (A) 2 1
(C)
2
2
(P2 – P1) (V2 – V1) 4
V V (B) 2 1
2
2
(D) (V2 – V1)2
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2.8
10 mole of ideal gas expand isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K. What is the largest mass which can lifted through a height of 100 meter ? (A) 31842 kg (B) 58.55 kg (C) 342.58 kg (D) None of these
2.9
A cyclic process ABCD is shown in PV diagram for an ideal gas. Which of the following diagram represents the same process?
(A)
(B)
(C)
(D)
2.10
A gas expands adiabatically at constant pressure such that T V–1/2. The value of (Cp,m/Cv,m) of the gas will be: (A) 1.30 (B) 1.50 (C) 1.70 (D) 2
2.11*
P-V plot for two gases (assuming ideal) during adiabatic processes are given in the figure. Plot A and plot B should correspond respectively to :
(A) He and H2 2.12
(B) H2 and He
(C) SO3and CO2
(D) N2 and Ar
A diatomic ideal gas initially at 273 K is given 100 cal heat due to which system did 209 J work. Molar heat capacity (Cm) of gas for the process is : (A)
3 R 2
(B)
5 R 2
(C)
5 R 4
(D) 5 R
2.13
One mole of an ideal monoatomic gas expanded irreversibly in two stage expansion. State-1 (8.0 bar, 4.0 litre, 300 K) State-2 (2.0 bar, 16 litre, 300 K) State-3 (1.0 bar, 32 litre, 300 K) Total heat absorbed by the gas in the process is : (A) 116 J (B) 40 J (C) 4000 J (D) None of these
2.14
1 mole of NH3 gas at 27° C is expanded in reversible adiabatic condition to make volume 8 times ( = 1.33). Final temperature and work done respectively are : (A) 150 K, 900 cal (B) 150 K, 400 cal (C) 250 K, 1000 cal (D) 200 K, 800 cal
2.15
For an ideal monoatomic gas during any process T = kV, find out the molar heat capacity of the gas during the process. (Assume vibrational degree of freedom to be active) (A)
2.16
5 R 2
(B) 3R
(C)
7 2
(D) 4 R
3
1 mole of an idal gas A (C v,m = 3R) and 2 mole of an ideal gas B are C v,m 2 R taken in a container and expanded reversible and adiabatically from 1 litre to 4 litre starting from initial temperature of 320 K. E or U for the process is : (A) –240 R (B) –240 R (C) 480 R (D) –960 R ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
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2.17
2.18
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be (A)
T1T2 (P1V1 P2 V2 ) P1V1T2 P2 V2 T1
(B)
P1V1T1 P2 V2 T2 P1V1 P2 V2
(C)
P1V1T2 P2 V2 T1 P1V1 P2 V2
(D)
T1T2 (P1V1 P2 V2 ) P1V1T1 P2 V2 T2
An ideal gas is taken around the cycle ABCA as shown in P-V diagram. The net work done by the gas during the cycle is equal to : (A) 12P1V1 (C) 5P1V1
2.19
(B) 6P1V1 (D) P1V1
What is the net work done (in calories) by 1 mole of monoatomic ideal gas in a process described by 1, 2, 3, 4 in given V–T graph.
Use : R = 2cal / mole K ln 2 = 0.7 (A) – 600 cal (B) – 660 cal
(C) + 660 cal
2.20
A heat engine carries one mole of an ideal mono-atomic gas around the cycle as shown in the figure, the amount of heat added in the process AB and heat removed in the process CA are : (A) qAB = 450 R and qCA = –450 R (B) qAB = 450 R and qCA = –225 R (C) qAB = 450 R and qCA = –375 R (D) qAB = 375 R and qCA = –450 R
2.21
Two moles of an ideal gas (CV =
(D) + 600 cal
5 R) was compressed adiabatically against constant pressure of 2 atm. 2
Which was initially at 350 K and 1 atm pressure. The work involve in the process is equal to (A) 250 R (B) 300 R (C) 400 R (D) 500 R 2.22
Two moles of Helium gas undergo a reversible cyclic process as showin in figure. Assuming gas to be ideal, what is the net work involved in the cyclic process c? (A) – 100 Rn4 (B) +100Rn4 (C) +200Rn4 (D) –200Rn4
3. Calculation of E, H, w and q 3.1
A gas (Cv,m =
5 R) behaving ideally was allowed to expand reversibly and adiabatically from 1 litre to 32 litre. 2
It's initial temperature was 327°C. The molar enthalpy change (in J/mole) for the process is (A) –1125 R (B) – 575 R (C) –1575 R (D) None of these ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
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3.2
5
One mole of an ideal gas Cv, m 2 R at 300 K and 5 atm is expanded adiabatically to a final pressure of 2 atm against a constant pressure of 2 atm. Final temperature of the gas is : (A) 270 K
(B) 273 K
(C)248.5 K
(D) 200 K
3.3
The work done in adiabatic compression of 2 mole of an ideal monoatomic gas by constant external pressure of 2 atm starting from intial pressure of 1 atm and initial temperature of 300 K (R = 2 cal/mol-degree) (A) 360 cal (B) 720 cal (C) 800 cal (D) 1000 cal
3.4
One mole of non-ideal gas undergoes a change of state (1.0 atm, 3.0 L, 200 K) to (4.0 atm, 5.0 L, 250 K) with a change in internal energy (U) = 40 L-atm. The change in enthalpy of the process in L-atm ; (A) 43 (B) 57 (C) 42 (D) None of these
3.5*
0.5 mole each of two ideal gases A Cv, m 2 R and B (Cv, m = 3R) are taken in a container and expanded
5
reversibly and adiabatically, during this process temperature of gaseous mixture decreased from 350 K and 250 K. Then for the process : (A) U = – 100 R (B) U = – 275 R (C) H = – 375 R (D) H = – 300 R 3.6
50 L of a certain liquid is confined in a piston system at the external pressure 100 atm. This pressure is suddenly released and liquid is expanded against the constant atmospheric pressure, volume of the liquid increases by 1 L and the final pressure on the liquid is 10 atm. Find the workdone. (A) 1L.atm (B) 5 L.atm (C) 500 L.atm (D) 50 L.atm
3.7
A vessel contains 100 litres of a liquid X. Heat is supplied to the liquid in such a fashion that, Heat given = change in enthalpy. The volume of the liquid increases by 2 litres. If the external pressure is one atm, and 202.6 Joules of heat were supplied then, [U - total internal energy] (A) U = 0 , H = 0 (B) U = + 202. 6J , H = + 202.6 J (C) U = – 202.6J, H = – 202.6J (D) U = 0, H = + 202.6J
3.8
For the real gases reaction 2CO (g) + O2 (g) 2CO2 (g) ; H = – 560 kJ. In 10 litre rigid vessel at 500 K, the initial pressure is 70 bar and after the reaction it becomes 40 bar. The change in internal energy is : (A) – 557 kJ (B) – 530 kJ (C) – 563 kJ (D) None of these
4.1
4.2
4. Thermochemistry For which of the following change H E? (A) H2 (g) + I2 (g) 2HI (g) (B) HCl (aq)+ NaOH(aq) NaCl(aq) + H2O(l) (C) C(s) + O2(g) CO2(g) (D) N2 (g)+ 3H2(g) 2NH3(g) In Haber's process of manufacturing of ammonia : N2(g) + 3H2(g) 2NH3(g) ; H025C = –92.2 kJ Molecule N2(g) H2(g) NH3(g) CP JK-1 mol–1 29.1 28.8 35.1 If CP is independent of temperature, then reaction at 100°C as compared to that of 25°C will be : (A) More endothermic (B) Less endothermic (C) More exothermic (D) Less exothermic
4.3*
Which of the reaction defines molar Hf°? (A) CaO(s) + CO2(g) CaCO3 (s) (C) N2 (g) + 2H2 (g) +
4.4
(B)
1 1 Br2 () + H2 (g) HBr(g) 2 2
3 1 1 O2 (g) NH4 NO3 (s) (D) 2 (s) + H2 (g) H (g) 2 2 2
For the allotropic change represented by the equation C (graphite) C (diamond), H = 1.9 kJ. If 6 g of diamond and 6 g of graphite are separately burnt to yield CO2, the enthalpy liberated in first case is (A) less than in the second case by 1.9 kJ (B) more than in the second case by 11.4 kJ (C) more than in the second case by 0.95 kJ (D) less than in the second case by 11.4 kJ ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
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4.5
In the reaction, CO2(g) + H2(g) CO(g) + H2O(g); H = 2.8 kJ, H represents (A) heat of reaction (B) heat of combustion (C) heat of formation (D) heat of solution
4.6
NH3(g) + 3Cl2(g) NCl3(g) + 3HCl (g); –H1 N2(g) + 3H2(g) 2NH3(g); H2 H2(g) + Cl2(g) 2HCl (g) ; H3 The enthalpy of formation of NCl3 (g) in the terms of H1, H2 and H3 is (A) Hf = – H1 +
(C) Hf = H1 – 4.7
H 2
2
H 2 2
3 H3 2
(B)
3 H3 2
Hf = H1 +
H 2 2
3 H3 2
(D) None
Given, H2(g) + Br2(g) 2HBr(g), H01 and standard enthalpy of condensation of bromine is H02, standard enthalpy of formation of HBr at 250C is (A) H01 / 2 (B) H01 / 2 + H02 (C) H01 / 2 H02
(D) (H01H02) / 2
4.8
For the following reaction,
C (diamond) + O2 CO2(g) ; H = –94.3 kcal C (graphite) + O2 CO2(g) ; H = –97.6 kcal The heat required to change 1 g of C (diamond) C (graphite) is (A) 1.59 kcal (B) 0.1375 kcal (C) 0.55 kcal (D) 0.275 kcal
4.9
