Question

At $T(K)$ consider the following equilibrium reaction \[ HgO(s) \rightleftharpoons Hg(g)+\frac{1}{2}O_2(g) \] The correct relation between $K_p$ and $P_{Total}(P_T)$ is}

• A. $K_p = \frac{2}{3^{1/2}}.P_T^{1/2}$
• B. $K_p = \frac{2}{3^{3/2}}.P_T^{3/2}$
• C. $K_p = \frac{2}{3^{2/3}}.P_T^{2/3}$
• D. $K_p = \frac{1}{3^{2/3}}.P_T$ \bigskip

Hint:

For gaseous equilibrium problems:
• Find mole fractions.
• Convert to partial pressures using \[ P_i=y_iP_T \]
• Substitute into the expression of \(K_p\). Always omit solids and pure liquids from equilibrium constant expressions.

Answer 1

The Correct Option is B

Concept: For heterogeneous equilibria, the activity of a pure solid is taken as unity and does not appear in the equilibrium constant expression. For the reaction \[ HgO(s)\rightleftharpoons Hg(g)+\frac12 O_2(g) \] \[ K_p=P_{Hg}\,(P_{O_2})^{1/2} \] We express the partial pressures in terms of total pressure.

Step 1: Assume decomposition of one mole of HgO.
Let the extent of decomposition be \(x\). Then \[ Hg(g)=x \] \[ O_2(g)=\frac{x}{2} \] Total gaseous moles \[ n_T=x+\frac{x}{2} =\frac{3x}{2} \]

Step 2: Calculate mole fractions.
For mercury vapor, \[ y_{Hg} = \frac{x}{3x/2} = \frac{2}{3} \] For oxygen, \[ y_{O_2} = \frac{x/2}{3x/2} = \frac13 \] Therefore, \[ P_{Hg} = \frac23 P_T \] \[ P_{O_2} = \frac13 P_T \]

Step 3: Substitute into the expression of \(K_p\).
\[ K_p = P_{Hg}(P_{O_2})^{1/2} \] \[ = \left(\frac23P_T\right) \left(\frac13P_T\right)^{1/2} \] \[ = \frac23 \cdot \frac{1}{\sqrt3} \cdot P_T^{3/2} \] \[ = \frac{2}{3\sqrt3} P_T^{3/2} \] \[ = \frac{2}{3^{3/2}} P_T^{3/2} \] Hence, \[ \boxed{ K_p= \frac{2}{3^{3/2}} P_T^{3/2} } \]