Question

Consider the equilibrium, $H _{2}+ I _{2} \rightleftharpoons 2 HI$
Calculate the equilibrium constant of the reverse reaction when the equilibrium concentration of $H _{2}, I _{2}$ and $HI$ are
$1.14 \times 10^{-2}, 0.12 \times 10^{-2}$ and $2.52 \times 10^{-2} mol \,L ^{-1}$
respectively

• A. 46.4
• B. 0.021
• C. 18.42
• D. 0.054

Answer 1

Answer: (B)

$H _{2}+ I _{2} \rightleftharpoons 2 HI$

Reverse of above reaction is

$2 HI \rightleftharpoons H _{2}+ I _{2} $

Equilibrium constant$(K)=\frac{\left[ H _{2}\right]\left[ I _{2}\right]}{[ HI ]^{2}}$

Given, Concentration of $H _{2}=1.14 \times 10^{-2} mol / L$

Concentration of $I _{2}=0.12 \times 10^{-2} mol / L$

Concentration of $HI =2.52 \times 10^{-2} mol / L$

$K=\frac{\left[1.14 \times 10^{-2}\right]\left[0.12 \times 10^{-2}\right]}{\left[252 \times 10^{-2}\right]^{2}} $

$K=0.021$