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 ${\text{H}}_{\text{2}}\text{+}{\text{I}}_{\text{2}}\rightleftharpoons \text{2HI}$.

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

Answer 1

The correct option is A $0.78M$

solution:

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

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 $\left(\right[HI]=2x=0.78$) and $(1.4$) M

Hence, the correct option is A