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JEE Advanced 2021 Paper 1

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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 : 26 of 57

Marks: +1, -0

\Delta H^{\Theta}

\mathrm{kJ} \mathrm{mol}^{-1}

Solution:

\Delta G^{\circ} = -R T \ln k_{eq} b ;[k_{eq} = \frac{p_z}{p^{\Theta}}

p^{\Theta} = 1]

\Delta G^{\circ} = -R T \ln \left( \frac{p_z}{1} \right) = \Delta H^{\Theta} - T \Delta S^{\Theta}

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

Slope

= - \frac{\Delta H^{\Theta}}{R} = 10^{4} \times \left( - \frac{4}{2} \right)

\Rightarrow \Delta H^{\Theta} = 2 \times 10^{4} \times R = 2 \times 10^{4} \times 8.314 = 166.28 \text{ kJ/mole}

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