Question

The reaction:

\[ \text{H}_2\text{O}(g) + \text{Cl}_2\text{O}(g) \rightleftharpoons 2 \text{HOCl}(g) \]

is allowed to attain equilibrium at 400K. At equilibrium, the partial pressures are given as:

• \( P_{\text{H}_2\text{O}} = 300 \) mm Hg
• \( P_{\text{Cl}_2\text{O}} = 20 \) mm Hg
• \( P_{\text{HOCl}} = 60 \) mm Hg

The value of \( K_p \) for the reaction at 400K is:

\[ K_p = \frac{P_{\text{HOCl}}^2}{P_{\text{H}_2\text{O}} \cdot P_{\text{Cl}_2\text{O}}} \]

• A. 36
• B. 6.0
• C. 60
• D. 3.6
• E. 0.60

Hint:

The equilibrium constant for partial pressures \( K_p \) is calculated by taking the ratio of the products of the partial pressures of the products to the reactants, raised to their respective stoichiometric coefficients.

Answer 1

Answer: (E)

The equilibrium constant \( K_p \) is given by the expression: \[ K_p = \frac{P_{{HOCl}}^2}{P_{{H}_2{O}} \times P_{{Cl}_2{O}}} \] Substituting the given values: \[ K_p = \frac{(60)^2}{300 \times 20} = \frac{3600}{6000} = 0.60 \] Thus, the value of \( K_p \) for the reaction at 300K is \( 0.60 \).