Hydrazine, a component of rocket fuel, undergoes combustion to yield N2 and H2O. N2H4 (l) + O2 (g) N2 (g) + 2H2O (l) What is the enthalpy combustion of N2H4 (kJ/mole) Given Reaction H/kJ 2NH3 (g) + 3N2O (g) 4N2 (g) + 3H2O (l) – 1011 kJ N2O (g) + 3H2 (g) N2H4 (l) + H2O (l) – 317 kJ 4NH3 (g) + O2 (g) 2N2H4 (l) + 2H2O (l) – 286 kJ
1 O (g) H2O (l) 2 2
H2 (g) +
(A) – 620.5 4.10
– 285 kJ
(B) – 622.75
C (s) + O2 (g) CO2, (g);
(C) 1167.5
(D) + 622.75
H = –94.3 kcal/mol
1 O (g) CO2 (g); H = – 67.4 kcal/mo! 2 2 O2(g) 2O (g); H = 117.4 kcal/mol CO (g) C (g) + O(g) ; H = 230.6 kcal/mol Calculate H for C (s) C (g) in kcal/mol. (A) 171 (B)154 (C)117 CO (g) +
4.11
(D)145
Find rU° for the reaction 4HCl (g) + O2 (g) 2Cl2(g) + 2H2O (g) at 300 K. Assume all gases are ideal. Given: H2(g) + Cl2(g) 2HCl (g) 2H2(g) + O2(g) 2H2O (g) (A) 111.5 kJ/mole
(B) –109.01 kJ/mole
o = – 184.5 kJ/mole r H 300 o r H 300 = – 483 kJ/mole (Use R = 8.3 J/mole)
(C) –111.5 kJ/mole
(D) None
4.12
The difference between H and E (on a molar basis) for the combustion of n-octane () at 25°C would be : (A) – 13.6 kJ (B) – 1.14 kJ (C) – 11.15 kJ (D) + 11.15 kJ
4.13
The standard heat of combustion of solid boron is equal to : (A) H°f (B2O3) (B) 1/2 H°f (B2O3) (C) 2H°f (B2O3)
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(D) 1/2 H°f (B2O3)
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4.14
The molar heat capacities at constant pressure (assume constant with respect to temperature) of A, B and C are in ratio of 1.5 : 3.0 : 2.0 . If enthalpy change for the exothermic reaction A + 2B 3C at 300 K is – 10 kJ/mol & Cp,m (B) is 300 J/mol then enthalpy change at 310 K is : (A) – 8.5 kJ/mol (B) 8.5 kJ/mol (C) – 11.5 kJ/mol (D) none of these
4.15
From the following data of H, of the following reactions,
1 O (g) CO (g) H = – 110 kJ 2 2 C(s) + H2O (g) CO (g) + H2(g) H = 132 kJ What is the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, to maintain constant temperature : (A) 0.5 : 1 (B) 0.6 : 1 (C) 0.8 : 1 (D) 1 : 1 C(s) +
4.16
2 mole of zinc is dissolved in HCl at 25ºC. The work done in open vessel is : (A) –2.477 kJ (B) –4.955 kJ (C) 0.0489 kJ (D) None
4.17
The lattice enthalpy of solid NaCl is 772 kJmol–1 and enthalpy of solution is 2 kJmol–1. If the hydration enthalpy of Na+ & Cl– ions are in the ratio of 3:2.5, what is the enthalpy of hydration of chloride ion? (A) –140 kJmol–1 (B) –350 kJmol–1 (C) –351.81 kJmol–1 (D) none
4.18
If heat of dissociation of CHCl2COOH is 0.7 kcal/mole then H for the reaction : CHCl2COOH + KOH CHCl2COOK + H2O (A) – 13 kcal
4.19
(B) + 13 kcal
(C) – 14.4 kcal
(D) – 13.7 kcal
H 0f of water is – 285.8 kJ mol –1. If enthalpy of neutralisation of monoacid strong base is –57.3 kJ mol–1,
H 0f of OH– ion will be
(A) – 228.5 kJ mol–1
(B) 228.5 kJ mol–1
(C) 114.25 kJ mol–1
(D) –114.25 kJ mol–1
4.20
A solution is 500 ml of 2 M KOH is added to 500 ml of 2 M HCl and the mixture is well shaken. The rise in temperature T1 is noted. The experiment is then repeated using 250 ml of each solution and rise in temperature T2 is again noted. Assume all heat is taken by the solution (A) T1 = T2 (B) T1 is 2 times as large as T2 (C) T2 is twice of T1 (D) T1 is 4 times as large as T2
4.21
H 0f of water is – 285.8 kJ mol –1. If enthalpy of neutralisation of monoacid strong base is –57.3 kJ mol–1,
H 0f of OH– ion will be
(A) – 228.5 kJ mol–1
(B) 228.5 kJ mol–1
(C) 114.25 kJ mol–1
(D) –114.25 kJ mol–1
4.22
One mole of anhydrous MgCl2 dissolves in water and librates 25 cal/mol of heat. Hhydration of MgCl2 = –30 cal/ mol. Heat of dissolution of MgCl2.H2O is (A) +5 cal/mol (B) –5 cal/mol (C) 55 cal/mol (D) –55 cal/mol
4.23
In the reaction CS2 () + 3O2 (g) CO2 (g) + 2SO2 (g) H = – 265 kcal The enthalpies of formation of CO2 and SO2 are both negative and are in the ratio 4 : 3. The enthalpy of formation of CS2 is +26kcal/mol. Calculate the enthalpy of formation of SO2. (A) – 90 kcal/mol (B) – 52 kcal/mol (C) – 78 kcal/mol (D) – 71.7 kcal/mol
4.24
Consider the Gºƒ and Hºƒ (kJ/mol) for the following oxides. Which oxide can be most easily decomposed to form the metal and oxygen gas ? (A) ZnO (Gº = –318.4, Hº = –348.3) (B) Cu2O (Gº = –146.0, Hº = –168.8) (C) HgO (Gº = –58.5, Hº = –90.8) (D) PbO (Gº = –187.9, Hº = –217.3)
4.25
If G = – 177 K cal for
3 O (g) Fe2O3 (s) 2 2 and G = – 19 K cal for (2) 4 Fe2O3 (s) + Fe(s) 3 Fe3O4 (s) What is the Gibbs free energy of formation of Fe3O4(s) ? (A) + 229.6 kcal/mol (B) – 242.3 kcal/mol (C) – 727 kcal/mol (D) – 229.6 kcal/mol (1) 2 Fe(s) +
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5.1
The reaction CH4(g) + Cl2(g)
5. Bond enthalpies CH3Cl(g) + HCl(g) has H = –25 kCal.
Bond
C—Cl H—Cl C—H Cl—Cl
Bond Energy kCal 84 103 x
y x:y= 9 :5
From the given data, what is the bond enthalpy of Cl—Cl bond (A) 70 kCal (B) 80 kCal (C) 67.75 kCal
(D) 57.75 kCal
5.2
The bond dissociation energy of gaseous H 2, Cl 2 and HCl are 104, 58 and 103 kcal mol –1 respectively. The enthalpy of formation for HCl gas will be (A) – 44.0 kcal (B) – 22.0 kcal (C) 22.0 kcal (D) 44.0 kcal
5.3
If x1, x2 and x3 are enthalpies of H–H, O=O and O–H bonds respectively, and x4 is the enthalpy of vaporisation of water, estimate the standard enthalpy of combustion of hydrogen (A) x1+
5.4
5.5
x2 2
–2x3+x4
(B) x1+
x2 2
–2x3–x4
x2 2
–x3+x4
The average O–H bond energy in H2O with the help of following data (1) H2O() H2O(g) ; H = + 40.6 KJ mol–1 (2) 2H(g) H2 (g) ; H = – 435.0 KJ mol–1 (3) O2(g) 2O(g) ; H = + 489.6 KJ mol–1 (4) 2H2 (g) + O2 (g) 2H2O() ; H = – 571.6 KJ mol–1 (A) 584.9 KJ mol–1 (B) 279.8 KJ mol–1 (C) 462.5 KJ mol–1
(D) 2x3–x1–
x2 2
–x4
(D) 925 KJ mol–1
What is the ratio of the enthalpy yield on combustion of hydrogen atoms to steam to the yield on combustion of an equal mass of hydrogen molecules to steam? Given : H2(g) +
1 O (g) H2O(g) 2 2
H = – 242 kJ
B.E. (H – H) = 436 kJ (A) 0.80 : 1 (B) 1 : 0.80 5.6
(C) x1+
(C) 1.80 : 1
(D) 2.80 : 1
Heat of hydrogenation of ethene is x1 and that of benzene is x2. Hence, resonance energy is : (A) x1 – x2 (B) x1 + x2 (C) 3x1 – x2 (D) x1 – 3x2
6. IInd Law of thermodynamics:Entropy calculation for different types of physical process on an ideal gas, solid and liquid 6.1
Predict which of the following reaction (s) has a positive entropy change ? I. Ag+ (aq) + Cl– (aq) AgCl (s) II. NH4Cl (s) NH3 (g) + HCl (g) III. 2NH3 (g) N2 (g) + 3H2 (g) (A) I and II (B) III (C) II and III (D) II
6.2
When two mole of an ideal gas Cp,m 2 R heated from 300 K to 600 K at constant pressure. The change
5
in entropy of gas (S) is : (A)
3 R ln 2 2
(B) –
3 R ln 2 2
(C) 5R ln 2
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(D)
5 R ln 2 2
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6.3
The entropy change when two moles of ideal monoatomic gas is heated from 200 to 300°C reversibly and isochorically (A)
3 300 R ln 2 200
(B)
5 573 R ln 2 273
573 473
(C) 3R ln
(D)
3 573 R ln 2 473
6.4
When one mole of an ideal gas is comressed to half of its initial volume and simultaneously heated to twice its initial temperature, the change in entropy of gas (S) is : (A) Cp, m ln 2 (B) Cv, m ln 2 (C) R ln 2 (D) (Cv, m – R) ln 2
6.5
The entropy change when two moles of ideal monoatomic gas is heated from 200 to 300ºC reversibly and isochorically ? (A)
6.6
300 3 R ln 200 2
(B)
573 5 R ln 273 2
If one mole of an ideal gas Cp,m
573
(C) 3R ln 473
(D)
573 3 R ln 473 2
5 R is expanded isothermally at 300 K until it’s volume is tripled, then 2
change in entropy of gas is : (A) zero
(B) infinity
(C)
5 R ln 3 2
(D) R ln 3
6.7
Two mole of an ideal gas is expanded irreversibly and isothermally at 37ºC until its volume is doubled and 3.41 kJ heat is absorbed from surrounding. Stotal (system + surrounding) is : (A) –0.52 J/K (B) 0.52 J/K (C) 22.52 J/K (D) 0
6.8
1 mole of an ideal gas at 25°C is subjected to expand reversibly and adiabatically to ten times of its initial volume. Calculate the change in entropy during expansion (in J k–1 mol–1) (A) 19.15 (B) – 19.15 (C) 4.7 (D) zero
6.9
One mole of an ideal diatomic gas (Cv = 5 cal) was transformed from initial 25°C and 1 L to the state when temperature is 100°C and volume 10 L. The entropy change of the process can be expressed as (R = 2 calories/mol/K) (A) 3 ln
298 + 2 ln 10 373
(B) 5 ln
373 + 2 ln 10 298
(C) 7 ln
373 1 + 2 ln 298 10
(D) 5 ln
