Jee 2014 Booklet5 Hwt Solutions Thermodynamics

Vidyamandir Classes

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Solutions to Home Practice Test-5/Chemistry Thermodynamics 1.

HWT - 1

Work done due to change in volume against constant pressure, W   p(V2  V1) = 1  105 Nm 2 (1 102  1  103 ) m3 = 900 Nm   900 J 1 Nm  1J 

2.

The reactions in which products has lesser energy than reactants, then energy is released in the reaction and such reactions are known as exothermic reactions e.g. N 2  3H 2   2NH3  92 k J

In this equation energy is released, so, it is an example of exothermic reaction. 3.

Bond breaking process or decomposition process are endothermic process.

4.

Macroscopic properties which determine the state of a system are referred as state functions. The change in the state properties depends only upon the initial and final state of the system. All thermodynamics functions are state functions except work and heat.

5.

Word done  W    pext  V2  V1  = 3   6  4    6 L atm = 6  101.32 J

 1 L atm

 101.32 J 

6.

= 607.92   608 J The properties of the system whose value is depends upon the amount of substance present in the system is called extensive property. Gibb’s free energy is an extensive property.

7.

In the adiabatic process no heat enters or leaves the system i.e., q = 0.

8.

Internal energy, enthalpy and entropy are state functions but work and heat are path functions.

9.

For exothermic reaction, H     for endothermic reaction, H     .

10.

For an isothermal process, E  0 As the process is taking place at constant T and p hence, from Equation, H  E  p.V We have, H  0  0  V  0 Hence, for the process, H  E  0

Thermodynamics

HWT - 2

1.

For an endothermic reactions H is positive because in endothermic reaction heat is always absorbed.

2.

For an isothermal process T  0 and E  0 and q  0 .

3.

TV 1  constant V    2 Tfinal  V1  T

 1

 5 / 3  1

2    Tfinal  1  T T  Tfinal 2 2 / 3 T

4.

 2

2 / 3

Endothermic reaction are those in which heat energy is absorbed.

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HWT-5/Chemistry

Vidyamandir Classes 5.

As we know that,

V2 . V1 Hence, V1 and V2 are in ratio in the relation. So, unit may be expressed in any one of m3, dm3 or cm3. Work done (W)  2.303 nRT log

6.

V1 = 100 mL ; V2 = 250 mL Pressure p  2 atm or 2 1.01 105 Nm 2 Work done by the gas W  pV or p  V2  V1 





Put the value in given formula W  2 1.01  105 0.250  103  0.100  103 = 2  1.01  105  0.15  103 = 30.30 J 7.

From first law of thermodynamic. E  q  W Given, q   300 cal

(  Heat is absorbed) (  Work done on surroundings)

W   500 cal

E  q  W  300    500  =  200cal

 8.

The heat of formation of CO is calculated by using Hess’s law. According to it, the total heat changes occurring during a chemical reaction are independent of path. 2CO (g)  O 2 (g)   2CO 2 (g) ;

H  135.2 kcal

1 O 2 (g) ; 2

I.

CO (g)   CO(g) 

II.

C(s)  O 2 (g)   CO 2 (g) ;

H 

135.2 kcal 2

H   94 kcal

1 C(s)  O 2 (g)   CO(g) ; 2

Required equation :

1  CO(g) ; Add Eqs. (I) and (II) C(s)  O 2 (g)  2

H  ? H   26.4 kcal

9.

Every system having some quantity of matter, is associated with a definite amount of energy. This energy is known as internal energy. It is sum of many type of energies, such as translation energy, rotational energy, vibrational energy, electronic energy and bonding energy of the molecule. E  E trans  E rot  E vib  E bonding  E electronic

10.

C(s)  O 2 (g)   CO 2 (g) ; H  r

CO(g) 

1 O 2 (g)   CO 2 (g); H  s 2

1 O 2 (g)   CO(g); H  ? 2 Subtract Eq. (ii) from Eq. (i) C(s) 

C(s)  O 2 (g)   CO 2 (g); H  r

CO(s) 

 C(s) 

1 O 2 (g)   CO 2 (g); H  s 2   

1 O 2 (g)   CO(g); H  r  s 2

Thermodynamics 1.

W

HWT - 3

1/ 2

1/1 pdV   p  V2  V1 

W  1 20  10   10 dm3 atm = 10dm3 

8.314 JK 1 mol1 0.0821 dm3 K 1 mol1

 1013 J

From, 1st law of thermodynamics U  q  W = 800 J   1013 J    213 J

2.

Hess’s law is based upon law of conservation of energy i.e., first law of thermodynamics.

