Question

$$\begin{gathered} \mathrm{Equilibrium}\mathrm{constant}\mathrm{of}\mathrm{reaction}2\mathrm{HI}\rightleftharpoons {\mathrm{H}}_2+{\mathrm{I}}_2\mathrm{at}\mathrm{a}\mathrm{given}\mathrm{temperature}\mathrm{is}0.01563.\mathrm{Calculate}\mathrm{the}\mathrm{equilibrium} \\ \mathrm{constants}\mathrm{for}\mathrm{the}\mathrm{following}\mathrm{reactions}. \\ \left(\mathrm{i}\right)\mathrm{HI}\rightleftharpoons \frac{1}{2}{\mathrm{H}}_2+\frac{1}{2}{\mathrm{I}}_2 \\ \left(\mathrm{ii}\right){\mathrm{H}}_2+{\mathrm{I}}_2\rightleftharpoons 2\mathrm{HI} \end{gathered}$$

Answer 1

The solution is as follows:

1. For the reaction:

HI $\leftrightharpoons$$\frac{1}{2}$ H₂ + ​$\frac{1}{2}$ I₂

We see that this equation is half of the given reaction. So,

K₂ = (K₁)$$\begin{gathered} {}^{\frac{1}{2}} \\ =\sqrt{0.01563}=0.12 \end{gathered}$$

2. As
H₂ + I₂ ​$\leftrightharpoons$ 2HI

K₂ = $\frac{1}{K_1}=\frac{1}{0.01563}=63.97$