373 1 + 2 ln 298 10
6.10
What is the change in entropy when 2.5 mole of water is heated from 27ºC to 87ºC ? Assume that the heat capacity is constant. (Cp,m (H2O) = 4.2 J/g-K ln (1.2) = 0.18) (A) 16.6 J/K (B) 9 J/K (C) 34.02 J/K (D) 1.89 J/K
6.11
Calculate the total entropy change for the transition at 368 K of 1 mol of sulphur from the monoclinic to the rhombic solid state and H = – 401.7 J mol–1 for the transition. Assume the surroundings to be an ice-water. Both at 0ºC : (A) – 1.09 JK–1 (B) 1.47 JK–1 (C) 0.38 JK–1 (D) None of these
6.12
Calculate standard entropy change in the reaction Fe2O3 (s) + 3H2 (g) 2Fe (s) + 3H2O () Given : Sºm (Fe2O3, S) = 87.4 , Sºm (Fe, S) = 27.3, Sºm (H2, g) = 130.7, Sºm (H2O, ) = 69.9 JK–1 mol–1. (A) –212.5 JK–1 mol–1 (B) –215.2 JK–1 mol–1 (C) –120.9 JK–1 mol–1 (D) None of these
6.13
Given rSº = –266 and the listed [Sºm values] calculate Sº for Fe3O4 (s) : 4Fe3O4 (s) [..............] + O2 (g) [205] 6Fe2O3 (s) [87] (A) +111.1 (B) +122.4 (C) 145.75 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
(D) 248.25
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6.14
C2H6 (g) + 3.5 O2 (g) 2CO2 (g) + 3H2O (g) Svap (H2O, ) = x1 cal K-1 (boiling point = T1) Hf (H2O, ) = x2 Hf (CO2) = x3 Hf (C2H6) = x4 Hence, H for the reaction is (A) 2x3 + 3x2 – x4 (C) 2x3 + 3x2 – x4 – 3x1T1
(B) 2x3 + 3x2 – x4 + 3x1T1 (D) x1T1 + X2 + X3 – x4
7. G calculation, spontaneity of chemical reaction, significance of G and IIIrd Law of thermodynamics 7.1
7.2*
For the reaction at 300 K A(g) + B(g) C (g)
E = –3.0 kcal value of G is
;
S = – 10.0 cal/K
(A) –600 cal
(B) –6600 cal
(C) –6000 cal
For isothermal expansion in case of an ideal gas : (A) H = 0 (B) E = 0 (C) G = –T.S
(D) None
(D) Tfinal = Tinitial
7.3
For the gas - phase decomposition, PCl5 (g) PCl3 (g) + Cl2 (g) : (A) H < 0, S < 0 (B) H > 0, S > 0 (C) H > 0, S < 0 (D) H < 0, S > 0
7.4
If Hvaporisation of substance X (l) (molar mass : 30 g/mol) is 300 J/g at it's boiling point 300 K, then molar entropy change for reversible condensation process is (A) 30 J/mol.K (B) – 300 J/mol.K (C) –30 J/mol.K (D) None of these
7.5
What is the free energy change (G) when 1.0 mole of water at 100ºC and 1 atm pressure is converted into steam at 100ºC and 1 atm pressure ? (A) 80 cal (B) 540 cal (C) 620 cal (D) Zero
7.6
What can be concluded about the values of H and S from this graph? (A) H > 0,
S > 0
(C) H < 0, S > 0
(B) H > 0,
S < 0
(D) H < 0, S < 0
7.7
The enthalpy change for a given reaction at 298 K is – x J mol–1 (x being positive). If the reaction occurs spontaneously at 298 K, the entropy change at that temperature (A) can be negative but numerically larger than x/298 (B) can be negative but numerically smaller than x/298 (C) cannot be negative (D) cannot be positive
7.8
The change in entropy of 2 moles of an ideal gas upon isothermal expansion at 243.6 K from 20 litre until the pressure becomes 1 atm, is : (A) 1.385 cal / K (B) –1.2 cal / K (C) 1.2 cal / K (D) 2.77 cal / K
7.9
A reaction has H = – 33 kJ and S = – 58
J . This reaction would be : K (A) spontaneous at all temperatures (B) non-spontaneous at all temperatures (C) spontaneous above a certain temperature (D) spontaneous below a certain temperature
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Single Correct Answer Type : 1.
A piece of zinc at a temperature of 20°C weighing 65.38 g is dropped into 180 g of boiling water (T = 100°C). The specific heat of zinc is 0.4 J g–1 C–1 and that of water is 4.2 J g–1°C–1. What is the final common temperature reached by both the zinc and water ? (A) 97.3 °C (B) 33.4 °C (C) 80.1 °C (D) 60.0°C
2.
The heat capacity of liquid water is 75.6 J / mol.K, while the enthalpy of fusion of ice is 6.0 kJ/mol. What is the smallest number of ice cubes at 0°C, each containing 9.0 g of water, needed to cool 500 g of liquid water from 20°C to 0°C? (A) 1 (B) 7 (C) 14 (D) None of these
3.
A certain mass of gas is expanded from (1L, 10 atm) to (4L, 5 atm) against a constant external pressure of 1 atm. If initial temperature of gas is 300 K and the heat capacity of process is 50 J/°C. Then the enthalpy change during the process is (1L atm ~ 100 J) (A) H = 15 kJ (B) H = 15.7 kJ (C) H = 14.4 kJ (D) H = 14.7 kJ
4.
A heating coil is immersed in a 100 g sample of H2O (l) at 1 atm and 100ºC in a closed vessel. In this heating process, 60% of the liquid is converted to the gaseous form at constant pressure of 1 atm. The densities of liquid and gaseous water under these conditions are 1000 kg/m3 and 0.60 kg/ m3 respectively. Magnitude of the work done for the process is : (A) 4997 J (B) 4970 J (C) 9994 J (D) None of these
5.
10 litres of a monoatomic ideal gas at 0°C and 10 atm pressure is suddenly released to 1 atm pressure and the gas expands adiabatically against this constant pressure. The final temperature and volume of the gas respectively are. (A) T = 174.9 K, V = 64 L (B) T = 153 K, V = 57 L (C) T = 165.4 K, V = 78.8 L (D) T = 161.2 K, V = 68.3 L
6.
Consider a classroom that is roughly 5 m × 10m × 3m. Initially t = 27°C and P = 1 atm. There are 50 people in an insulated class loosing energy to the room at the average rate of 150 watt per person. How long can they remain in class if the body temperature is 42°C and person feels uncomfortable above this temperature. Heat capacity of air = (7/2) R. (A) 4.34 minutes (B) 5.91 minutes (C) 6.86 minutes (D) 7.79 minutes
7.
At 5 × 105 bar pressure, density of diamond and graphite are 3 g/cc and 2 g/cc respectively, at certain temperature ‘T’. Find the value of U – H for the conversion of 1 mole of graphite to 1 mole of diamond at temperature ‘T’ : (A) 100 kJ/mol (B) 50 kJ/mol (C) – 100 kJ/mol (D) None of these
8.
A new flurocarbon of molar mass 102 g mol1 was placed in an electrically heated vessel. When the pressure was 650 torr, the liquid boiled at 770C. After the boiling point had been reached, it was found that a current of 0.25 A from a 12.0 volt supply passed for 600 sec vaporises 1.8g of the sample. The molar enthalpy & internal energy of vaporisation of new flourocarbon will be : (A)H = 102 kJ/mol, E = 99.1 kJ/mol (C) H = 107 kJ/mol, E = 105.1 kJ/mol
(B) H = 95 kJ/mol, E = 100.3 kJ/mol (D) H = 92.7 kJ/mol, E = 97.4 kJ/mol
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9.
Two rigid adiabatic vessels A and B which initially, contain two gases at different temperatures are connected
by pipe line with valve of negligible volume. The vessel 'A' contain 2 moles Ne gas C p , m
5 R at 300 K, 2
vessel 'B' contain 3 moles of SO2 gas (Cp,m = 4 R) at 400 K. The volume of A & B vessel is 4 and 6 litre respectively. The final total pressure (in atm) when valve is opened and 12 Kcal heat supplied through it to vessels. [Use : R = 2 cal / mol. K and R = 0.08 L . atm / mol K as per desire]
A Ne
(A) 3.5 atm
(B) 7 atm
B SO2 (C) 35 atm
(D) 70 atm
10.
The maximum efficiency of a heat engine operating between 100°C and 25°C is (A) 20.11% (B) 22.2% (C) 25.17% (D) None
11.
A heat engine operating between 227°C and 27°C absorbs 2 Kcal of heat from the 227°C reservoir reversibly per cycle. The amount of work done in one cycle is (A) 0.4 Kcal (B) 0.8 Kcal (C) 4 Kcal (D) 8 Kcal
12.
A reversible heat engine A (based on carnot cycle) absorbs heat from a reservoir at 1000K and rejects heat to a reservoir at T2. A second reversible engine B absorbs, the same amount of heat as rejected by the engine A, from the reservoir at T2 and rejects energy to a reservoir at 360K. If the efficiencies of engines A and B are the same then the temperature T2 is (A) 680 K (B) 640 K (C) 600 K (D) 550 K
13.
Pressure of 10 moles of an ideal gas is changed from 2 atm to 1 atm against constant external pressure without change in temperature. If surrounding temperature (300 K) and pressure (1 atm) always remains constant then calculate total entropy change (Ssystem + Ssurrounding) for given process. [Given : n2 = 0.70 and R = 8.0 J/mol/K] (A) 56 J/K (B) 14 J/K
(C) 16 J/K
(D) None of these
14.
For a perfectly crystalline solid Cp,m = aT3 + bT, where a and b constant. If Cp,m is 0.40 J/K mol at 10 K and 0.92 J/K mol at 20 K, then molar entropy at 20 K is : (A) 0.92 J/K mol (B) 8.66 J/K mol (C) 0.813 J/K mol (D) None of these
15.
