Question

In the reaction
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
the amounts of $H_2$, $I_2$ and $HI$ are $0.2g,9.2525g$ and $44.8g$ respectively at equilibrium at a certain temperature. Calculate the equilibrium constant of the reaction.

Answer 1

$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
Applying law of mass action
$K_c=\frac{{\left[HI\right]}^2}{\left[I_2\right]{\left[H_2\right]}^{}}$
Let the total volume be $V$ litre, then
$\left[H_2\right]=\frac{0.2g}{2g/mol\times VL}=\frac{0.1}{V}M;$
$\left[I_2\right]=\frac{9.2525g}{254g/mol\times VL}=\frac{0.0364}{V}M$
$\left[HI\right]=\frac{44.8g}{128g/mol\times VL}=\frac{0.35}{V}mol{L}^{-1}$.
so, $K_c=\frac{{\left(\frac{0.35}{V}\right)}^2}{\frac{0.1}{V}\times \frac{0.0364}{V}}=33.65$