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Vidyamandir Classes 3.

According to Hess’s law, the total heat changes occurring during a chemical reaction are independent of path.

4.

The first law of thermodynamics can be expressed as: E  q  W q  E  W

5.

According to Hess’s law total heat changes during a chemical reaction are independent of path of reaction. Given, …(i) I 2 (s)  I 2 (g), H1  57.3 kJ / mol I 2 (s)  I 2 (g), H 2   15.5 kJ / mol

…(ii)

Required equation I2 (l )  I 2 (g), H  ? subtract Eq. (ii) from Eq. (i)

 6.

7.

I2 (l )   I 2 (g), H  57.3  ( 15.5)   41.8 kJ / mol

CH 4 (g)  2O 2 (g)   CO 2 (g)  2H 2O(l ) n g  1  3   2 We know that,

E  H  n g RT



H   885389    2   8.314  298 =  885389  4955.1440 =  880433.86 J mol1

According to second law of thermo chemistry the law states that the total heat change ( H ) accompanying a chemical reaction is the same whether the reaction takes place in one or more steps. It means that heat of a reaction depends only on the initial reactants and final products and not on intermediate products that may be formed. Now, H  H1  H 2  H3 Enthalpy change in a reaction is always constant and independent of the path followed.

8.

Internal energy of a gas depends upon its pressure and temperature. Thus, if a gas expands at constant temperature and pressure, then its internal energy remains same.

9.

Relation between H (enthalpy change) and E (internal energy change) is : H  E   n g RT Where,  n g = moles of gaseous products-moles of gaseous reactants = 2  1

 1366.5  E 1  8.314  103  300 E   1364.0 kJ mol1 10.

4NO 2 (g)  O 2 (g)  2N 2O5 (g); 111 kJ

2N 2O5 (g)  2N 2O5 (s); s H   54  2  kJ 4NO 2 (g)  O 2 (g)  2N 2O5 (s);  r H   219 kJ

Note: H of sublimation of N 2O5 (s) is  54 kJ mol1 . Thus, for reverse process, it is  54 kJ mol1

Thermodynamics 1.

NaOH  20  0.5

2.

HCl

100  0.1

 

HWT - 4

NaCl  H 2O 10 millimole produced

Hf of elements in their standard state is taken to be zero. Cl2 is gas, Br2 is liquid and I2 is solid at room temperature.

3.

Given,

1 3 N 2 (g)  H 2 (g)  NH3 (g) 2 2

Hf  46.0 kJ mol1

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Vidyamandir Classes 2H(g)  H 2 (g) Hf   436 kJ mol1 2N(g)  N 2 (g) Hf   712 kJ mol1



Assuming X is the bond energy of N – H bond in kJ mol1



1 3   712     436   3X   46.0 2 2

3X  1056 kJ mol1 So, 4.

X  352 kJ mol1 .

In neutralization reaction, when acid and base both are weak, a large amount of heat is utilized to ionise them. Thus, for such reactions, enthalpy of the reaction is least. Hence, enthalpy is least for HCN  NH 4OH   NH 4CN  H 2O

5.

A  B, H   24 kJ / mol



H B  H A   24

…(i)

B  C, H  18 kJ / mol



H C  H B   18



H B  H C   18

…(ii)

From Equations (i) and (ii), we have

HC  H A  6 H B  HC  H A

6.

5 O 2  2CO 2  H 2O; H  310 kcal 2

I.

C 2 H5 

II.

C  O 2  CO 2 ;

H   94 kcal

1 O 2  H 2O; H   68 kcal 2 (Equation (II) × ) + Equation (I) gives

III.

H2 

2C  H 2   C2H 2 ;

H   94  2    68    310   = + 54 kcal

7.

2NH3 (g)  N 2 (g)  3H 2 (g)

H r    2  enthalpy of formation of NH3 

=   2   46   92 kJ 8.

Given,

H 2O(g)  C(g)   CO(g)  H 2 (g);

H   131kJ

…(i)

1 O 2 (g)   CO 2 (g); H   282 kJ 2 1 H 2 (g)  O 2 (g)   H 2O(g); H   242 kJ 2

CO(g) 

…(ii) …(iii)

C(g)  O 2 (g)   CO 2 (g); H  ? On adding Equations (i), (ii) and (iii), we get Equation (iv)

…(iv)

H 2O(g)  C(g)   VO(g)  H 2 (g); H  131 kJ

1 O 2 (g)   CO 2 (g); H   282 kJ 2 1 H 2 (g)  O 2 (g)   H 2O(g); H  242 kJ 2 C(g)  O 2 (g)   CO 2 (g) H  131 282  242  KJ

CO(g) 

=  393 kJ 9.

