Question

In the following reaction,
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
the amounts of $H_2,I_2$ and $HI$ are $7.8g,203.2g$ and $1638.4g$ respectively at equilibrium at a certain temperature. Calculate the equilibrium constant of the reaction.

Answer 1

The molecular weights of HI, $I_2$ and $H_2$ are 128 g/mol, 256 g/mol and 2 g/mol respectively
The number of moles of $HI=\frac{\text{1638.4 g}}{\text{128 g/mol}}=\text{12.8 mol}$
The number of moles of $I_2=\frac{\text{203.2g}}{\text{256 g/mol}}=\text{0.794 mol}$
The number of moles of $H_2=\frac{\text{7.8g g}}{\text{2 g/mol}}=\text{3.9 mol}$
Let V L be the total volume.
The equilibrium constant
$K_c=\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}$
$K_c=\frac{\frac{\text{3.9 mol}}{\text{V L}}\times \frac{\text{0.794 mol}}{\text{V L}}}{(\frac{\text{12.8 mol}}{\text{V L}}{)}^2}$
$K_c=0.019$