Question

Equilibrium constant $\left(K_c\right)$ for the following reaction $2HI\rightleftharpoons H_2+I_2$ is $\frac{1}{36}$ Then, percentage of HI left at equilibrium will be?

• A. 25%
• B. 50%
• C. 75%
• D. 12.5%

Answer 1

The correct option is C 75%

solution:

$2HI\rightleftharpoons H_2+I_2$

(1-x ) x\2 x\2 at equilibium

$K_C=\frac{\left[I_2\right]\left[H_2\right]}{[HI{]}^2}=\frac{1}{36}$

$K_C=\frac{\left[x/2\right]\left[x/2\right]}{\left[\right(1-x){]}^2}=\frac{1}{36}$

$\frac{[x{]}^2}{\left[\right(1-x){]}^2}=\frac{1}{9}$

$\frac{\left[x\right]}{\left[\right(1-x\left)\right]}=\frac{+1}{3}or\frac{-1}{3}$

$\frac{\left[3x\right]}{}=\frac{+1(1-x)}{}$ or $\frac{\left[3x\right]}{}=\frac{-1(1-x)}{}$

$x=\frac{1}{4};x=\frac{1}{2}$

$(1-x)=\frac{3}{4}or\frac{1}{2}$

only taking first value because 2nd one is at time =0

$\%\left[HI\right]=\frac{(1-x)}{x/2+x/2+(1-x)}=(1-x)$

which is 75%

hence the correct option : C