Question

The equilibrium constant for the following reaction is 49 at ${450}^{o}C.$ If $0.22mol$ of $I_2$, $0.22mol$ of $H_2$ and $0.66mol$ of $HI$ were put in a evacuated 1 litre container.
$H_2+I_2\rightleftharpoons 2HI$
Then, the correct statement will be:

• A. Reaction will proceed in forward direction until equilibrium is established.
• B. Reaction will proceed in backward direction until equilibrium is established.
• C. Reaction is already at equilibrium.
• D. None of the above

Answer 1

The correct option is A Reaction will proceed in forward direction until equilibrium is established.

Given, $n_{H_2}=0.22,n_{I_2}=0.2,n_{HI}=0.66V=1L$

$\therefore \left[H_2\right]=0.22,\left[I_2\right]=0.22,\left[HI\right]=0.66$

The concentration given in question are not necessarily the equilibrium concentrations.
Calculating reaction quotient $Q_c$ from these concentrations :

The reaction is
$H_2+I_2\rightleftharpoons 2HI$

For this reaction, $Q_c=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$

putting the value,
$Q_c=\frac{(0.66{)}^2}{\left(0.22\right)\left(0.22\right)}$
$Q_c=9.0$ but $K_c=49$ (Given), which is greater than $Q_c$

i.e. $K_c>Q_c$

Hence, the reaction will move in forward direction.