Question

On a given condition, the equilibrium concentration of $HI$, $H_2$ and $I_2$ are

0.80, 0.10 and 0.10 mole/litre. The equilibrium constant for the reaction

$H_2+I_2\rightleftharpoons 2HI$ will be

• A. 64
• B. 12
• C. 8
• D. 0.8

Answer 1

The correct option is A

64

$H_2+I_2\rightleftharpoons 2HI;\left[HI\right]=0.80,\left[H_2\right]=0.10,\left[I_2\right]=0.10$

$K$ = $\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}=\frac{0.80\times 0.80}{0.10\times 0.10}$ = $64$