# For the reaction

Question:

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 K with 22 millimoles of A, the amount of B is the equilibrium mixture is _______ millimoles. (Round off to the Nearest Integer).

$\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ; \ell \mathrm{n} 10=2.303\right]$

Solution:

$\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ell \mathrm{n} \mathrm{K}_{\mathrm{eq}}$

Given $\Delta \mathrm{G}^{\circ}=-9.478 \mathrm{KJ} / \mathrm{mole}$

$\mathrm{T}=495 \mathrm{~K} \quad \mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1}$

So $-9.478 \times 10^{3}=-495 \times 8.314 \times \ell \mathrm{n} \mathrm{K}_{\text {eq }}$

\begin{aligned} \ell \mathrm{n} \mathrm{K}_{\mathrm{eq}} &=2.303 \\ &=\ell \mathrm{n} 10 \end{aligned}

So $\mathrm{K}_{\mathrm{eq}}=10$

Now $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})$

$\begin{array}{lcc}\mathrm{t}=0 & 22 & 0 \\ \mathrm{t}=\mathrm{t} & 22-\mathrm{x} & \mathrm{x}\end{array}$

$\mathrm{K}_{\mathrm{eq}}=\frac{[\mathrm{B}]}{[\mathrm{C}]}=\frac{\mathrm{x}}{22-\mathrm{x}}=10$

or $x=20$

So millmoles of $\mathrm{B}=20$