JEE PYQ: Thermodynamics Question 2
Question 2 - 2021 (24 Feb Shift 2)
Assuming ideal behaviour, the magnitude of $\log K$ for the following reaction at $25^\circ$C is $x \times 10^{-1}$. The value of $x$ is ___. (Integer answer)
$3\text{HC} \equiv \text{CH}_{(g)} \rightleftharpoons \text{C}6\text{H}{6(\ell)}$
[Given: $\Delta_f G^\circ (\text{HC} \equiv \text{CH}) = -2.04 \times 10^5$ mol$^{-1}$; $\Delta_f G^\circ (\text{C}_6\text{H}_6) = -1.24 \times 10^5$ J mol$^{-1}$. $R = 8.314$ J K$^{-1}$ mol$^{-1}$]
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Answer: 855
Solution
$\Delta G_r^\circ = \Delta G_f^\circ[\text{C}6\text{H}6(\ell)] - 3 \times \Delta G_f^\circ[\text{HC} \equiv \text{CH}] = [-1.24 \times 10^5 - 3 \times (-2.04 \times 10^5)] = 4.88 \times 10^5$ J/mol. $\Delta G_r^\circ = -RT\ln(K{eq})$. $\log(K{eq}) = \dfrac{-\Delta G_r^\circ}{2.303RT} = \dfrac{-4.88 \times 10^5}{2.303 \times 8.314 \times 298} = -8.55 \times 10^1 = 855 \times 10^{-1}$.
BETA
