JEE Advanced 2023 — Paper 2 — CHEM-IV solution

JEE Advanced2023Paper 2CHEM-IV

PARAGRAPH I

The entropy versus temperature plot for phases $α$ and $β$ at $1$ bar pressure is given. $S_{T}$ and $S_{0}$ are entropies of the phases at temperatures $T$ and $0$ K, respectively.

The transition temperature for $α$ to $β$ phase change is $600$ K and $C_{P, β} - C_{P, α} = 1$ J mol $^{- 1}$ K $^{- 1}$ . Assume $(C_{P, β} - C_{P, α})$ is independent of temperature in the range of $200$ to $700$ K. $C_{P, α}$ and $C_{P, β}$ are heat capacities of $α$ and $β$ phases, respectively.

Q1. The value of entropy change, $S_{β} - S_{α}$ (in J mol $^{- 1}$ K $^{- 1}$ ), at $300$ K is __.

[Use: $ln 2 = 0.69$ . Given: $S_{β} - S_{α} = 0$ at $0$ K]

Your answer

Solution

Use the temperature dependence of entropy at constant pressure, $d S = (C_{P} / T) d T$ . Integrating from $300$ K to $600$ K for each phase,

$S_{α}^{\circ} (600) = S_{α}^{\circ} (300) + C_{P, α} ln \frac{600}{300}$ ,

$S_{β}^{\circ} (600) = S_{β}^{\circ} (300) + C_{P, β} ln \frac{600}{300}$ .

Subtracting,

$[S_{β}^{\circ} (600) - S_{α}^{\circ} (600)] = [S_{β}^{\circ} (300) - S_{α}^{\circ} (300)] + (C_{P, β} - C_{P, α}) ln 2$ .

From the graph, $S_{β}^{\circ} (600) - S_{α}^{\circ} (600) = 6 - 5 = 1$ J mol $^{- 1}$ K $^{- 1}$ . Using $C_{P, β} - C_{P, α} = 1$ and $ln 2 = 0.69$ ,

$1 = (S_{β}^{\circ} (300) - S_{α}^{\circ} (300)) + 0.69$ .

$∴ S_{β} - S_{α} = 1 - 0.69 = 0.31$ J mol $^{- 1}$ K $^{- 1}$ .

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