When two equal sized pieces of the same metal at different temperatures Th (hot piece) and Tc(cold piece) are brought into contact into thermal contact and isolated from it's surrounding. The total change in entropy of system is given by
T (A) C ln c
Th
2Tc
16.
T (B) C ln 2 T1
(C) C ln
(Tc Th ) 2 2Th .Tc
(D) C ln
(Tc Th ) 2 4Th .Tc
Reactions involving gold have been of particular interest to a chemist . Consider the following reactions, Au(OH)3 + 4 HCl HAuCl4 + 3 H2O , H = 28 kCal Au(OH)3 + 4 HBr HAuBr4 + 3 H2O , H = 36.8 kCal In an experiment there was an absorption of 0.44 kCal when one mole of HAuBr4 was mixed with 4 moles of HCl . What is the percentage conversion of HAuBr4 into HAuCl4 ? (A) 0.5 % (B) 0.6 % (C) 5 % (D) 50 %
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17.
Ethanol can undergoes decomposition to form two sets of products
C2H5OH (g)
if the molar ratio of C2H4 to CH3CHO is 8 : 1 in a set of product gases, then the enthalpy involved in the decomposition of 1 mole of ethanol is (A) 65.98 kJ (B) 48.137 kJ (C) 48.46 kJ (D) 57.22 kJ 18.
19.
(i) Cis2 butene trans 2 butene, H1 (ii) Cis 2 butene 1 butene, H2 (iii) Trans 2 butene is more stable than cis 2 butene. (iv) Enthalpy of combustion of 1butene, H = 649.8 kCal/mol (v) 9H1 + 5 H2 = 0 (vi) Enthalpy of combustion of trans 2 butene, H = 647.0 kCal/mol. The value of H1 & H2 in KCal/mole are (A) 1.0 , 1.8 (B) 1.8, -1.0 (C) –5, 9
(D) –2, 3.6
The enthalpy changes of the following reactions at 27°C are
1 Cl (g) NaCl (s) rH = –411 kJ/mol 2 2 H2(g) + S (s) + 2O2 (g) H2SO4 (l) rH = –811 kJ/mol 2Na(s) + S(s) + 2O2 (g) Na2SO4 (s) rH = –1382 kJ/mol Na(s) +
1 1 H2(g) + Cl2(g) HCl (g) 2 2
rH = –92 kJ/mol;
R = 8.3 J/K-mol
from these data, the heat change of reaction at constant volume ( in kJ/mol) at 27°C for the process 2NaCl (s) + H2SO4 (l) Na2SO4 (s) + 2HCl (g) is (A) 67 (B) 62.02 (C) 71.98 (D) None 20.
The standard enthalpy of formation of FeO & Fe2O3 is 65 kcal mol1 and 197kcalmol1 respectively. A mixture of two oxides contains FeO & Fe2O3 in the mole ratio 2 : 1. If by oxidation, it is changed into a 1 : 2 mole ratio mixture, how much of thermal energy will be released per mole of the initial mixture ? (A) 13.4 kcal/mole (B) 14.6 kcal/mole (C) 15.7 kcal/mole (D) 16.8 kcal/mole
21.
An athelete is given 100 g of glucose (C6H12O6) of energy equivalent to 1560 kJ. He utilises 50 percent of this gained energy in the event. In order to avoid storage of energy in the body, the weight of water he would need to perspire is- (The enthalpy of evaporation of water is 44 kJ/mole.) (A) 319 gm (B) 422 gm (C) 293 gm (D) 378 gm
22.
The heat of formation of C2H5OH() is 66 kcal/mole. The heat of combustion of CH3OCH3 (g) is – 348 kcal/mole. Hf for H2O and CO2 are 68 kcal/mole and 94 kcal/mole respectively. Then, the H for the isomerisation reaction C2H5OH () CH3OCH3(g), and E for the same are (A)H = 18 kcal/mole, E = 17.301 kcal/mole (B) H = 22 kcal/mole, E = 21.408 kcal/mole (C) H = 26 kcal/mole, E = 25.709 kcal/mole (D) H = 30 kcal/mole, E = 28.522 kcal/mole
23.
AB, A 2 and B 2 are diatomic molecules. If the bond enthalpies of A 2, AB & B 2 are in the ratio 1 : 1 : 0.5 and enthalpy of formation of AB from A2 and B2 is – 100 kJ/mol–1. What is the bond enthalpy of A2. (A) 400 kJ/mol (B) 200 kJ/mol (C) 100 kJ/mol (D) 300 kJ/mol
24.
Enthalpy of polymerisation of ethylene, as represented by the reaction, nCH2=CH2 CH2CH2)n is 100 kJ per mole of ethylene. Given bond enthalpy of C=C bond is 600 kJ mol1, enthalpy of CC bond (in kJ mol) will be : (A) 116.7 (B) 350 (C) 700 (D) indeterminate ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
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25.
Substance A2B(g) can undergoes decomposition to form two set of products :
If the molar ratio of A2(g) to A(g) is 5 : 3 in a set of product gases, then the energy involved in the decomposition of 1 mole of A2B2 , is : (A) 48.75 kJ/mol (B) 43.73 kJ/mol (C) 46.25 kJ/mol (D) None of these 26.
For the hypothetical reaction A2(g) + B2(g) 2AB(g) If rGº and rSº are 20 kJ/mol and –20 JK–1 mol–1 respectively at 200 K. rCp is 20 JK–1 mol–1 then rHº at 400 K is : (A) 20 kJ/mol (B) 7.98 kJ/mol (C) 28 kJ/mol
(D) None of these
27.
Enthalpy of neutralization of H3PO3 acid is – 106.68 kJ/mol using NaOH. If enthalpy of neutralization of HCl by NaOH is –55.84 kJ/mol. Calculate Hionization of H3PO3 into its ions : (A) 50.84 kJ/mol (B) 5 kJ/mol (C) 2.5 kJ/mol (D) None of these
28.
The enthalpy of neutralisation of a weak acid in 1 M solution with a strong base is – 56.1 kJ mol–1. If the enthalpy of ionization of the acid is 1.5 kJ mol–1 and enthalpy of neutralization of the strong acid with a strong base is – 57.3 kJ equiv–1 , what is the % ionization of the weak acid in molar solution (assume the acid to be monobasic)? (A) 10 (B) 15 (C) 20 (D) 25
29.
The heat of formation of HCl at 348 K from the following data, will be 0.5 H2(g) + 0.5 Cl2 (g) HCl
H°298 = – 22060 cal
The mean heat capacities over this temperature range are, H2(g), CP = 6.82 cal mol–1 K–1 ; Cl2 (g), Cp = 7.71 cal mol–1 K–1 ; HCl (g), CP = 6.81 cal mol–1 K–1 (A) – 20095 cal (B) – 32758 cal (C) – 37725 cal (D) – 22083 cal 30.
The average XeF bond energy is 34 kcal/mol, first I.E. of Xe is 279 kcal/mol, electron affinity of F is 85 kcal/mol & bond dissociation energy of F2 is 38 kcal/mol. Then, the enthalpy change for the reaction XeF4 Xe+ + F + F2 + F will be (A) 367 kcal/mole
(B) 425 kcal/mole
(C) 292 kcal/mole
(D) 392 kcal/mole
31.
Combustion of sucrose is used by aerobic organisms for providing energy for the life sustaining processes. If all the capturing of energy from the reaction is done through electrical process (non P–V work) then calculate maximum available energy which can be captured by combustion of 34.2 g of sucrose Given : Hcombustion (sucrose) = – 6000 kJ mol–1 Scombustion = 180 J/K - mol and body temperature is 300 K (A) 600 kJ (B) 594.6 kJ (C) 5.4 kJ (D) 605.4 kJ
32.
The enthalpy of tetramerization of X in gas phase (4X(g) X 4(g)) is – 100 kJ/mol at 300 K. The enthalpy of vaporisation for liquid X and X4 are respectively 30 kJ/mol and 72 kJ/mol respectively. S for tetramerization of X in liquid phase is – 125 J / K mol at 300 K. What is the G at 300 K for tetramerization of X in liquid phase ? (A) –52 kJ/mol (B) –89.5 kJ/mol (C) –14.5 kJ/mol (D) None of these
One or More Than One Correct Answer Type : 33.
The normal boiling point of a liquid 'A' is 350 K.Hvap at normal boiling point is 35 kJ/mole. Pick out the correct statement(s). (Assume DHvap to be independent of pressure). (A) Svaporisation > 100 KJ/mole at 350 K and 0.5 atm (B) Svaporisation < 100 KJ/mole at 350 K and 0.5 atm (C) Svaporisation < 100 KJ/mole at 350 K and 2 atm (D) Svaporisation = 100 KJ/mole at 350 K and 2 atm ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
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34.
Which of the following statement(s) is/are false : (A) rS for
1 N (g) N(g) is positive 2 2
(B) Gsystem is always zero for a reversible process in a closed system (C) G° for an ideal gas is a function of temperature and pressure (D) entropy of a closed system is always maximized at equilibrium 35.
Which statement is / are correct: (A) Final temperature in reversible adiabatic expansion is lesser than in irreversible adiabatic expansion. (B) When heat is supplied to an ideal gas in an isothermal process, kinetic energy of gas will increase (C) When an ideal gas is subjected to adiabatic expansion it gets cooled (D) Entropy increases in atomisation of dihydrogen
36.
Which statement is / are correct: (A) wadiabatic > wisothermal in an ideal gas compression from same initial state to same final state (B) The value of
C p remains constant for diatomic gas at all temperature Cv
(C) Entropy increases when an ideal gas expanded isothermally. (D) rH & rS both are + ve for the decomposition of MgCO3(s) . 37. 38.
In isothermal ideal gas compression: (A) w is +ve (B) H is zero
(C) Sgas is +ve
(D) G is +ve
Which of the following is true for reversible adiabatic process involving an ideal gas? (A) Gas with higher has high magnitude of slope in a P (y-axis) v/s T (x-axis) curve (B) Gas with higher has high magnitude of slope in a V (y-axis) v/s T (x-axis) curve (C) Gas with higher has high magnitude of slope in a P (y-axis) v/s V (x-axis) curve (D) Gas with higher has low magnitude of slope in a P (y-axis) v/s T (x-axis) curve
39.
