Question

In the following reaction
$H_2+I_2\rightleftharpoons HI$
the amount of $H_2,I_2$ and $HI$ are $0.2g$, $9.525g$ and $44.8g$ respectively at equilibrium at a certain temperature. Calculate the equilibrium constant of the reaction.

Answer 1

$H_2+I_2\rightleftharpoons 2HI$

Let volume of container=V

Concentration of $H_2$=$\frac{\text{no. of moles}}{volume}=\frac{\frac{0.2}{2}}{V}=\frac{0.1}{V}$

Concentration of $I_2$=$\frac{\text{no. of moles}}{volume}=\frac{\frac{9.525}{254}}{V}=\frac{0.0375}{V}$

Concentration of $HI$=$\frac{\text{no. of moles}}{volume}=\frac{\frac{44.89}{128}}{V}=\frac{0.35}{V}$

$K_c=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}=\frac{{\left(0.35\right)}^2}{\left(0.0375\right)\left(0.1\right)}=32.667$