online-solutiohn-of-assignment-on-chemical-thermodynamics-and-energetic.pdf

Chemistry - Chemical Thermodynamics and energetic [Time : 90 Mins]

Note : 1

SOLUTION OF ASSIGNMENT

[Max. Marks : 35]

Chemical Thermodynamics and energetic. Properties chapter Weightage total 8 marks.

2. All questions are compulsory. 3. Draw neat and labelled diagram wherever necessary. 4. Figure to the right indicates full marks. 5. Answer to every new question must begin on a new page.

I.

Objective Type Questions

I.

Choose the correct answer:

1.

Thermochemistry is the study of relationship between heat energy and (a) Chemical energy

(b) Binding energy

(c) Gravitational energy

(d) all of these.

Solution :

(a)

2.

An exothermic reaction is one in which heat content of (a) Products is more than that of reactants (b) Reactants is more than that of products (c) Reactants and products is same (d) None of these

Solution :

(b)

3.

The standard heat of combustion of C is -94.0 Kcals. What is ∆Hf of CO? (a) -94.0 Kcals

Solution :

(a)

4.

In an adiabatic process

(b) -46.0 Kcals

(c) +94.0 Kcals

(d) -26.0 Kcals

(a) pressure is maintained constant (b) the gas is isothermally expanded Ideal 21st Century Competitions

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(c) there is perfect heat insulation (d) the system exchanges heat with surroundings Solution :

(c)

5.

Which of the following processes is reversible? (a) Evaporation of water at 100°C and 1 atm pressure (b) Melting of ice at 10°C (c) Mixing of two gases by diffusion (d) None of these

Solution :

(d)

6.

Enthalpy is same as (a) Entropy

(b) Heat

(c) Heat content

(d) Volume

Solution :

(c)

7.

In an isothermal expansion of an ideal gas (a) w = 0

(b) ∆U = 0

(c) q = 0

(d) ∆V = 0

Solution :

(b)

II.

Short Answer Type Questions-II

1.

What do you mean by the law of conservation of energy? Derive the mathematical relationship of heat, internal energy and work.

Solution :

Law of conservation of energy states that energy can neither be created nor be destroyed. It can change from one form to another. The total energy of universe remains constant. Let E1 be initial internal energy, 'q' is heat supplied to the system, 'w' is work done on the system, then final internal energy U2 = U1 + q + w U2 - U1 = q + w

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2.

What is the basic difference between enthalpy of formation and enthalpy of reaction? Illustrate with suitable examples.

Solution :

Enthalpy of formation is defined as enthalpy change when 1 mole of substance is formed from the constituting elements in their standard states, e.g., H2(g) + - O2(g)→→H2O(l) ∆f H = —286kJ/mol

[1 Mark]

Enthalpy of reaction is defined as enthalpy change when reactants react completely to form products according to balanced chemical reaction, e.g., N2(g) + 3H2(g) → 2NH3(g) ∆H = - 92.0 kJ-1

3.

[1 Mark]

Using the bond energy of H2 = 435 kJ mol-1, Br2 = 192 kJ mol-1, H-Br = 368 kJ mol-1 Calculate enthalpy change for the reaction : H2(g) + Br2(g)→ 2HBr(g)

Solution :

∆rH°

= ∑ bond enthalpiesreactants - ∑ bond enthalpiesprocducts = [E0H-H + E0Br-Br] - [2[E0H-Br] = (435 kJ mol-1 + 192 kJ mol-1) - 2 ∆ 368 kJ mol-1 = 627 mol-1 - 736 kJmol-1 = -109 kJmol-1

4.

[2 Marks]

Calculate the heat of reaction of the following reaction : CO2(g) + H2(g) → CO(g) + H2O(g) Given that the ∆Hf0 CO = - 110.5 kJ, ∆Hf0 CO2 = - 393.8 kJ, ∆Hf0 H2O(g) = - 241.8 kJ respectively.

Solution :

CO2(g) + H2(g) → CO(g) + H2O(g) ∆H

= ∑∆Hf product -∑∆Hf reactants = ∆HfCO(g) + ∆HfH2O(g) - ∆CO2(g) - ∆HfH2(g) = - 110.5 kJ - 241.8 kJ - (- 393.8 kJ = - 352.3 kJ + 393.8 kJ = 41.5 kJ.

5.

[2 Marks]

Define 1.

Isothermal Process

2.

Adiabatic process

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Solution :

1. Isothermal Process: When temperature of the system remains constant (i.e., ∆T = 0) throughout the process whether heat enters or leaves the system is called as isothermal process. Change of state e.g., freezing, melting, evaporation and condensation are the examples of isothermal process. [1 Mark] 2. Adiabatic process: When heat change is zero (∆Q = 0) i.e., no heat enters or leaves the system during any step of the process is known as adiabatic process. A reaction carried out in an isolated [1 Mark]

system is an example of adiabatic process.

6.

