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Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

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What is the entropy change for the reaction given below,

2H₂ (g) + O₂ (g) $\to$ 2H₂O(l)

at temperature 300 K? Standard entropies of H₂ (g), O₂(g) and H₂O(l) are 126.6, 201.20 and 68.0 JK^-1mol^-1 respectively.

-318.4 JK^-1mol^-1

318.4 JK^-1mol^-1

31.84 JK^-1mol^-1

none of these

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Explanation

(a) $∆ S_{reaction} = {ΣS}_{product} - {ΣS}_{reactant} = 2 x S_{H_{2} O} - [2 {xS}_{H_{2}} + S_{O_{2}}]$

= 2 x 68-[2 x 126.6 + 201.20]

= -318.4 JK^-1mol^-1

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If S $°$ for H₂, Cl₂ and HCl are 0.13, 0.22 and 0.19 kJ K^-1mol^-1 respectively. The total change in standard entropy for the reaction, H₂ + Cl₂ $\to$ 2HCl is:

30 J K^-1mol^-1

40 J K^-1mol^-1

60 J K^-1mol^-1

20 J K^-1mol^-1

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Explanation

(a) $∆ S$ =S_P-S_R

=(2 x0.19) - 0.13-0.22

= 0.03 kJ K^-1mol^-1

= 30 JK^-1mol^-1

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In a flask, colourless N₂O₄(g) is in equilibrium with brown coloured NO₂(g). At equilibrium when the flask is heated to 100 $°$ C, the brown colour deepens and on cooling it becomes coloured. Which statement is incorrect about this observation?

The $∆$ H for the reaction N₂O₄(g) $⇌$ 2NO₂(g) is +ve

Paramagnetism increases on heating

The $∆$ H- $∆$ U at 100 $°$ C is equal to 200 cal

Dimerisation is reduced on heating

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Explanation

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Enthalpy of the reaction,

CH₄(g) + 1/2 O₂(g) $\to$ CH₃OH (l), is negative. If enthalpy of combustion of CH₄ and CH₃OH are x and y respectively, then which relation is correct?

x>y

x<y

x=y

x $⩾$ y

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Explanation

(b) CH₄(g) + 1/2 O₂(g) $\to$ CH₃OH (l)

.^.. $∆ H = [(∆ H_{c} of {CH}_{3} OH) - (∆ H_{c} of {CH}_{4})]$

= -[(-y)-(-x)]

=-[y+x] =y-x

.^.. x<y

Alternate explanation:

${CH}_{4} + 2 O_{2} \to {CO}_{2} + 2 H_{2} O ∆ H = x {CH}_{3} OH + \frac{3}{2} O_{2} \to {CO}_{2} + 2 H_{2} O ∆ H = y$

- (2)

${CH}_{4} + \frac{1}{2} O_{2} \to {CH}_{3} OH \Rightarrow x - y$

As x - y < 0

x < y

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In the reaction, $∆$ H and $∆$ S both are positive. In which of the following cases, the reaction would not be spontaneous?

$∆$ H>T $∆$ S

$∆$ S= $∆$ H/T

$∆$ H=T $∆$ S

All of these

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Explanation

(d) $∆ G$ =+ ve in a, and ZERO in b,c

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When an ideal gas is compressed adiabatically and reversibly, the final temperature is:

higher than the initial temperature

lower than the initial temperature

the same as the initial temperature

dependent on the rate of compression

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Explanation

(a) q= $∆$ U-W, if adiabatic process q=0, then - $∆$ U = -W, i.e, a decrease in free energy brings in work done by the system (-W).

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The entropy change in the fusion of one mole of a solid melting at 27 $°$ C (latent heat of fusion is 2930 J mol^-1) is

9.77 JK^-1mol^-1

10.73 JK^-1mol^-1

2930 JK^-1mol^-1

108.5 JK^-1mol^-1

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Explanation

(a) Entropy, $∆ S_{f} = ∆ H_{f} / T_{f}$ = Fusion enthalpy/Temperature

$∆ S_{f}$ = 2930 J mol^-1/300 K = 9.77 JK^-1mol^-1

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The maximum work done in expanding 16 g oxygen at 300 K and occupying a volume of 5 dm³ isothermally until the volume becomes 25 dm³ is:

2.01 x 10³ J

+2.81 x 10³ J

2.01 x 10^-3 J

+2.01 x 10^-6 J

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Explanation

(a) -W = +2.303nRTlog $\frac{V_{2}}{V_{1}}$

-W = 2.303 x 16/32 x 300 x 8.314log(25/5)

-W=2.01 x 10³ J

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1 mole of an ideal gas at 25 $° C$ is subjected to expand reversibly ten times of its initial volume. The change in entropy of expansion is:

19.15 JK^-1mol^-1

16.15 JK^-1mol^-1

22.15 JK^-1mol^-1

none of these

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Explanation

(a) $∆$ S=q/T = $\frac{2.303 nRT}{T} \log \frac{V_{2}}{V_{1}}$ =2.303 x 1 x 8.314log10

= 19.15 JK^-1mol^-1

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During an adiabatic process:

pressure is maintained constant

gas is isothermally expanded

there is perfect heat insulation

the system changes heat with surroundings

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Explanation

(c) An adiabatic process is one in which exchange of heat is not taking place in between system and surroundings. This can be made by putting insulation at the boundaries of system.

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