One mole of an ideal diatomic gas (Cv = 5 cal) was transformed from initial 25°C and 1 L to the state when temperature is 100°C and volume 10 L. Then for this process(R = 2 calories/mol/K) (take calories as unit of energy and kelvin for temp) (A) H = 525
(B) S = 5 ln
373 + 2 ln 10 298
(C) E = 525
(D) G of the process can not be calculated using given information. 40.
Consider the reactions (i) S (rhombic) + 3/2 O2(g) SO3(g), H1
(ii) S (monoclinic) + 3/2O2(g) SO3(g), H2
(iii) S (rhombic) + O3(g) SO3(g), H3 (A) H1 < H2 < H4 (magnitude only) (C) H1 < H2 = H4 (magnitude only)
(iv) S (monoclinic) + O3(g) SO3(g), H4 (B) H1 < H3 < H4 (magnitude only) (D) H1 + H4 = H3
41.
The normal boiling point of a liquid `X` is 400 K. Which of the following statement is true about the process X (l) X(g)? (A) at 400 K and 1 atm pressure G = 0 (B) at 400 K and 2 atm pressure G = + ve (C) at 400 K and 0.1 atm presure G = – ve (D) at 410 K and 1 atm pressure G = + ve
42.
100 ml 0.5 N H2SO4 (strong acid) is neutralised with 200 ml 0.2M NH4OH in a constant pressure Calorimeter which results in temperature rise of 1.4 °C. If heat capacity of Calorimeter content is 1.5 kJ/°C. Which statement is/are correct Given : HCl + NaOH NaCl + H2O + 57 kJ CH3COOH + NH4OH CH3COONH4 + H2O + 48.1 kJ (A) Enthalpy of neutralisation of HCl v/s NH4OH is – 52.5 kJ/mol (B) Enthalpy of dissociation (ionization) of NH4OH is 4.5 kJ/mol (C) Enthalpy of dissociation of CH3COOH is 4.6 kJ/mol (D) H for 2H2O(l) 2H+ (aq.) + 2OH¯(aq.) is 114 kJ ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
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43.
From the following data at 25°C Reaction
1 1 H2(g) + O2(g) OH (g) 2 2 1 H2(g) + O2(g) H2O(g) 2 H2(g) 2H(g) O2(g) 2O(g)
rH° kJ/mol 42
–242 436 495
Which of the following statement(s) is/are correct: (A) rH° for the reaction H2O (g) 2H(g) + O(g) is 925.5 kJ/mol (B) rH° for the reaction OH(g) H(g) + O(g) is 502 kJ/mol (C) Enthalpy of formation of H(g) is –218 kJ/mol (D) Enthalpy of formation of OH(g) is 42 kJ/mol 44.
Which of the following is true ? (A) For the reaction CaCO3(calcite) CaCO3 (aragonite) given : (B)
45.
46.
f G o298 (calcite) = –1128.8 kJ/mol, f G o298 (aragonite) = – 1127.75 kJ/mol,
then calcite form is thermodynamically more stable at standard conditions. For the reaction, (a) C(diamond) + 2H2 (g) CH4(g) H1
(C)
(b) C(g) + 4H(g) CH4(g) then more heat is evolved in reaction (b). fH° (I2, g) = subH [I2, s] at 25°C.
H2
(D)
For the exothermic reaction 2Ag(s) + 1/2 O2(g) 2Ag2O(s) at 298 K. H < U
Which of the following do(es) not represent H formation of the product. (A)
1 H (g) + (aq) H+(aq) 2 2
(B)
2 O (g) O2(g) + e¯ 3 3
(C)
NH 4 (g) + Cl–(g) NH4Cl(s)
(D) P4(black) + 5O2 (g) P4O10(s)
Which of the following statement(s) is/are true? (A) When (Gsystem)T,P < 0; the reaction must be exothermic (B)
f H° (S, monoclinic) 0
(C)
If dissociation enthalpy of CH4(g) is 1656 kJ/mole and C2H6 (g) is 2812 kJ/mole, then value of C–C bond enthalpy will be 328 kJ/mole
(D)
If H+(aq) + OH–(aq) H2O(l) rH° = –56 kJ/mol
f H°(H2O, g) = –242 kJ/mole; Enthalpy of vaporization of liquid water = 44 kJ/mol then, f H°(OH–, aq) will be –142 kJ/mole 47.
Select the option in which heat evolved is maximum. Given : fH° (CO2,g) = – 75 kCal/mol ; fH° (CO,g) = – 25 kCal/mol The product will be CO if excess amount of carbon is present and CO2 if excess O2 is present (A) 10 moles of carbon and 4.5 moles of O2 (B) 24 gm of carbon and 64 gm of O2 (C) 4 moles of carbon and 3.5 moles of O2 (D) 30 gm of carbon and 80 gm of O2
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1. MATCH THE COLUMN 1.1
Columm-I
Columm-II
(A) Reversible isothermal expansion of an ideal gas
V2 (p) w = –2.303 nRT log V
(B) Reversible adiabatic compression of an ideal gas
(q) PV = constant
(C) Irreversible adiabatic expansion of an ideal gas (D) Irreversible isothermal compression of an ideal gas
(r) w = ( 1) (T2 – T1) (s) H = 0
Columm - I (A) C (s, graphite) + O2(g) CO2(g) (B) C(s, graphite) C(g)
Columm - II (p) Hºcombustion (q) Hºformation
1.2
1 O (g) CO2(g) 2 2 (D) CH4(g) C(g) + 4H(g) (C) CO(g) +
1.3
1.4
1
nR
(r) Hºatomization (s) Hºsublimation
Column-I, may have more than one matching options in column-II Column-I Column-II (A) Reversible condensation of vapour (p) Ssystem > 0 (B) Irreversible adiabatic compression of an ideal gas
(q)
(C) Isothermal free expansion of an ideal gas in vacuum
(r) Ssurounding < 0
(D) Reversible adiabatic expansion of an ideal gas
(s) Ssurounding = 0
Column–I (Related to process) (A) Fusion at melting point (B) Vapourisation at boiling point (C) Condensation at triple point
Column–II (Related to system) (p)G = 0 (q) G < 0 (r) S > 0
(D) Melting at normal boiling point
(s) H ~ U
1.5 (A) (B) (C)
(D)
Ssystem < 0
Column I
Column II
For the process A(l) A(s),H & V may be A2(s) + B2(g) C2(s) + D2(s) H & G may be For the given reaction A2(g) B2(g) + C2(g), Ea(forward)= 50 kJ/mol and Ea(backward) = 40 kJ/mol , at very high temperature H & G are For the given reaction A(g) B(g), at very low temperature H & G may be
(p) – ve, + ve (q) + ve, – ve (r) + ve, + ve
(s) – ve, – ve
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2. COMPREHENSION Read the following comprehension carefully and answer the questions : Comprehension-1 Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram. 5
2×10
A
B
D
C
P(Pa) 5
1×10
300K
500K
T
2.1
Assuming the gas to be ideal the work done by the gas in taking it from A to B is – A to B is (A) 200 R (B) 300 R (C) 400 R (D) 500 R
2.2
The work done on the gas in taking it from D to A is – (A) – 414 R (B) + 414 R (C) – 690 R
(D) + 690 R
The net work done by the gas in the cycle ABCDA is – (A) Zero (B) 276 R (C) 1076 R
(D) 1904 R
2.3
Comprehension # 2 A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages of expansion/compression the states of gases are not defined). The work done can be calculated using dw = – Pext dV while in case of reversible process the work done can be calculated using dw = – PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process. Since P =
nRT , so V vf
w=
dw = –
nRT Vf . dV = –nRT ln V Vi vi
Since dw = – PdV so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram. 2.4
An ideal gaseous sample at initial state i (P0 , V0,T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2 = K1 and in second process the equation of the process is PV = K2 . Then, (A) work done in first process will be greater than work in second process (magnitude wise) (B) The order of values of work done can not be compared unless we know the value of K1 and K2. (C) value of work done (magnitude) in second process is greater in above expansion irrespective of the value of K1 and K2 . (D) st process is not possible
2.5
There are two samples of same gas initially under similar initial state. Gases of both the samples are expanded. st sample using reversible isothermal process and nd sample using reversible adiabatic process till final pressures of both the samples becomes half of initial pressure, then (A) Final volume of st sample < final volume of nd sample (B) Final volume of nd sample < final volume of st sample (C) final volumes will be equal (D) Information is insufficient
2.6
In the above problem (A) work done by gas in st sample > work done by gas in nd sample (B) work done by gas in nd sample > work done by gas in st sample (C) work done by gas in st sample = work done by gas in nd sample (D) none of these ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
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2.7
If four identical samples of an ideal gas initially at similar state (P0, V0, T0) are allowed to expand to double their volumes by four different process. : by isothermal irreversible process : by reversible process having equation P2 V = constant : by reversible adiabatic process V : by irreversible adiabatic expansion against constant external pressure. Then, in the graph shown the final state is represented by four different points then, the correct match can be (A) 1 – , 2 - , 3 – , 4 - V (B) 1 – , 2 - , 3 – V, 4 - (C) 2 – , 3 - , 4 – , 1 - V (D) 3 – , 1 - , 3 – V, 4 -
2.8
Two samples (initially under same states) of an ideal gas are first allowed to expand to double their volume using irreversible isothermal expansion against constant external pressure, then samples are returned back to their original volume first by reversible adiabatic process and second by reversible process having equation PV2 = constant then (A) final temperature of both samples will be equal (B) final temperature of first sample will be greater than of second sample (C) final temperature of second sample will be greater than of first sample (D) none of these.
Comprehension-3 A cylindrical container of volume 44.8 litres is containing equal no. of moles (in integer no.) of an ideal monoatomic gas in two sections A and B separated by an adiabatic frictionless piston as shown in figure. The initial temperature and pressure of gas in both section is 27.3 K and 1 atm. Now gas in section 'A' is slowly heated till the volume of section B becomes (1/8)th of initial volume.
Given : R = 2 cal/mol-K, Cv,m of monoatomic gas =
3 R, 2
At 1 atm & 0°C ideal gas occupy 22.4 litre. 2.9 2.10 2.11
What will be the final pressure in container B : (A) 2 atm (B) 8 atm
(C) 16 atm
(D) 32 atm
Find temperature in container A will be (A) 1638 K (B) 6988 K
(C)3274 K
(D) 51 K
Change in enthaly for section A in Kcal. (A) 48.3 (B) 80.53
(C) 4.83
(D) 8.05
Comprehension 4 Using the data (all values are in kJ/mol at 25º C) given below : Hºcombustion (ethane) = 1559.8 ; Hºcombustion (ethene) = 1410.9 Hºcombustion (acetylene) = 1299.7 ; Hºcombustion (acetaldehyde) = 1192.3 Hºf CO2(g) = 393.5 ; Hºf of H2O(l) = 285.8 Hº for C(s) (graphite) C(g) = + 716.68 ; Bond enthalpy of H H = 435.94 Bond enthalpy of O = O = 498.94 Calculate the following bond energies : 2.12 2.13
CC (A) 97.81 kJ
(B) 195.62 kJ
(C) 48.9 kJ
(D) 434.3 kJ
CH (A) 227.32 kJ
(B) 454.64 kJ
(C) 151.54 kJ
(D) 909.28 kJ
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2.14
2.15
2.16
C=O (A) 402.13 kJ
(B) 201.06 kJ
(C) 804.26 kJ
(D) 1608.52 kJ
C=C (A) 97.81 kJ
(B) 195.62 kJ
(C) 217.15 kJ
(D) 434.3 kJ
CC (A) 733.48 kJ
(B) 366.74 kJ
(C) 1466.96 kJ
(D) None of these
Comprehension # 5 For an ideal monoatomic gas, an illustration of three different paths A, (B + C) and (D + E) from an initial state P1, V1, T1 to a final state P2, V2, T1 is shown in the given figure.
Path A represents a reversible isothermal expansion from P1, V1 to P2, V2, path (B + C) represents a reversible adiabatic expansion (B) from P1, V1, T1 to P3, V2, T2 followed by reversible heating of the gas at constant volume (C) from P3, V2, T2 to P2, V2, T1 . Path (D + E) represents a reversible expansion at constant pressure P1 (D) from P1, V1, T1 to P1, V2, T3 followed by a reversible cooling at constant volume V2 (E) from P1, V2, T3 to P2 , V2, T1. 2.17
What is qrev, for path (A) ?
V2 (B) – nR ln V 1
(A) Zero 2.18
V2 (D) nRT1 ln V 1
(C) Zero
(D) nR(V2 – V1)
V2 (C) – nR ln V 1
V2 (D) nR ln V 1
What is S for path (A) ?
V2 (A) nR ln V 1 2.19
V2 (C) – nRT1 ln V 1
V2 (B) – nR ln V 1
What is S for (D + E) T1
(A) Zero
(B)
T3
C V (T ) dT T
3. ASSERTION / REASONING DIRECTIONS : Each question has 4 choices (A), (B), (C), (D) out of which ONLY ONE is correct. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 3.1
Statement -1 : Statement -2 :
There is no change in enthalpy of an ideal gas during compression at constant temperature. Enthalpy of an ideal gas is a function of temperature and pressure.
3.2
Statement-1: Due to adiabatic free expansion, temperature of a real gas always increases. Statement-2: If a real gas is at inversion temperature then no change in temperature is observed in adiabatic free expansion.
3.3
Statement-1: The enthalpy of formation of H2O() is greater than of H2O (g) in magnitude. Statement-2: Enthalpy change is negative for the condensation reaction H2O (g) H2O() ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
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3.4
Statement-1 : Heat of neutralisation of perchloric acid, HClO4, with NaOH is same as that of HCl with NaOH. Statement-2 : Both HCl and HClO4 are strong acids.
3.5
Statement-1 : In the following reaction : C(s) + O2 (g) CO2 (g) ;
H = U – RT
Statement-2 : H is related to U by the equation, H = U + ng RT 3.6
Statement-1 : The magnitude of the work involved in an isothermal expansion is greater than that involved in an adiabatic expansion. Statement-2 : P–V curve (P on y-axis and V on x-axis) decrease more rapidly for reversible adiabatic expansion compared to reversible isothermal expansion starting from same initial state.
3.7
Statement-1 : Entropy change in reversible adiabatic expansion of an ideal gas is zero. Statement-2 : The increase in entropy due to volume increase just componsate the decrease in entropy due to fall in temperature.
3.8
Statement-1 : The standard free energy changes of all spontaneously occuring reactions are negative. Statement-2 : The standard free energies of the elements in their standard states at 1 bar and 298 K are taken as zero.
3.9
Statement-1 : A reaction which is spontaneous and accompanied by decrease of randomness, must be exothermic. Statement-2 : All exothermic reactions are accompanied by decrease of randomness.
3.10
Statement-1 : Many endothermic reactions that are not spontaneous at room temperature become spontaneous at high temperature. Statement-2 : Hº of the endothermic reaction increases with increase in temperature.
4. TRUE / FALSE 4.1
Pressure is an intensive property.
4.2
Like U and H, S is also a state function.
4.3
When a system undergoes a change at constant pressure, it is referred to an isothermal process.
4.4
A reversible process is always quasi-static.
4.5
The workdone by a gas during free expansion is equal to zero.
4.6.
First law of T.D. is applicable to all processes irrespective to whether they are reversible or irreversible.
4.7
All spontaneous processes proceed in one direction only.
4.8
Whenever a system undergoes a cyclic change
4.9
Positive value of Ssystem during the process can be taken as sole criterion of spontaneity.
4.10
A real crystal has higher entropy than the ideal crystal.
dQ 0 T
5. FILL IN THE BLANKS 5.1
According to IUPAC conventions work done on the surroundings is________ .
5.2
A system is said to be _______ if it can neither exchange matter nor energy with surrounding.
5.3
A carnot cycle uses only _______ thermal reservoir.
5.4
A carnot cycle consists of only _______ processes.
5.5
The efficiency of a carnot engine can be increased by _________ sink temperature when the source temperature is held constant.
5.6
For a reversible adiabatic process, S = constant and hence it is called as an ______ process.
5.7
Entropy change of a system is determined by the ________ and ______ states only, irrespective of how the system has changed its states.
5.8
Solidification of liquid shows ________ in entropy.
5.9
When Fe(s) is dissolved in a aqueous HCl in a closed rigid vessel, the work done is ______.
5.10
For Non-spontaneous process at constant T & P, G is _________. ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
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PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS) * Marked Questions are having more than one correct option. 1.
In thermodynamics, a process is called reversible when (A) surroundings and system change into each other (B) there is no boundary between system and surroundings (C) the surroundings are always in equilibrium with the system (D) the system changes into the surroundings spontaneously
[JEE 2001, 1/35]
2.
Which one of the following statement is false : [JEE 2001, 1/35] (A) work is a state function (B) temperature is a state function (C) change in the state is completely defined when the initial and final states are specified (D) work appears at the boundary of the system.
3.
One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) (4.0 atm, 5.0 L, 245 K) with a change in internal energy, U = 30.0 L atm. The change in enthalpy (H) of the process in L atm is (A) 40.0 (B) 42.3 [JEE 2002, 3/90] (C) 44.0 (D) not defined, because pressure is not constant
4.
Two moles of a perfect gas undergo the following process : [JEE 2002, 5/60] (a) a reversible isobaric expansion from (1.0 atm, 20.0 L) to (1.0 atm, 40.0 L). (b) a reversible isochoric change of state from (1.0 atm, 40.0 L) to (0.5 atm, 40.0 L). (c) a reversible isothermal compression from (0.5 atm, 40.0 L) to (1.0 atm, 20.0 L). (i) Sketch with labels each of the processes on the same P-V diagram. (ii) Calculate the total work (w) and the total heat change (q) involved in the above processes. (iii) What will be the values of U, H and S for the overall process ?
5.
Which of the following equation gives the values of heat of formation (Hf0)
[JEE 2003, 3/84]
(A) C (diamond) + O2(g) CO2 (g)
1 1 (B) H2(g) + F2(g) HF (g) 2 2
(C) N2 (g) + 3H2 (g) 2NH3 (g)
(D) H2(g) + F2 (g) 2HF (g)
6.
The molar heat capacity, Cv, of helium gas is 3/2 R and is independent of temperature. For hydrogen gas, Cv approaches 3/2 R at very low temperatures, equal 5/2 R at moderate temperatures and is higher than 5/2 R at high temperatures. Give a reason for the temperature dependence of Cv in case of hydrogen, in not more than two or three sentences. [JEE 2003, 2/60]
7.
2 moles of ideal gas is expanded isothermally & reversibly from 1 litre to 10 litre. Find the enthalpy change in kJ mol–1. [JEE 2004, 3/84] (A) 0 (B) 11.7 (C) – 11.7 (D) 25
8.
H = 30 kJ mol–1 , S = 75 J / k / mol. Find boiling temperature at 1 atm. [JEE 2004, 3/84] (A) 400 K (B) 300 K (C) 150 K (D) 425 K
9.
Spontaneous adsorption of a gas on a solid surface is exothermic process because [JEE 2004, 3/84] (A) enthalpy of the system increases. (B) entropy increases. (C) entropy decreases. (D) free energy change increases.
10.
There is 1 mol liquid (molar volume 100 ml) in an adiabatic container initial, pressure being 1 bar Now the pressure is steeply increased to 100 bar, and the volume decreased by 1 ml under constant pressure of 100 bar. Calculate H and E. [Given 1 bar = 105 N/m2] [JEE 2004, 2/60]
11.
For the reaction : N2O4(g) 2NO2 (g) (i) In a mixture of 5 mol NO2 and 5 mol N2O4 and pressure of 20 bar. Calculate the value of G for the reaction. Given Gf° (NO2) = 50 kJ / mol, Gf° (N2O4) = 100 kJ / mol and T = 298 K. (ii) Predict the direction in which the reaction will shift, in order to attain equilibrium [Given at T = 298 K, 2.303 RT = 5.7 kJ / mol.] [JEE 2004, 2/60] ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 50
12.
One mole of an ideal monoatomic gas at temperature T and volume 1L expands to 2L against a constant external pressure of one atm under adiabatic conditions, then final temperature of gas will be : [JEE 2005, 3/84] (A) T +
13.
(B) T –
2 3 0.0821
(C)
T 2
5 / 3 1
(D)
T 2
5 / 3 1
The ratio of P to V at any instant is constant and is equal to 1, for a monoatomic ideal gas under going a process. What is the molar heat capacity of the gas. [JEE 2006, 3/184] (A)
14.
2 3 0.0821
3R 2
(B)
4R 2
(C)
5R 2
(D) 0
The direct conversion of A to B is difficult, hence it is carried out by the following shown path:
S(A C) = 50 ;
S(C D) = 30;
S(B D) = – 20
The entropy change for the process A B is (A) 100
(B) – 60
[JEE 2006, 3/184] (C) – 100
(D) + 60
15.
The given reaction 2CO2 2CO + O2 H = – 560 kJ 2moles 1 mole is carried and in one litre container, if the pressure in the container gets changes from 70 atm to 40 atm as reaction gets completed. Calculate U of the reaction. [1L atm = 0.1 kJ] [JEE 2006, 6/184]
16.
The value of log10K for a reaction A
B is : (Given : r H298K = –54.07 kJ mol–1, r S 298K = 10 JK–1 mol–1 and
R = 8.314 JK–1 mol–1 ; 2.303 × 8.314 × 298 = 5705) (A) 5 (B) 10 (C) 95
[JEE 2007, 3/162] (D) 100
17.
For the process H2O() (1 bar, 373 K) H2O(g) (1 bar, 373 K), the correct set of thermodynamic parameters is : [JEE 2007, 3/162] (A) G = 0, S = +ve (B) G = 0, S = –ve (C) G = +ve, S = 0 (D) G = –ve, S = +ve
18.
STATEMENT-1 : For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero. STATEMENT-2 : At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy. [JEE 2008, 3/163] (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True
19.
STATEMENT-1 : There is a natural asymmetry between converting work to heat and converting heat to work. STATEMENT-2 : No process is possible in which the sole result is the absorption of heat form a reservoir and its complete conversion into work. [JEE 2008, 3/163] (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True
20.*
Among the following, the state function(s) is(are) : (A) Internal energy (B) Irreversible expansion work (C) Reversible expansion work (D) Molar enthalpy
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[JEE 2009, 4/160]
THERMODYNAMICS_JEE MAINS # 51
21.
In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increases from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K–1, the numerical value for the enthalpy of combustion of the gas in kJ mol–1 is. [JEE 2009, 4/160]
22.
The species which by definition has ZERO standard molar enthalpy of formation at 298 K is : [JEE 2010, 3/163] (A) Br2(g) (B) Cl2(g) (C) H2O(g) (D) CH4(g)
23.
The bond energy (in kcal mol–1) of a C–C single bond is approximately : [JEE 2010, 3/163] (A) 1 (B) 10 (C) 100 (D) 1000
24.*
Among the following, the intensive property is (properties are) : (A) molar conductivity (B) electromotive force (C) resistance
25.
One mole of an ideal gas is taken from a and b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is ws and that along the dotted line path is wd, then the integer closest to the ratio wd / ws is : [JEE 2010, 3/163]
26.
Match the transformations in column I with appropriate options in column II. [JEE 2011, 8/240] Column-I Column-II (A) CO2(s) CO2(g) (p) phase transition (B) CaCO3(s) CaO(s) + CO2(g) (q) allotropic change (C) 2H• H2(g) (r) H is positive (D) P(white, solid) P(red, solid) (s) S is positive (t) S is negative For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct? [Take S as change in entropy and w as work done] [JEE 2012, 4/198]
27.
(A) Sx z = Sx y + Sy z (C) wx y z = wx y
[JEE 2010, 3/163] (D) heat capacity
(B) wx z = wx y + wy z (D) Sx y z = Sx y
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THERMODYNAMICS_JEE MAINS # 52
28.
Using the data provided, calculate the multiple bond energy (kJ mol–1) of a C C bond in C2H2. That energy is (take the bond energy of a C – H bond as 350 kJ mol–1) [JEE 2012, 3/198] 2C(s) + H2(g) C2H2(g) 2C(s) 2C(g)
H = 225 kJ mol–1 H = 1410 kJ mol–1
H2(g) 2H(g) H = 330 kJ mol–1 (A) 1165 (B) 837 (C) 865 29.
(D) 815
The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statements(s) is (are) correct ? [JEE 2012, 4/198]
(A) T1 = T2 (C) wisothermal > wadiabatic
(B) T3 > T1 (D) Uisothermal > Uadiabatic Paragraph for Question Nos. 30 to 31
A fixed mass 'm' of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure.
30.
The pair of isochoric processes among the transformation of states is : (A) K to L and L to M (B) L to M and N to K (C) L to M and M to N
[JEE Advanced 2013] (D) M to N and N to K
31.
The succeeding operations that enable this transformation of states, are : [JEE Advanced 2013] (A) Heating, cooling, heating, cooling (B) Cooling, heating, cooling, heating (C) Heating, cooling, cooling, heating (D) Cooling, heating, heating, cooling
PART - II : AIEEE PROBLEMS (PREVIOUS YEARS) 1.
An ideal gas expands in volume from 1 × 10–3 m3 to 1 × 10–2 m3 at 300 K against a constant pressure of 1 × 105 Nm–2 . The work is : [AIEEE 2004] (1) – 900 J (2) – 900 kJ (3) 270 kJ (4) + 900 kJ
2.
The enthalpies of combustion of carbon and carbon monoxide are – 393.5 and – 283 kJ mol–1 respectively. The enthalpy of formation of carbon monoxide per mole is : [AIEEE 2004] (1) 110.5 kJ (2) 676.5 kJ (3) – 676.5 kJ (4) – 110.5 kJ
3.
For a spontaneous reaction the G, equilibrium constant (K) and E°cell will be respectively : [AIEEE 2005] (1) – ve, > 1, + ve (2) + ve, > 1, – ve (3) – ve, < 1, –ve (4) – ve, > 1, – ve ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 53
4.
If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of 1 : 1 : 0.5 and Hf for the formation of XY is - 200 KJ mol–1. The bond dissociation energy of X2 will be : [AIEEE 2005] (1) 800 KJ mol–1 (2) 200 KJ mol–1 (3) 300 KJ mol–1 (4) 400 KJ mol–1
5.
Consider the reaction, N2 (g) + 3H2 (g) 2NH3 (g); carried out at constant temperature and pressure. If H and U are enthalpy change and internal energy change respectively, which of the following expressions is true ? [AIEEE 2005] (1) H = 0 (2) H = U (3) H < U (4) H > U
6.
An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti is the initial temperature and Tf is the final temperature, which of the following statements is correct ? [AIEEE 2006] (1) Tf > Ti for reversible process but Tf = Ti for irreversible process (2) (Tf)rev = (Tf)irrev (3) Tf = Ti for both reversible and irreversible processes (4) (Tf)irrev > (Tf)rev
7.
The enthalpy changes for the following processes are listed below: Cl2(g) 2Cl(g); 242.3 kJ mol1 ; I2(g) 2I(g); 151.0 kJ mol1
[AIEEE 2006]
ICl(g) I(g) + Cl(g); 211.3 kJ mol1 ; I2(s) I2(g); 62.76 kJ mol1 Given that the standard state for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) is (1) 16.8 kJ mol1 (2) +16.8 kJ mol1 (3) +244.8 kJ mol1 (4) 14.6 kJ mol1 8.
(HU) for the formation of carbon monoxide (CO) from its elements at 298 K is [AIEEE 2006] 1 1 (R = 8.314 J K mol ) (1) 1238.78 J mol1 (2) 2477.57 J mol1 (3) 2477.57 J mol1 (4) 1238.78 J mol1
9.
Identify the correct statement regarding a spontaneous process : [AIEEE 2007, 3/120] (1) Exothermic processes are always spontaneous. (2) Lowering of energy in the reaction process is the only criterion for spontaneity. (3) For a spontaneous process in an isolated system, the change in entropy is positive. (4) Endothermic processes are never spotaneous.
10.
In conversion of lime-stone to lime, CaCO3 (s) CaO(s) + CO2 (g) the values of H0 and S0 are + 179.1 kJ mol–1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that H° and S° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is : [AIEEE 2007, 3/120] (1) 845 K (2) 1118 K (3) 1008 (4) 1200 K
11.
Assuming that water vapour is an ideal gas, the internal energy change (U) when 1 mol of water is vapourised at 1 bar pressure and 100°C, (Given : Molar enthalpy of vapourization of water at 1 bar and 373 K = 41 kJ mol– 1 and R = 8.3 J mol–1 K–1) will be : [AIEEE 2007, 3/120] (1) 37.904 kJ mol–1 (2) 41.00 kJ mol–1 (3) 4.100 kJ mol–1 (4) 3.7904 mol–1
12.
The standard enthalpy of formation (Hf°) at 398 K for methane, CH4(g) is 74.8 kJ mol–1. The additional information required to determine the average energy for C - H bond formation would be : [AIEEE 2007, 3/120] (1) the dissociation energy of H2 and enthalpy of sublimation of carbon (2) latent heat of vapourisation of methane (3) the first four ionization energies of carbon and electron gain enthalpy of hydrogen (4) the dissociation energy of hydrogen molecule, H2
13.
Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK–1 mol–1, respectively. For the reaction, (1) 500 K
[AIEEE 2008, 3/105]
1 3 X + Y2 XY3 H = – 30 kJ. To be at equilibrium the temperature will be : 2 2 2
(2) 750 K
(3) 1000 K
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(4) 1250 K
THERMODYNAMICS_JEE MAINS # 54
14.
On the basis of the following thermochemical data : (ƒGºH+(aq) = 0) H2O() H+ (aq) + OH– (aq.) ; H = 57.32 kJ H2(g) +
1 O2 (g) H2O () ; H = –286.20 kJ 2
The value of enthalpy of formation of OH– ion at 25ºC is : (1) –228.88 kJ (2) +228.88 kJ (3) –343.52 kJ 15.
[AIEEE 2009, 8/144]
(4) –22.88 kJ
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH3OH() +
3 O2 (g) CO2(g) + 2H2O() 2
At 298 K, standard Gibb’s energies of formation for CH3OH(), H2O() and CO2 (g) are –166.2, –237.2 and –394.4 kJ mol–1 respectively. If standard enthalpy of combustion of methanol is –726kJ mol–1, efficiency of the fuel cell will be : [AIEEE 2009, 8/144] (1) 87% (2) 90% (3) 97% (4) 80% 16.
The standard enthalpy of formation of NH3 is – 46.0 kJ mol–1. If the enthalpy of formation of H2 from its atoms is – 436 kJ mol–1 and that of N2 is – 712 kJ mol–1, the average bond enthalpy of N – H bond in NH3 is : [AIEEE 2010, 4/144] (1) – 964 kJ mol–1 (2) + 352 kJ mol–1 (3) + 1056 kJ mol–1 (4) – 1102 kJ mol–1
17.
