State the second law of thermodynamics.

The second law of thermodynamics can be expressed in terms of entropy. It can be stated as “the entropy of an isolated system increases during a spontaneous process”. The total entropy change ( ∆ S total ) for the system and the surroundings of a spontaneous process is given by, ∆ S total = ∆ S system + ∆ S surrounding > 0 When a system is in equilibrium, the entropy is maximum and the change in entropy, ∆ S = 0 . Hence, entropy of a spontaneous process increases till it reaches a maximum and at equilibrium the change in entropy is zero.

The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at 25 ° C is

298 × 10 7 × 8 . 31 × 2 . 303 log 2 .

2 . 303 × 298 × 0 . 082 log 2 .

2 . 303 × 298 × 0 . 082 log 0 . 5 .

8 . 31 × 10 7 × 298 × 2 . 303 log 0 . 5 .

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by:

with increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases.

with increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive.

with increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative.

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The enthalpy change for a certain reaction at $300 K$ is $- 15.0 K cal {mol}^{- 1} .$ The entropy change under these conditions is $- 7.2 cal K^{- 1} {mol}^{- 1} .$ The free energy change for the reaction and its spontaneous/non-spontaneous character will be:

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Important Questions on Thermodynamics

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-2>Q 48

Look at the following diagram:

Question Image

The enthalpy change for the reaction $A \to B$ will be

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-2>Q 49

In Haber's process of manufacturing of ammonia:

$N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g); H_{25 ° C}^{0} = - 92.2 kJ$

Molecule	$N_{2} (g)$	$H_{2} (g)$	${NH}_{3} (g)$
$C_{p} {JK}^{- 1} {mol}^{- 1}$	$29.1$	$28.8$	$35.1$

If $C_{p}$ is independent of temperature, then reaction at $100 ° C$ as compared to that of $25 ° C$ will be:

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-2>Q 50

In the reaction

{AB}_{2} (l) + 3 X_{2} (g) ⇌ {AX}_{2} (g) + 2 {BX}_{2} (g) ΔH = - 270 kcal {mol}^{- 1} .

The enthalpies of formation of

{AX}_{2} (g)

and

{BX}_{2} (g)

are in the ratio of

4 : 3

and have opposite sign. The value of

{ΔH}_{f}^{0} ({AB}_{2} (l)) = + 30 kcal {mol}^{- 1} .

Then

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-2>Q 51

The heat of formation of $C_{2} H_{5} OH (l)$ is $- 66 kcal {mol}^{- 1}$ . The heat of combustion of ${CH}_{3} {OCH}_{3} (g)$ is $- 348 kcal {mol}^{- 1}$ . ${ΔH}_{f}$ for $H_{2} O$ and ${CO}_{2}$ are $- 68 kcal {mol}^{- 1}$ and $- 94 kcal {mol}^{- 1}$ respectively. Then, the $ΔH$ for the isomerisation reaction $C_{2} H_{5} OH (l) ⟶ {CH}_{3} {OCH}_{3} (g)$ and $ΔE$ for the same are at $T = 25 º C$

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-2>Q 52

Cesium chlorides is formed according to the following equation

Cs (s) + 0.5 {Cl}_{2} (g) \to CsCl (s) .

The enthalpy of sublimation of

Cs,

enthalpy of dissociation of chlorine, ionization energy of

Cs

and electron affinity of chlorine are

81.2, 243.0, 375.7

and

- 348.3 kJ {mol}^{- 1} .

The energy change involved in the formation of

CsCl

- 388.6 kJ {mol}^{- 1} .

Calculate the lattice energy of

CsCl .

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-2>Q 53

The enthalpies of neutralization of a weak base AOH and a strong base BOH by HCl are -12250 cal/mol and -13000 cal/mol respectively. When one mole of HCl is added to a solution containing 1 mole of AOH and 1 mole of BOH, the enthalpy change was -12500 cal/ mol. In what ratio is the acid distribution between AOH and BOH?

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-2>Q 54

The enthalpy of neutralisation of

40.0 g

NaOH

60.0 g

{CH}_{3} COOH

will be:

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-2>Q 57

Enthalpy of polymerisation of ethylene, as represented by the reaction,

{nCH}_{2} = {CH}_{2} ⟶ {(- {CH}_{2} - {CH}_{2} -)}_{n}

- 100 kJ {mol}^{- 1}

of ethylene. Given bond enthalpy of

C = C

bond is

600 kJ {mol}^{- 1},

enthalpy of

C - C

bond (in

kJ {mol}^{- 1}

) will be:

The enthalpy change for a certain reaction at 300 K is -15.0K cal mol-1. The entropy change under these conditions is -7.2 cal K-1 mol-1. The free energy change for the reaction and its spontaneous/non-spontaneous character will be:

Important Questions on Thermodynamics

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The enthalpy change for a certain reaction at $300 K$ is $- 15.0 K cal {mol}^{- 1} .$ The entropy change under these conditions is $- 7.2 cal K^{- 1} {mol}^{- 1} .$ The free energy change for the reaction and its spontaneous/non-spontaneous character will be: