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JEE Advanced 2021 Paper 1

Paragraph for next 2 questions

For the reaction

X(s) \rightleftharpoons Y(s)+Z(g)

, the plot of

\ln\;\frac{p_{Z}}{p^{\Theta}}

versus

\;\frac{10^{4}}{T}

is given below (in solid line), where

p_{z}

is the pressure (in bar) of the gas Z at temperature T and

P^{\Theta}=1

bar.

(Given,

\;\frac{d(\ln K)}{d\left(\frac{1}{T}\right)} = -\frac{\Delta H^{\Theta}}{R}

, where the equilibrium constant,

K =\;\frac{p_{Z}}{p^{\Theta}}

and the gas constant,

R = 8.314 \; \mathrm{J} \; \mathrm{K}^{-1} \; \mathrm{mol}^{-1}

)

Section: Chemistry

Question : 27 of 57

Marks: +1, -0

The value of

\Delta S^{\Theta}

\mathrm{J}\ \mathrm{K}^{-1}\ \mathrm{mol}^{-1}

1000\ \mathrm{K}

Solution:

Form the plot when,

\;\frac{10^{4}}{T}=10\;\; [T=1000\ \mathrm{K}]

\ln \left(\;\frac{p_{z}}{1}\right)=-3

Substituting in following equation:

\ln \left(\;\frac{p_{z}}{1}\right)=-\;\frac{\Delta H^{\Theta}}{R T}+\;\frac{\Delta S^{\Theta}}{R}

We get,

\;\; -3=-\;\frac{2 \times 10^{4} \times R}{R \times 1000}+\;\frac{\Delta S^{\Theta}}{R} \Rightarrow \;\; \Delta S^{\Theta}=17R

\Rightarrow \Delta S^{\Theta}=17 \times 8.314 \ \mathrm{J} / \mathrm{K} - \mathrm{mol} \Rightarrow \Delta S^{\Theta}=141.34 \ \mathrm{J} / \mathrm{K} - \mathrm{mol}

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