Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\mathrm{N}{\mathrm{H}}_4^{+}\rightleftharpoons \mathrm{N}{\mathrm{H}}_3+{\mathrm{H}}^{+};{\mathrm{k}}_1=5.6\times 10^{10}$…(i)
${\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}^{+}+\mathrm{O}\mathrm{H};{\mathrm{k}}_2=1\times 10^{14}$…(ii)
$\mathrm{K}=\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{5.6\times 10^{-10}}{1\times 10^{-14}}$
$=5.6\times 10^4$
By subtracting equation (ii) from equation (i) we get
$\mathrm{N}{\mathrm{H}}_4^{+}+\mathrm{O}\mathrm{H}\rightleftharpoons \mathrm{N}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{O}$
From equation
${\mathrm{k}}_{\mathrm{f}}=3.4\times 10^{10}/\mathrm{m}\mathrm{i}{\mathrm{l}}^{-1}{\mathrm{S}}^{-1}$
$\mathrm{N}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{N}{\mathrm{H}}_4^{+}+\mathrm{O}{\mathrm{H}}^{-}$
${\mathrm{k}}_{\mathrm{r}}=?$
$\mathrm{K}=\frac{{\mathrm{k}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{r}}}$
So${\mathrm{k}}_{\mathrm{r}}=\frac{{\mathrm{k}}_{\mathrm{f}}}{\mathrm{K}}=\frac{3.4\times 10^{10}}{5.6\times 10^4}$
$=6.07\times 10^5$