# Solve the following

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 $495 \mathrm{~K}$ with 22 millimoles of $\mathrm{A}$, the amount of $\mathrm{B}$ is the equilibrium mixture is....millimoles. (Round off to the Nearest Integer).

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

Solution:

(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 $B=20$