JEE Advanced 2021 Paper 1 · Q07 · Chemical Equilibrium & Le Chatelier

Correct answer:166.28

Solution

For the equilibrium X(s) $\rightleftharpoons$ Y(s) + Z(g) the equilibrium constant K$_p$ = p$_Z$/p$^\circ$. Given $\Delta G^\circ$ = -RT ln K$_p$ and $\Delta G^\circ$ = $\Delta H^\circ$ - T$\Delta S^\circ$:

ln(p$_Z$/p$^\circ$) = -$\Delta H^\circ$/(RT) + $\Delta S^\circ$/R.

From the plot of ln(p$_Z$/p$^\circ$) versus 10$^4$/T, the slope of the straight line = -$\Delta H^\circ$/(R \* 10$^4$). The plot passes through (10$^4$/T = 10, ln = -3) and (10$^4$/T = 12, ln = -7), giving slope = (-7 - (-3))/(12 - 10) = -4/2 = -2.

Thus -$\Delta H^\circ$/(R \* 10$^4$) = -2 \* 10$^{-4}$ \* 10$^4$ rearranged correctly: -$\Delta H^\circ$/R = -2 \* 10$^4$, so $\Delta H^\circ$ = 2 \* 10$^4$ \* R = 2 \* 10$^4$ \* 8.314 J mol$^{-1}$ = 166 280 J mol$^{-1}$ = 166.28 kJ mol$^{-1}$.

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