Question

$15$ moles of $H_2$ and $5.2$ moles of $I_2$ are mixed and allowed to attain equilibrium at $500^\circ$C. At equilibrium, the concentration of $HI$ is found to be $10$ moles. The equilibrium constant for the formation of $HI$ is

• A. 50
• B. 15
• C. 100
• D. 25

Answer 1

Answer: (A)

Equilibrium constant

$\left(K_{c}\right) =\frac{( HI )^{2}}{\left[ H _{2}\right]\left[ I _{2}\right]}$

$=\frac{10 \times 10}{10 \times 0.2}$

$=50$