Question:
The equilibrium pressure for the reaction MSO\(_4\cdot 2\)H\(_2\)O(s) \( \rightleftharpoons \) MSO\(_4\)(s) + 2H\(_2\)O(g) is \( \frac{\pi}{4} \) atm at 400 K. The \( K_p \) for the given reaction (in atm\(^2\)) is
The equilibrium pressure for the reaction MSO\(_4\cdot 2\)H\(_2\)O(s) \( \rightleftharpoons \) MSO\(_4\)(s) + 2H\(_2\)O(g) is \( \frac{\pi}{4} \) atm at 400 K. The \( K_p \) for the given reaction (in atm\(^2\)) is
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In heterogeneous equilibrium, solids are ignored in $K_p$.
Updated On: May 2, 2026
- $\frac{\pi^2}{4}$
- $\frac{\pi}{6}$
- $\frac{\pi^2}{16}$
- $\frac{\pi}{16}$
- $\frac{16}{\pi}$
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The Correct Option is C
Solution and Explanation
Concept: Equilibrium involving solids and gases
• Solids do not appear in equilibrium expression
• Only gaseous components contribute to $K_p$ ---
Step 1: Write equilibrium expression
\[ K_p = (P_{\text{H}_2\text{O}})^2 \] ---
Step 2: Given pressure
Total pressure = $\frac{\pi}{4}$ atm Only gas present = H$_2$O \[ P_{\text{H}_2\text{O}} = \frac{\pi}{4} \] ---
Step 3: Calculate $K_p$
\[ K_p = \left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{16} \] --- Final Answer: \[ \boxed{\frac{\pi^2}{16}} \]
• Solids do not appear in equilibrium expression
• Only gaseous components contribute to $K_p$ ---
Step 1: Write equilibrium expression
\[ K_p = (P_{\text{H}_2\text{O}})^2 \] ---
Step 2: Given pressure
Total pressure = $\frac{\pi}{4}$ atm Only gas present = H$_2$O \[ P_{\text{H}_2\text{O}} = \frac{\pi}{4} \] ---
Step 3: Calculate $K_p$
\[ K_p = \left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{16} \] --- Final Answer: \[ \boxed{\frac{\pi^2}{16}} \]
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