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JEE Advanced 2023 · Paper 2, Chemistry question 15

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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 enthalpy change, Hβ−HαH_\beta - H_\alpha (in J mol−1\mathrm{J\,mol^{-1}}), at 300 K is _____.

Answer and worked solution

Official answer: 300

At the transition temperature. At 600 K the phases are in equilibrium, so ΔG=0\Delta G=0:
ΔH(600)=TΔS=600×1=600 J mol−1.\Delta H(600)=T\Delta S=600\times1=600\ \text{J mol}^{-1}.

Kirchhoff's law with constant ΔCp=1\Delta C_p=1:
ΔH(300)=600+1×(300−600)=300 J mol−1.\Delta H(300)=600+1\times(300-600)=300\ \text{J mol}^{-1}.

Topics:First law and enthalpyGibbs energy and spontaneity
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