Question

When 1 mole of $H_2\left(g\right)$ is heated with one mole of $I_2\left(g\right)$ , it was found that 1.48 moles of HI(g) is formed at equilibrium. Its $K_c$ is:

• A. 16
• B. 32
• C. 8
• D. 24

Answer 1

The correct option is B 32

The equilibrium reaction is $H_2\left(g\right)+I_2\left(g\right)\iff 2HI\left(g\right)$.
Assume that the total volume of the reaction mixture is 0.
Initial concentrations of $H_2,I_{2}andHI$ are1,1 and 0 M respectively.
At equilibrium, the concentrations of $H_2,I_{2}andHI$ are$(1-0.74=0.26),(1-0.74=0.26)and1.48$ M respectively.
The expression for the equilibrium constant is $K_c=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}=\frac{(1.48{)}^2}{0.26\times 0.26}=32$.