To solve the problem, we need to match the reactions in List I with the appropriate statements in List II. Let's analyze each reaction step by step.
### Step 1: Analyze Reaction A
**Reaction A:** \( \text{CaCO}_3 \rightleftharpoons \text{CaO} + \text{CO}_2 \)
- **Type of Equilibrium:** This is a heterogeneous equilibrium because it involves solids (CaCO₃ and CaO) and a gas (CO₂).
- **Change in Moles of Gas (Δn):**
- Products: 1 mole of CO₂ (gas)
- Reactants: 0 moles of gas
- Δn = 1 - 0 = 1
- **Kp and Kc Relationship:** Since there is a change in the number of moles of gas, we have:
\[
K_p = K_c (RT)^{\Delta n} \quad \text{(where Δn = 1)}
\]
- **Effect of Pressure:** High pressure favors the backward reaction because it shifts the equilibrium towards the side with fewer moles of gas.
**Conclusion for A:**
- Matches with (Q) "High pressure favours backward reaction", (R) \( K_p = K_c (RT) \), and (S) "Heterogeneous equilibrium".
### Step 2: Analyze Reaction B
**Reaction B:** \( \text{NH}_4\text{HS} \rightleftharpoons \text{NH}_3 + \text{H}_2\text{S} \)
- **Type of Equilibrium:** This is a heterogeneous equilibrium because NH₄HS is a liquid, while NH₃ and H₂S are gases.
- **Change in Moles of Gas (Δn):**
- Products: 2 moles of gas (NH₃ and H₂S)
- Reactants: 0 moles of gas
- Δn = 2 - 0 = 2
- **Kp and Kc Relationship:**
\[
K_p = K_c (RT)^{\Delta n} \quad \text{(where Δn = 2)}
\]
- **Effect of Pressure:** High pressure favors the backward reaction because there are more moles of gas on the product side.
**Conclusion for B:**
- Matches with (Q) "High pressure favours backward reaction" and (S) "Heterogeneous equilibrium".
### Step 3: Analyze Reaction C
**Reaction C:** \( \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \)
- **Type of Equilibrium:** This is a heterogeneous equilibrium because PCl₅ is a solid, while PCl₃ and Cl₂ are gases.
- **Change in Moles of Gas (Δn):**
- Products: 1 mole of Cl₂ (gas)
- Reactants: 0 moles of gas
- Δn = 1 - 0 = 1
- **Kp and Kc Relationship:**
\[
K_p = K_c (RT)^{\Delta n} \quad \text{(where Δn = 1)}
\]
- **Effect of Pressure:** High pressure favors the backward reaction because it shifts the equilibrium towards the side with fewer moles of gas.
**Conclusion for C:**
- Matches with (Q) "High pressure favours backward reaction", (R) \( K_p = K_c (RT) \), and (S) "Heterogeneous equilibrium".
### Step 4: Analyze Reaction D
**Reaction D:** \( 3\text{Fe} + 4\text{H}_2 \rightleftharpoons \text{Fe}_3\text{O}_4 + 4\text{H}_2 \)
- **Type of Equilibrium:** This is a heterogeneous equilibrium because it involves solids (Fe and Fe₃O₄) and gases (H₂).
- **Change in Moles of Gas (Δn):**
- Products: 4 moles of H₂ (gas)
- Reactants: 4 moles of H₂ (gas)
- Δn = 4 - 4 = 0
- **Kp and Kc Relationship:**
\[
K_p = K_c \quad \text{(because Δn = 0)}
\]
- **Effect of Pressure:** Since Δn = 0, pressure has no effect on the equilibrium.
**Conclusion for D:**
- Matches with (P) \( K_p = K_c \) and (S) "Heterogeneous equilibrium".
### Final Matching:
- **A** corresponds to (Q), (R), and (S)
- **B** corresponds to (Q) and (S)
- **C** corresponds to (Q), (R), and (S)
- **D** corresponds to (P) and (S)
### Summary of Matches:
- A → (Q), (R), (S)
- B → (Q), (S)
- C → (Q), (R), (S)
- D → (P), (S)
