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JEE Advanced 2023 · Paper 2, Chemistry question 14

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PARAGRAPH "I"

The entropy versus temperature plot for phases α and β at 1 bar pressure is given. STS_\mathrm T and S0S_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 Cp,β−Cp,α=1 J mol−1 K−1C_{\mathrm p,\beta} - C_{\mathrm p,\alpha} = 1\ \mathrm{J\,mol^{-1}\,K^{-1}}. Assume (Cp,β−Cp,α)(C_{\mathrm p,\beta} - C_{\mathrm p,\alpha}) is independent of temperature in the range of 200 to 700 K. Cp,αC_{\mathrm p,\alpha} and Cp,βC_{\mathrm p,\beta} are heat capacities of α and β phases, respectively.

S_T - S_0 in J mol^-1 K^-1 (y-axis) against temperature in K (x-axis) for phases α and β: two rising, concave-down curves, with β above α. At 600 K (dashed vertical guide) β has S_T - S_0 = 6 and α has 5 (dashed horizontal guides).

The value of entropy change, Sβ−SαS_\beta - S_\alpha (in J mol−1 K−1\mathrm{J\,mol^{-1}\,K^{-1}}), at 300 K is _____.

[Use: ln⁡2=0.69\ln 2 = 0.69. Given: Sβ−Sα=0S_\beta - S_\alpha = 0 at 0 K]

Answer and worked solution

Official answer: 0.3

At 600 K. Both phases have the same S0S_0, so the plot gives directly
ΔS(600)=Sβ−Sα=6−5=1 J mol−1K−1.\Delta S(600)=S_\beta-S_\alpha=6-5=1\ \text{J mol}^{-1}\text{K}^{-1}.

Correct to 300 K. ΔCp=1\Delta C_p=1 is constant:
ΔS(300)=ΔS(600)+ΔCpln⁡300600=1−0.69=0.31 J mol−1K−1.\Delta S(300)=\Delta S(600)+\Delta C_p\ln\frac{300}{600}=1-0.69=0.31\ \text{J mol}^{-1}\text{K}^{-1}.

Topics:EntropyGibbs energy and spontaneity
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