Chemistry · Thermodynamics & Thermochemistry

JEE Advanced 2023 — Paper 2 — Question 44

The entropy versus temperature plot for phases α\alpha and β\beta at 1 bar pressure is given. ST\mathrm{S}_{\mathrm{T}} and S0\mathrm{S}_{0} are entropies of the phases at temperatures T and 0 K , respectively.

figure

The transition temperature for α\alpha to β\beta phase change is 600 K and Cp,β−Cp,α=1 J mol−1 K−1\mathrm{C}_{\mathrm{p}, \beta}-\mathrm{C}_{\mathrm{p}, \alpha}=1 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}. Assume

(Cp,β−Cp,α)\left(\mathrm{C}_{\mathrm{p}, \beta}-\mathrm{C}_{\mathrm{p}, \alpha}\right) is independent of temperature in the range of 200 to 700 K.Cp,α700 \mathrm{~K} . \mathrm{C}_{\mathrm{p}, \alpha} and Cp,β\mathrm{C}_{\mathrm{p}, \beta} are heat

capacities of α\alpha and β\beta phases, respectively.

The value of enthalpy change, Hβ−Hα(\mathrm{H}_{\beta}-\mathrm{H}_{\alpha}\left(\right. in Jmol−1)\left.\mathrm{J} \mathrm{mol}^{-1}\right), at 300 K is ____\_\_\_\_ .

Answer: 300

Numerical answer — enter this value.

Step-by-step solution

Transition : α⇌β;ΔG=0\alpha \rightleftharpoons \beta ; \Delta \mathrm{G}=0 So, ΔH=TΔS\Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S} ΔH600=600×1∵Δ S=1\Delta \mathrm{H}_{600}=600 \times 1 \because \Delta \mathrm{~S}=1 =600 J mol−1=600 \mathrm{~J} \mathrm{~mol}^{-1} From Krichoff's law ΔCp=ΔH600−ΔH300600−300\Delta \mathrm{C}_{\mathrm{p}}=\frac{\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}}{600-300} 1=600−ΔH3003001=\frac{600-\Delta \mathrm{H}_{300}}{300} ΔH300=300 J mol−1\Delta \mathrm{H}_{300}=300 \mathrm{~J} \mathrm{~mol}^{-1}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 2
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Entropy Change in Different Processes
The entropy versus temperature plot for phases α and β at 1 bar… | JEE Advanced 2023 PYQ with Solution · DhiX AI