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An irreversible process occuring isothermally in an isolated system leads to

Zero entropy

An increase in the total entropy of the system

A decrease in the total entropy of the system

None of these

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Explanation

Entropy of the system increases as the process occur irreversibly and isothermally in an isolated system.

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The entropy values (in JK^-1 mol^-1) of H_2(g) = 130.6, Cl_2(g) = 223.0 and HCl_(g) = 186.7 at 298 K and 1 atm pressure. Then entropy change for the reaction $H_{2 (g)} + {Cl}_{2 (g)} \to 2 {HCl}_{(g)}$ is [BHU 2005]

+ 540.3

+ 727.3

– 166.9

+ 19.8

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Explanation

$Δ S^{o} = 2 S_{H C l}^{o} - (S_{H_{2}}^{o} + S_{C l_{2}}^{o})$

$= 2 \times 186 .7 - (130 .6 + 223 .0) = 19 .8 J K^{- 1} m o l^{- 1}$

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The ΔS for the vaporisation of 1 mol of water is 88.3 J/mole K. The value of ΔS for the condensation of 1 mol of vapour will be

88.3 J/mol K

(88.3)² J/mol K

– 88.3 J/mol K

$\frac{1}{88 .3}$ J/mol K

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Explanation

For reverse reaction sign will be change.

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The occurrence of a reaction is impossible if [AIIMS 1982, 91; MP PET 1997; MP PMT 1999; DPMT 2002]

ΔH is +ve ; ΔS is also + ve but ΔH < TΔS

ΔH is – ve ; ΔS is also – ve but ΔH > TΔS

ΔH is – ve ; ΔS is + ve

ΔH is + ve ; ΔS is – ve

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Explanation

+ve ΔH and –ve ΔS both oppose the reaction.

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When enthalpy and entropy change for a chemical reaction are –2.5 × 10³ cal and 7.4 cal deg^–1 respectively. Predict the reaction at 298 K is [AFMC 1998; MH CET 1999; CBSE PMT 2000]

Spontaneous

Reversible

Irreversible

Non-spontaneous

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Explanation

When ΔH = –ve, ΔS = +ve and ΔG = –ve than reaction is spontaneous.

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Which of the following is true for the reaction $H_{2} O (l) ⇌_{}^{} H_{2} O (g)$ at 100°C and 1 atmosphere [KCET 1991; AIIMS 1996]

ΔE = 0

ΔH = 0

ΔH = ΔE

ΔH = TΔS

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Explanation

At equilibrium, ΔG = 0

Hence 0 = ΔH – TΔS or ΔH = TΔS.

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The enthalpy of vapourization water is 386 kJ. What is the entropy of water [BHU 1997]

0.5 kJ

1.03 kJ

1.5 kJ

22.05 kJ

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Explanation

$∆ S = ∆ H_{v a p \div} T ∆ S = 386 \times 1000 \div 373 = 1.03 K J$

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Identify the correct statement regarding entropy [CBSE PMT 1998; BHU 2001]

At 0°C, the entropy of a perfectly crystalline substance is taken to be zero

At absolute zero of temperature, the entropy of a perfectly crystalline substance is +ve

At absolute zero of temperature, the entropy of all crystalline substances is taken to be zero

At absolute zero of temperature, the entropy of a perfectly crystalline substance is taken to be zero

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Explanation

This is the statement of third law of thermodynamics.

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An engine operating between 150°C and 25°C takes 500 J heat from a higher temperature reservoir if there are no frictional losses, then work done by engine is [MH CET 1999]

147.7 J

157.75 J

165.85 J

169.95 J

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Explanation

$T_{2} = 150 + 273 = 423 K$

$T_{1} = 25 + 273 = 298 K$

Q = 500 K

$\frac{W}{Q} = \frac{T_{2} - T_{1}}{T_{2}}$ ; $W = 500 (\frac{423 - 298}{423}) = 147 .7 J$ .

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The standard entropies of CO₂(g), C(s) and O₂(g) are 213.5, 5.690 and 205 JK^–1 respectively. The standard entropy of formation of CO₂(g) is [CPMT 2001]

1.86 JK^–1

1.96 JK^–1

2.81 JK^–1

2.86 JK^–1

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Explanation

Formation of CO₂ is,

$C_{(s)} + O_{2 (g)} \to C O_{2 (g)}$

$Δ S^{o} = 213 .5 - 5 .690 - 205 = 2 .81 J K^{- 1}$ .

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