H reaction   H f product   H f reactant H f of a single element O2, H2 Cl2 is considered equal to zero.

CH 4  2O 2   CO 2  2H 2O H reaction  H f (CO 2 )  2H f · (H 2O)  H f (CH 4 )  2H f (O 2 )

= 400  2(280)   70)  0  =  800 kJ mol1

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Vidyamandir Classes 10.

Given,

S(rh) 

3 O 2 (g)   SO3 (g); 4

1 SO 2 (g) O 2 (g)   SO3 (g); 2 S(s)  O 2 (g)   SO 2 (g);

H   2x kJ mol1

…(i)

H   y kJ mol1

…(ii)

H  ?

Subtract Equation (ii) from Equation (i) S(rh( 

3 O 2 (g)   SO3 (g); 2

H  2x kJ mol1

…(i)

1 SO 2 (g)  O 2 (g)   SO3 (g); H   y kJ mol1 2

…(ii)



   3 1   S(rh)     O 2 (g)   SO 2 (g); H   2x  y  kJ mol1 2 2

S(rh)  O 2 (g)   SO 2 (g); H   y  2x  kJ mol1

Thermodynamics 1.

HWT - 5

Hf is the enthalpy change when 1 mol of the substance is formed from its elements in the standard state. Reaction (a) does not represent Hf because standard state of carbon is graphite and not diamond.

2.

H  U  n g RT H = enthalpy change (at constant pressure)

U = internal energy change (at constant volume) given reaction is exothermic) n g = mole of (gaseous products – gaseous reactants) =  ve Thus, H  U 3.

The relationship between H and U is H  U  nRT

4.

The bond energy is the energy required to break one mole of similar bonds.

5.

The heat of reaction for an ideal gas, at constant pressure and constant volume are related as  H  E  nRT

 6.

q p  q v  nRT

The molar heat capacity for any process is given by a following expression C  C v 

 Cp R  when p V  constant and C 1  V

1 p  1 i.e. p V  constant V 3 R 3 R 4 C R  R  R 2 1  1 2 2 2

Here,

7.

A(g)  2B(g)   2C(g)  3D(g) n  5  3  2



H  E  nRT 19000  E  2  2  300 E  17800 cal = 17.8 kcal

8.

H  E  pV

For isochoric process V  0



H  E

49.

The enthalpy of H2O(l) is less than that of H2O (g), hence more energy will be released when H2O(l) is formed, therefore H1  H 2 .

50.

H  E for CO(g) 

1 O 2 (g)   CO 2 (g) 2

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Thermodynamics 1.

HWT - 6

The amount of heat evolved or absorbed in the formation of one mole of product from its elements. So, the correct answer is

C  O 2 1 atm    CO 2 1 atm  . 2.

The formation of one mole of HCl can be represented as,

1 1  H  H    Cl  Cl   H  Cl; H   90 kJ 2 2

Hence, for the reaction, Heat evolved = heat evolved in bond formation – heat required for bond breaking



1 1  90 KJ   H  Cl bond energy     430   240  2 2  



 H  Cl bond energy   90   215  120  = 425 kJ mol1

3.

4.

Given, 20 mL, of 0.5 M NaOH 100 mL of 0.1 M HCl Milliequivalents (or millimoles) of NaOH = 20 × 0.5 = 10 Milliequivalents (or millimoles) of HCl = 100 × 0.1 = 10 10 milliequivalents on neutralisation give =  x kJ heat 



lg-eq (i.e., 1000 milliequivalents) will give =  x  100 =  100 x kJ heat

Given,

H   OH    H 2O; H   55.9 kJ

HCN  OH    H 2O  CN  H  12.1 kJ On reversing (i) we get :

…(ii)

H 2O   H   OH  ; H   55.9 kJ On adding (iii) to (ii) we get :

…(iii)

HCN   H   CN  ;

5.

…(i)

H   43.8 kJ X 2  Y2   2XY

Formation of XY is known as :

H   BE X  X   BE Y  Y  2  BE X  Y

If (BE) of X  Y  a Then (BE) of (X  X)  a



a 2 H f (X  Y)   200 kJ



400(for 2 mol XY)  a 

And (BE) of  Y  Y  

a  2a 2

a 2 a = + 800 kJ

400  

The bond dissociation energy of X 2  800 kJ mol1 . 6.

