Question

The equilibrium constant for the reaction $H_2\left(g\right)+I_2\left(g\right)\iff 2HI\left(g\right)$ is $81$ at a certain temperature. If the concentrations of $H_2$ and $I_2$ are $3$ mole/lit each at equilibrium, the equilibrium concentration of $HI$ is:

• A. $17$ mol/litre
• B. $27$ mol/litre
• C. $34$ mol/litre
• D. $51$ mol/litre

Answer 1

Answer: (B)

$27$ mol/litre

The given equilibrium reaction is :-

$H_{2\left(g\right)}+I_{2\left(g\right)}\rightleftharpoons 2{HI}_{\left(g\right)};K_C=81$

$K_C=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$

Given: $\left[H_2\right]=3M,\left[I_2\right]=3M$

$\Rightarrow \left[HI\right]=\sqrt{81\times 3\times 3}=9\times 3=27mol/litre$
Option B is correct.