Question

Assuming ideal behaviour, the magnitude of $\log K$ for the following reaction at $25°\mathrm{C}$ is $\mathrm{x}\times {10}^{-1}$. The value of $\mathrm{x}$ is (integer answer) -

$3\mathrm{CH}\equiv \mathrm{CH}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)$

[given-$∆{}_{\mathrm{f}}\mathrm{G}^0\left(\mathrm{CH}\equiv \mathrm{CH}\right)=-2.04\times {10}^5\mathrm{J}{\mathrm{mol}}^{-1}$, $∆{}_{\mathrm{f}}\mathrm{G}^0\left({\mathrm{C}}_6{\mathrm{H}}_6\right)=-1.24\times {10}^5\mathrm{J}{\mathrm{mol}}^{-1}$, $\mathrm{R}=8.314\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$]

Answer 1

Step 1: Formulae used- $∆{\mathrm{G}}_{\mathrm{rxn}}^0=-2.303\mathrm{RT}\log \mathrm{K}$, where $∆{\mathrm{G}}_{\mathrm{rxn}}^0$is the Gibbs free energy of the reaction, $\mathrm{R}$ is the gas constant, $\mathrm{T}$is the temperature and $\mathrm{K}$ is the equilibrium constant of the reaction.

$∆{\mathrm{G}}_{\mathrm{rxn}}^0=∆{\mathrm{G}}_{\mathrm{f}}^0\left({\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)\right)-3∆{\mathrm{G}}_{\mathrm{f}}^0\left(\mathrm{CH}\equiv \mathrm{CH}\left(\mathrm{g}\right)\right)$, where, $∆{\mathrm{G}}_{\mathrm{rxn}}^0$ is the Gibbs free energy of the reaction.

Step 2: Calculation of the $∆{\mathrm{G}}_{\mathrm{rxn}}^0$-

$∆{\mathrm{G}}_{\mathrm{rxn}}^0=\left(-1.24\times {10}^5\right)-\left(-3\times 2.04\times {10}^5\right)\mathrm{J}{\mathrm{mol}}^{-1}$

$∆{\mathrm{G}}_{\mathrm{rxn}}^0=4.88\times {10}^5\mathrm{J}{\mathrm{mol}}^{-1}$

Step 3: Calculation of $\log K$-

$\log K=\frac{-∆{\mathrm{G}}_{\mathrm{rxn}}^0}{2.303\mathrm{RT}}$

$\log K=\frac{-4.88\times {10}^5}{2.303\times 8.314\times 298}$

On solving the question:

$\log K=855\times {10}^{-1}$

So, for the integer value, the value of $\mathrm{x}$ is $855$.