For a particular reversible reaction at temperature T, H and S were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when. [AIEEE 2010, 4/144] (1) Te > T (2) T > Te (3) Te is 5 times T (4) T = Te
18.
The entropy change involved in the isothermal reversible expansion of 2moles of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27°C is : [AIEEE 2011, 4/120] (1) 38.3 J mol–1 K–1 (2) 35.8 J mol–1 K–1 (3) 32.3 J mol–1 K–1 (4) 42.3 J mol–1 K–1
19.
The incorrect expression among the following is : (1)
Gsystem S total
(3) lnK = 20.
–T
H – TS RT
[AIEEE 2012, 4/120] (2) In isothermal process, wreversible = – nRT n
Vf Vi
(4) K = e–G°/RT
A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperture of 37.0°C. As it does so, it absorbs 208J heat. The values of q and w for the proces will be : (R = 8.314 J/mol K) (In 7.5 = 2.01) [JEE Mains 2013] (1) q = + 208 J, w = – 208 J (2) q = – 208 J, w = – 208 J (3) q = – 208 J, w = + 208 J (4) q = + 208 J, w = + 208 J
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THERMODYNAMICS_JEE MAINS # 55
5 NCERT QUESTIONS 1.
Choose the correct answer. A thermodynamic state function is a quantity : (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine pressure volume work (iv) whose value depends on temperature only.
2.
For the process to occur under adiabatic conditions, the correct condition is : (i) T = 0 (ii) p = 0 (iii) q = 0 (iv) w = 0
3.
The enthalpies of all elements in their standard states are : (i) unity (ii) zero (iii) < 0 (iv) different for each element
4.
U of combustion of methane is – X kJ mol–1. The value of H is : (i) = U (ii) > U (iii) < U (iv) = 0
5.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, – 890.3 kJ mol–1 – 393.5 kJ mol–1, and – 285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be : (i) – 74.8 kJ mol–1 (ii) – 52.27 kJ mol–1 (iii) + 74.8 kJ mol–1 (iv) + 52.26 kJ mol–1
6.
A reaction, A + B C + D + q is found to have a positive entropy change. The reaction will be : (i) possible at high temperature (ii) possible only at low temperature (iii) not possible at any temperature (v) possible at any temperature
7.
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
8.
The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and U was found to be – 742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH2 CN(g)
3 O2 (g) N2 (g) CO(g) H2 O(I) 2
9.
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35 C to 55 C. Molar heat capacity of Al is 24 J mol–1 K–1.
10.
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 C to ice – 10.0 C. fusH = 6.03 kJ mol–1 at 0 C. Cp [H2O(l)] = 75.3 J mol–1 K–1 Cp [H2O(s)] = 36.8 J mol–1 K–1
11.
Enthalpy of combustion of carbon to CO2 is – 393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.
12.
Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are – 110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of rH for the reaction : N2O4(g) + 3CO(g) N2O(g) + 3CO2(g)
13.
Given rH = – 92.4 kJ mol–1. N2(g) + 3H2(g) 2NH3(g) ; What is the standard enthalpy of formation of NH3 gas?
14.
Calculate the standard enthalpy of formation of CH3OH(l) from the following data :
CH3 OH(l)
3 O2 (g) CO2 (g) 2H2 O(l) ; r H 726 kJmol1 2
C (graphite) + O 2 (g) CO2 (g) ; cH 393 kJ mol-1
H2 (g)
1 O2 (g) H2 O(l) ; f H 286 kJ mol-1 2 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 56
15.
Calculate the enthalpy change for the process CCl4(g) C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g). vap H (CCl4) = 30.5 kJ mol–1. f H (CCl4) = – 135.5 kJ mol–1. a H (C) = 715.0 kJ mol–1 , where a H is enthalpy of atomisation a H (Cl2) = 242 kJ mol–1
16.
For an isolated system, U = 0, what will be S ?
17.
For the reaction at 298 K, 2A + B C H = 400 kJ mol–1 and S = 0.2 kJ K–1 mol–1. At what temperature will the reaction become spontaneous considering H and S to be constant over the temperature range.
17.
For the reaction, 2 Cl(g) Cl2(g), what are the signs of H and S ?
18.
For the reaction, 2 A(g) + B(g) 2D(g)
U 10.5 kJ and S 44.1 Jk -1 Calculate G for the reaction, and predict whether the reaction may occur spontaneously.. 19.
The equilibrium constant for a reaction is 10. What will be the value of G ? R = 8.314 JK–1 mol–1, T = 300 K.
20.
Comment on the thermodynamic stability of NO(g), given
1 1 N2 (g) O2 (g) NO(g) ; r H 90 kJ mol-1 2 2 N2 (g) 21.
1 O2 (g) NO2 (g) ; r H 74 kJ mol-1 2
Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. r H 286 kJ mol-1
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THERMODYNAMICS_JEE MAINS # 57
EXERCISE # 1 1.1 2.6 2.12 2.19
(C) (A) (D) (C)
1.2 2.7 2.13 2.20
(C) (C) (C) (C)
2.1 2.8 2.14 2.21
(B) (B) (A) (D)
2.2 2.9 2.15 2.22
(A) (C) (A) (A)
2.3 2.10 2.16 3.1
(C) (B) (D) (C)
2.4 2.11 2.17 3.2
(B) 2.5 (B) (C) (D) (A) 2.18 (C) 3.3
(D)
3.4
(B)
3.5
(B)(C)
3.6
(A)
3.7
(D)
3.8
(B)
4.1
(D)
4.2
(C)
4.3
(B)(C)(D)
4.4
(C)
4.5
(A)
4.6
(A)
4.7
(D)
4.8
(D)
4.9
(A)
4.10
(D)
4.11
(C)
4.12
(C)
4.13
(B)
4.14
(C)
4.15
(B)
4.16
(B)
4.17
(B)
4.18
(A)
4.19
(A)
4.20
(A)
4.21
(A)
4.22
(A)
4.23
(D)
4.24
(C)
4.25
(B)
5.1
(D)
5.2
(B)
5.3
(B)
5.4
(C)
5.5 6.6 6.13 7.5
(D) (D) (C) (D)
5.6 6.7 6.14 7.6
(C) (B) (B) (A)
6.1 6.8 7.1 7.7
(C) (D) (A) (B)
6.2 6.9 7.2 7.8
(C) 6.3 (B) 6.10 (A)(B)(C)(D) (D) 7.9
(C) (C)
6.4 6.11 7.3
(D) (C) (B)
6.5 6.12 7.4
(C) (B) (C)
(C) (B)
(D)
EXERCISE # 2 1.
(A)
2.
(C)
3.
(B)
4.
(C)
5.
(A)
6.
(B)
7.
(A)
8. 15. 22.
(A) (D) (B)
9. 16. 23.
(C) (C) (A)
10. 17. 24.
(A) (B) (B)
11. 18. 25.
(B) (A) (B)
12. 19. 26.
(C) (B) (A)
13. 20. 27.
(C) (A) (B)
14. 21. 28.
(D) (A) (C)
29.
(D)
30.
(C)
31.
(D)
32.
(C)
33.
(A) (C) 34.
35. 39.
36. 40.
(A)(C)(D) (A)(B)(D)
37. 41.
(A)(B)(D) (A)(B)(C)
43.
(A) (C) (D) (A) (B)(D) (A)(B)(D) (A)(D) 44.
47.
(C)
(A)(B)(C)(D)
45.
(A)(B) (C)(D)
(B)(C)(D) 38. 42.
46.
(C)(D)
(B)(C)
EXERCISE # 3 1.1
(A) (p, s) ; (B) (q, r) ; (C) (r) ; (D) (s)
1.2
(A) (p, q) ; (B) (q, r, s) ; (C) (p) ; (D) (r)
1.3
(A) q (B) (p,s) (C) (p,s) (D) s s)
1.4
(A) (p, s, r) (B) (p, r) (C) (p) (D) (q, r,
1.5
(A) (p,s) (B) (p,r,s) (C) (q) (D) (r,s)
2.1
(C)
2.2
(B)
2.3
(B)
2.4
(C)
2.5
(B)
2.6
(A)
2.7
(B)
2.8
(C)
2.9
(D)
2.10
(A)
2.11
(B)
2.12
(A)
2.13
(B)
2.14
(C)
2.15
(D)
2.16
(A)
2.17
(D)
2.18
(A)
2.19
(D)
3.1
(D)
3.2
(D)
3.3
(A)
3.4
(A)
3.5
(D)
3.6
(A)
3.7
(A)
3.8
(B)
3.9
(C)
3.10
(B)
4.1
T
4.2
T
4.3
F
4.4
T
4.5
T
4.6.
T
4.7
T
4.8
T
4.9
F
4.10
T
5.1
negative
5.2
isolated
5.3
two
5.4 reversible
5.5
decresing
5.6
isentropic
5.7
Initial, final
5.8
decrease
5.9
zero
5.10
positive
ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 58
EXERCISE # 4 PART - I 1.
(C)
2.
(A)
5.
(B)
6.
Hydrogen is diatomic so at high temperature rotational and vibrational motion also counts.
7.
(A)
8.
(A)
11.
(i) G = – 5700 J/mol
(ii) backward shifting
12.
(B)
14.
15.
DH = DU + D(PV)
13.
(B)
3.
9.
(C)
(C)
4.
10.
(c) (ii) w = – 620.77, q = + 620.77 (iii) U = H = S = 0
H = 0 J, E = 10 J
(D)
so,
DU = DH – D(PV) = – 560 – [40 – 70] (L atm) = (– 560 + 30 × 0.1) kJ = – 557 kJ
16.
(B)
17.
(A)
23.
(C)
24.
27.
(A) (C) 28.
18.
(D)
19.
(A)
(A) (B) 25.
2
26.
(A) (p,r,s) (B) (r,s) (C) (t) (D) (q,t)
(D)
(A) (D) 30.
29.
(B)
20.
(A) (D) 21.
31.
(C)
9
22.
(B)
PART - II 1.
(1)
2.
(4)
3.
(1)
4.
(1)
5.
(3)
6.
(4)
7.
(2)
8.
(1)
9.
(3)
10.
(2)
11.
(1)
12.
(1)
13.
(2)
14.
(1)
15.
(3)
16.
(2)
17.
(2)
18.
(1)
19.
(3)
20.
(1)
ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
THERMODYNAMICS_JEE MAINS # 59
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