Reaction in which yield of product will increase with increase in pressure is

Options

A. `H_(2)(g)+I_(2)(g) hArr 2HI(g)`
B. `H_(2)O(g)+CO(g) hArr CO_(2)(g)+H_(2)(g)`
C. `H_(2)O(g)+C(s)hArr CO(g)+H_(2)(g)`
D. `CO(g)+3H_(2)(g) hArr CH_(4)(g)+H_(2)O(g)`

Answer

To determine which reaction's yield of product will increase with an increase in pressure, we can apply Le Chatelier's principle. Here’s a step-by-step solution:

### Step 1: Understand Le Chatelier's Principle
Le Chatelier's principle states that if an external change is applied to a system at equilibrium, the system will adjust itself to counteract that change. In the case of pressure, if the pressure is increased, the equilibrium will shift towards the side with fewer moles of gas.

**Hint:** Remember that the side with fewer moles of gas will be favored when pressure is increased.

### Step 2: Analyze the Given Reactions
We need to evaluate each reaction option to determine the number of moles of gas on both the reactant and product sides.

1. **Option A:**
\[ \text{H}_2(g) + \text{I}_2(g) \rightarrow 2\text{HI}(g) \]
- Reactants: 1 mole of H₂ + 1 mole of I₂ = 2 moles
- Products: 2 moles of HI
- **Conclusion:** No change in moles (2 = 2).

2. **Option B:**
\[ \text{H}_2O(g) + \text{CO}(g) \rightarrow \text{CO}_2(g) + \text{H}_2(g) \]
- Reactants: 1 mole of H₂O + 1 mole of CO = 2 moles
- Products: 1 mole of CO₂ + 1 mole of H₂ = 2 moles
- **Conclusion:** No change in moles (2 = 2).

3. **Option C:**
\[ \text{H}_2O(g) + \text{C(s)} \rightarrow \text{CO(g)} + \text{H}_2(g) \]
- Reactants: 1 mole of H₂O + 1 mole of C (solid) = 1 mole of gas (we don’t count solids)
- Products: 1 mole of CO + 1 mole of H₂ = 2 moles
- **Conclusion:** Increase in moles (1 < 2).

4. **Option D:**
\[ \text{CO(g)} + 3\text{H}_2(g) \rightarrow \text{CH}_4(g) + \text{H}_2O(g) \]
- Reactants: 1 mole of CO + 3 moles of H₂ = 4 moles
- Products: 1 mole of CH₄ + 1 mole of H₂O = 2 moles
- **Conclusion:** Decrease in moles (4 > 2).

### Step 3: Identify the Correct Reaction
From the analysis:
- **Option D** is the only reaction where the number of moles of products (2 moles) is less than the number of moles of reactants (4 moles). Therefore, if the pressure is increased, the equilibrium will shift to the right (towards the products).

### Final Answer
The reaction in which the yield of the product will increase with an increase in pressure is:
**D: CO(g) + 3H₂(g) → CH₄(g) + H₂O(g)**

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