Question

The equilibrium constant for the given reaction is 81 at a certain temperature. If the concentrations of $H_2$ and $I_2$ are 3 moles/ lit each at equilibrium, the equilibrium concentration of HI is:

$H_2\left(g\right)+I_2\left(g\right)\leftrightharpoons 2HI\left(g\right)$

• A. 3 mole / lit
• B. 27 mole / lit
• C. 9 mole / lit
• D. 13.5 mole / lit

Answer 1

Answer: (B)

27 mole / lit

The equilibrium reaction is $H_2\left(g\right)+I_2\left(g\right)\leftrightharpoons 2HI\left(g\right)$.

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

Substitute values in the above expression.

$81=\frac{[HI{]}^2}{3\times 3}$.

Thus $\left[HI\right]=27mole/lit$.

Hence, the correct option is $\text{B}$