CH3 | 13 CH3  C H  CH3  CH3  O 2   4CO 2  5H 2 O 2 H  E  nRT

n = mole of (gaseous products–gaseous reactants) = + ve Thus, H  E 7.

This is based on Joule-Thomson effect.

8.

0.5 mol HNO3  0.5 mol H  and 0.3 mol OH  0.5 H  (aq)  0.3 OH  (aq)   0.3 H 2O(l )

(0.3 mole of OH  is neutralised by 0.3 mole of H+) = 17.13 kJ H  0.3  57.1 = 17.1 kJ

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Vidyamandir Classes 9.

Enthalpy of formation of H2SO4 can be represented by the following equation: H 2  S  2O 2   H 2SO 4

Given,

S  O 2   SO 2 ,

SO 2 

H  298.2 kJ

1 O 2   SO3 , 2

H   98.7 kJ

SO3  H 2O   H 2SO 4 ,

H  130.2 kJ

1 O 2   H 2O, H   287.3 kJ 2 On adding all the four equations, we get the equation (i), H2 

Hence, 10.

H f   298.2    98.7     130.2    287.3 = 814.4 kJ

The heat of neutralisation of strong acid and strong base is constant because it is infact heat of formation of water by H  and OH  . Its value is almost equal to 57.3 kJ .

Thermodynamics 1.

HWT - 7

Molar heat capacity Heat required to raise the temperature of 1 mole of a substance by 1 C is called molar heat capacity.

2.

HF is more stable than HCl. The reason is that in the formation of HF, more energy is produced. It means, HF has less energy than HCl and hence is more stable. In other words, we need more energy to break H–F bond, hence HF is more stable.

3.

According to question, 3 2Al  O 2   Al2O3 ; 2

H  1596 kJ

3 O 2   Cr2O3; H  1134 kJ 2 Reversing Equation (ii) and adding both reactions, we get : 2Cr 

2Al  Cr2O3   2Cr2  Al2O3;

4.

…(i) …(ii)

H  462 kJ

Standard heat of formation of substance is the amount of heat evolved or absorbed when one mole of substance is formed from its elements in their standard states. Graphite is the standard state of carbon and hydrogen is found in form of H2. 



 CH 4 (g) Standard heat of formation of methane is C  graphite   2H 2 (g) 

5.

CH4 has four C – H bonds. So, 4 × 416 kJ is required to break CH4 into C and 4H.

6.

We know, H  U  nRT

n = [total number of moles of gaseous products] – [Total number of moles of gaseous reactants]  4C(g)  D(l ) For this reaction A(s)  3B(g) 

n   4  3 = 1 H  U  RT

7.

We know that, H  E  n g RT For reaction,

N 2O 4   2NO 2

n g  2  1  1  8.

H  E  RT

or

H  E

H  E  pV  E  n g RT PCl5 (g)   PCl3 (g)  Cl2 (g)

n g  n (p)g  n (r)g  2 1  1  9.

H  E  1  RT



H  E

The structure of N2H4 is as follows

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HWT-5/Chemistry

Vidyamandir Classes Hence, in the reaction, N 2H 4 (g)   2N(g)  4H (g); H   N  N bond energy    N  H bond energy  4 

 10.

1724  N  N bond energy = 1724  1564 = 160 kJ/mol

The amount of heat either evolved or absorbed when one gram mole of a substance is formed form its constituent elements, is known as the standard heat of formation (Hf ) . For standard state temperature is 25 C or 298 K and pressure of gaseous substance is one atmosphere. Thus, in given thermochemical equation 'b' represents the standard heat of formation of HF. 1 1 H 2 (g)  F2 (g)   HF(g); 2 2

Hf  ? [standard heat of formation of HF (g)].

Thermodynamics 1.

2. 3.

4.

HWT - 8

H vap

 88 JK 1 mol1 Tb Trouton’s rule is relationship between enthalpy of vaporization and boiling point. 1  573.2  H 2 (g)  O 2 (g) H   286.6 kJ / mol Heat of decomposition of water is H 2O(g)  2 2 According to Trouton’s rule

Bond energy O – H bond = 109 kcal/mol



Energy absorbed in the dissociation of 1 mol of water (H2O) = 2 × 109 = 218 kcal



Energy released in the formation of 1 mole of water = 218 kcal.

G    RT ln K or, H  TS   RT ln K H S  RT R Comparing with y = m.x + c

or, ln K 

 5.

y intercept is

S R

G  H  TS

At equilibrium, G  0 For a reaction to be spontaneous G should be negative, so T should be greater than Te. 6.

