Question

For the reaction $A\left(g\right)\rightleftharpoons B\left(g\right)$ at $495\text{K},\Delta G^{\circ }=–9.478{\text{kJ mol}}^{-1}$. If we start the reaction in a closed container at $495\text{K}$ with $22$ millimoles of $A$, the amount of $B$ in the equilibrium mixture is millimoles. (Round off to the nearest Integer).
$[R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1};ln10=2.303]$

Answer 1

By using the relation,
$\Delta G^{\circ }=-RTln{K}_{eq}\text{Where},\Delta G^{\circ }=–9.478{\text{kJ mol}}^{-1}T=\text{Temperature}=495\text{K}R=\text{Universal gas constant}=8.314{\text{J mol}}^{-1}{\text{K}}^{-1}\text{Substituting the values, we get,}-9.478\times {10}^3=-495\times 8.314ln{K}_{eq}ln{K}_{eq}=2.303=ln10$
$\Rightarrow K_{eq}=10$
Now,
$$\begin{array}{ccccc}& & A\left(g\right)& \rightleftharpoons & B\left(g\right)\\ t=0& & 22& & 0\\ t=t& & 22-x& & x\end{array}$$
$K_{eq}=\frac{\left[B\right]}{\left[A\right]}=\frac{x}{(22-x)}=10\Rightarrow x=20$