Question

The reaction $2HI\rightleftharpoons H_2+I_2$, at equilibrium contained $7.8g,203.2g$ and $1638.4g$ of $H_2,I_2$ and $HI$, respectively. Calculate $K_c$.

Answer 1

$2HI\to H_2+I_2$

Moles at equilibrium $\frac{1638.4}{128}=12.8;\frac{7.8}{2}=3.9;\frac{203.2}{254}=0.8,$ respectively.

Let the volume of the container by $V$ litre.

$\left[H_2\right]=\frac{3.9}{V};\left[HI\right]=\frac{12.8}{V};\left[I_2\right]=\frac{0.8}{V}$

$\therefore K_c=\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}=\frac{3.9\times 0.8}{V\times V\times {\left(\frac{12.8}{V}\right)}^2}=0.019$

$\therefore K_c=0.019$.