For the reaction $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})$ at $495 \mathrm{~K}$, $\Delta_{\mathrm{r}} \mathrm{G}^{\mathrm{o}}=-9.478 \mathrm{~kJ} \mathrm{~mol}^{-1}$
If we start the reaction in a closed container at $495 \mathrm{~K}$ with 22 millimoles of $\mathrm{A}$, the amount of $\mathrm{B}$ is the equilibrium mixture is....millimoles. (Round off to the Nearest Integer).
$\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ; \ell \ln 10=2.303\right]$
(20)
Sol. $\Delta G^{\circ}=-R T \ell \mathrm{K}_{e q}$
Given $\Delta G^{\circ}=-9.478 \mathrm{KJ} /$ mole
$\mathrm{T}=495 \mathrm{~K} \quad \mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1}$
So $-9.478 \times 10^{3}=-495 \times 8.314 \times \ell \mathrm{nK}_{\mathrm{eq}}$
$\ell \mathrm{nK}_{\mathrm{eq}}=2.303$
$=\ell \mathrm{n} 10$
So $\mathrm{K}_{\mathrm{eq}}=10$
Now $\quad \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})$
$\mathrm{t}=0 \quad 22 \quad 0$
$\mathrm{t}=\mathrm{t} \quad 22-\mathrm{x} \quad \mathrm{x}$
$K_{e q}=\frac{[B]}{[C]}=\frac{x}{22-x}=10$
or $x=20$
So millmoles of $B=20$
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