Standard Gibb’s energy of formation (G f ) of a substance is defined as the Gibb’s energy change when mole of the substance is formed from its elements in their standard states. The following reaction define G f , 1 1 H 2 (g)  F2 (g)   HF(g) 2 2

7.

Gibbs-Helmholtz equation is a follows: G  H  TS From Gibbs-Helmholtz equation, it is clear that G will always be negative, if H is negative and TS is positive.

8.

1 3 X 2  Y2   XY3 2 2 Sreaction  Sproducts  Sreactants

1 3  Sreaction  50    40   60  =  40 J mol1 2 2  G  H  TS At equilibrium as S  0

 9.

H  T  S



T

H 30 103   750 K S 40

Entropy is a measure of randomness of the system, so entropy of a substance in different physical states is in the order Gas > liquid > solid

S  Sproduct  Sreactant

VMC/Solutions/Thermodynamics

In option (c) solid is converted into gaseous state, so S will be positive.

8

HWT-5/Chemistry

Vidyamandir Classes 10.

In going from initial to final state, the entropy change, S for an ideal gas is given by the following reactions

T V S  nC v ln 2  nR ln 2 T1 V1 But at constant temperature, V S  nR ln 2 V1

or

T p S  nCp ln 2  nR ln 1 T1 p2

i.e.,

p S  nR ln 1 p2

Thermodynamics 1.

HWT - 9

H   ve (combustion reaction) S   ve (spontaneous at all temperature) G   ve (because reaction is spontaneous)

2.

The standard free energy change (G  ) is related to equilibrium constant K as

G    RT ln K G    2.303 RT log K 3.

C  D 



A

B

Given,

S(A C)  50eu

S(CD)  30eu S(BD)  20 eu Where, eu is entropy unit S(A  B)  S(A  C)  S(C  D)  S(D  B) Therefore, = S(A C)  S(C  D)  S(B  D) = 50  30  20  60 eu 4.

For spontaneous reaction, the energy change (G) should be negative.

5.



6.



G   H  T · S = 29.8  298   0.1 = 29.8  29.8 = 0 G   nFE G    2.303RT log K E 

2.303 RT log K nF

E 

2.303  8.314  298 log K n  96500

E 

0.0591 log K n

At 298 K,

7.

The unit of entropy is J mol1 K 1 .

8.

Variation of Keq with temperature t is given by van’t Hoff equation.

log K eq  

H S  2.303RT R B A

Slope of the given line is positive indicating that term A is positive thus H is negative. Thus, reaction is exothermic. 9.

For a reaction to be spontaneous G must be negative. According to relationship of G , G  H  T · S

If H and S both are positive, then term T. S will be greater than H at high temperature and consequently G will be negative at high temperature. (boiling point of water) and reaction becomes feasible. 10.

For a system in equilibrium, G  0 , when all the reactants and products are in the standard state [at constant temperature and pressure] and Kc = 1.

VMC/Solutions/Thermodynamics

9

HWT-5/Chemistry

Vidyamandir Classes

Thermodynamics 1.

S  16 J mol1 K 1 H v  6 kJ mol1 Tbp 

2.

HWT - 10

H vapour Svapour



6  1000  375K 16

H 2 (g)  Cl2 (g)   2HCl(g) S   Sp   SR = 2 187  131  223 = 374  354 = 20 JK 1 mol1

3.

Given, H  31400 cal

S  32 cal K 1 mol1 T  1000  273  1273K G  ?

From Gibb’s-Helmholtz equation G  H  T · S = 31400 1273  32 = 31400  40736   9336 cal 4.

For spontaneous adsorption process, standard Gibb’s free energy (G) must be negative as well as the degree of randomness of gas molecule on the surface of solid decreases. For exothermic process, H must be negative. Hence, with the help of following equation. H   G  T ·  S T · S is negative.

Thus, during adsorption S decreases. 5.

Given that, H vaps  30 kJ / mol  30  103 J / mol Svaps  75 JK 1 mol1

We know that, G  H  TS

At equilibrium,

G  0



H  T  S

T

6.

 vapS   vapS 

or

30  103  T  75

30  103  400 K 75

 vap H T 40.8 103  109.4 JK 1 mol1 373

7.

For spontaneity of a cell, G is always negative and E is positive.

8.

As we know that, G  H  TS Hence, G may be greater, lesser or equal to H .

9.

For a reversible process Ssystem  Ssurroundings  0

10.

For spontaneous process, G   ve G   H  T  S



For G to be negative H  0

(i.e., negative)

S  0

(i.e., positive)

VMC/Solutions/Thermodynamics

10

HWT-5/Chemistry