Write Hess’s Law of Constant Heat Summation

Solution :

Hess’s Law of Constant Heat Summation When a chemical reaction gets completed directly in one step or indirectly in two or more steps, the total energy change in the reaction remains same. i.e., the change does not depend on the path of a [1 Mark]

chemical reaction. Reactant , Example – Combustion of carbon Method 1:–

C(s) + O2 (g) CO2 (g) + 94 Kcal

Method 2:–

C(s) +1/2 O2(g) CO(g) + 16.4 Kcal

CO(s) + ½ O2 (g) CO2 (g) + 67.6 Kcal

∆H = –94 Kcal .......... [1] ........ [2] [1 Marks]

On summing up equations [1] and [2], ∆H = –94 Kcal

III.

Four sub-questions of 3 marks each (Attempt any THREE)

1.

Explain the concept of maximum work.

Solution :

Concept of Maximum work (Wmax) Maximum work can be obtained only during isothermal and reversible process. It is denoted by Wmax In isothermal process, temperature of system remains constant at every stage. Hence for such process change in internal energy of system remains constant i.e ∆E = 0. Hence first law of thermodynamics which is mathematically expressed as, q = ∆E - W becomes q = -W. Thus all the heat supplied in isothermal process is used to perform work, hence work is maximum. The value of work done by the system also depends on the magnitude of external opposing pressure. The external opposing pressure has maximum possible value only in case of reversible

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process because ecause in such process, the driving and opposing forces differ by an infinitesimally small [2 Marks]

value. Graphical representation of maximum work done is as follow.

From the above graph it is evident that that work done is maximum if the process is carried out in infinite [1 Mark]

steps.

2.

Derive the expression for maximum work done in an isothermal reversible process.

Solution :

Mathematical expression for maximum work done in an isothermal reversible process : The a small amount of work done in a single step of small expansion of gas is dw = PdV In reversible process, an infinite number of such steps are involved. Hence, Henc the totel work done in these steps is the maximum work. It is obtained by intergrating abovs equation in limits of initial volume V1 to final volume V2.

(As per IUPAC conventions, work done in expansion is shown by negative sign) Where, n = No. of moles of gas.  weight of subs tance  n =  molecular weight  

R = Molar gas constant. T = Constant temperature in K. V1 = Initial volume of the gas. V2 = Final volume of the gas. The unit of work depends upon the unit of molar gas constant, R.

[2 Mark]

In the compression of gas, work is positive Wmax = 2.303nRt log10

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V1 V2

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In terms of initial pressure P1 and final pressure, P2 the expression for maximum work is, Wmax = 2.303 nRt log10

P2 P1

(In expansion, work has -ve sign.) In compression of a gas, work has positive sign. Wmax = 2.303 nRt log10

P2 P1

3.

Derive the Mathematical expression of first law of thermodynamics.

Solution :

Mathematical form of first law of thermodynamics ∆E = q + W

i.e.,

[1 Mark]

change in internal energy = heat change + work done

The following sign conventions are used for this equation

(a)

(i)

Heat absorbed by the system = +q

(ii)

Heat released by the system = –q

(iii)

Work done on the system by the surroundings = +W

(iv)

Work done by the system on the surroundings = –W

[1 Mark]

For isothermal process ∆T = 0, hence ∆E = 0 0=q+W q = –W

(b)

For isochoric process ∆V=0, hence W = 0 ∆E = q i.e., heat given to system at constant volume changes into internal energy.

(c)

[1 Mark]

For adiabatic process q = 0 ∆E = 0 + W ∆E = W i.e., work done on system changes into internal energy

(d)

For isobaric process (i)

When a gas expands against a constant pressure P such that P ≠ 0, the work done by the system. – W = – P∆V ∆E = q – PV

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(ii)

If a gas expands in vacuum, the work done, W = 0, because P = 0. [1 Mark]

∆E = q

IV.

Two sub-questions of 7 marks each (Attempt any ONE)

1.

Derive a relation, qp = qv + ∆H = ∆U + ∆nRT.

Solution :

Consider a reaction in which n1 moles of gaseous reactant in initial state change to n2 moles of gaseous product in the final state. Let H1, U1, P1, V1 and H2, U2, P2, V2 represent enthalpies, internal energies, pressures and volumes in the initial and final states respectively then, T n1 A( g )  → n2 B( g )

H1, U1, P1, V1

[1 Mark]

H2, U2, P2, V2

The heat of reaction is given by enthalpy change ∆H as, ∆H = H2 – H1 By definition, H = U + PV

∴ H1 = U1 + P1V1 and H1 = U2 + P2V2 ∴ ∆H = (U2 + P2V2) – (U1 + P1V1)

[2 Mark]

Now, ∆U = U2 – U1 Since PV = nRT, For initial state, P1V1 = n1RT For Final state, P2V2 = n2RT

∴ P2V2 - P1V1 = n2RT - n1RT = (n2 – n2) RT [2 Marks]

= ∆ nRT

 Number of moles 

 Number of moles



Where ∆n =   -  of gaseous products  of gaseous reactants 

∴ ∆H = ∆U + ∆nRT If qp and qv are the heats involved in the reaction at constant pressure and constant volume respectively, then since qp = ∆H and qv = ∆U.

∴ qp = qv ∆nRT

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