Question

A 20.0 litre vessel initially contains 0.50 mole each of $H_2$ and $I_2$ gases. These substances react and finally reach an equilibrium condition. Calculate the equilibrium concentration of $HI$ if $K_{eq}=49$ for the reaction $H_2+I_2\rightleftharpoons 2HI.$

• A. 0.78 M
• B. 0.039 M
• C. 0.033 M
• D. 0.021 M

Answer 1

The correct option is A 0.78 M

The equilibrium reaction is,

$\underset{0.50.5-x}{H_2}+\underset{0.50.5-x}{I_2}\rightleftharpoons \underset{02x}{2}HI.$

The expression for the equilibrium constant is $K_{eq}=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$

Conc. at eqm.-

$\left[HI\right]=\frac{2x}{20},\left[I_2\right]=\frac{0.5-x}{20}and\left[H_2\right]=\frac{0.5-x}{20}$

Substitute values in the above expression
$49=\frac{(\frac{2x}{20}{)}^2}{\left(\frac{0.5-x}{20}\right)\left(\frac{0.5-x}{20}\right)}$

Thus $x=0.388$ and $0.7$ or $\left[HI\right]=2x=0.78$ and $1.4$ M.

Hence, the correct option is A.