Consider the following reactions .In which case the formation of product is favoured by decreasein pressure?
(1)`CO_(2)(g)+C(s)hArr2CO(g),DeltaH^(@)=+172.5Kj`
(2)`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) hArr2NH_(3)(g),DeltaH^(@)=-91.8KJ`
(3) `N_(2)(g)+O_(2)(g)hArr2NO(g),DeltaH^(@)=181KJ`
(4) `2H_(2)O(g)hArr2H_(2)(g)+O_(2)(g),DeltaH^(@)=484.6KJ`

Options

A. `2,3`
B. `3,4`
C. `2,4`
D. `1,4`

Answer

To determine in which reaction the formation of products is favored by a decrease in pressure, we can apply Le Chatelier's principle. According to this principle, if the pressure of a system at equilibrium is decreased, the equilibrium will shift in the direction that produces more moles of gas.

Let's analyze each reaction step by step:

### Step 1: Analyze Reaction 1
**Reaction**: \( CO_2(g) + C(s) \rightleftharpoons 2CO(g) \)
- **Moles of Reactants**: 1 (from \( CO_2 \)) + 1 (from \( C \), solid does not count) = 1 mole of gas
- **Moles of Products**: 2 (from \( 2CO \)) = 2 moles of gas
- **Conclusion**: Since the number of gaseous moles increases from 1 to 2, decreasing pressure will favor the formation of products.

### Step 2: Analyze Reaction 2
**Reaction**: \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \)
- **Moles of Reactants**: 1 (from \( N_2 \)) + 3 (from \( H_2 \)) = 4 moles of gas
- **Moles of Products**: 2 (from \( 2NH_3 \)) = 2 moles of gas
- **Conclusion**: Since the number of gaseous moles decreases from 4 to 2, decreasing pressure will not favor the formation of products.

### Step 3: Analyze Reaction 3
**Reaction**: \( N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \)
- **Moles of Reactants**: 1 (from \( N_2 \)) + 1 (from \( O_2 \)) = 2 moles of gas
- **Moles of Products**: 2 (from \( 2NO \)) = 2 moles of gas
- **Conclusion**: Since the number of gaseous moles is equal (2 on both sides), a decrease in pressure will not favor either the reactants or products.

### Step 4: Analyze Reaction 4
**Reaction**: \( 2H_2O(g) \rightleftharpoons 2H_2(g) + O_2(g) \)
- **Moles of Reactants**: 2 (from \( 2H_2O \)) = 2 moles of gas
- **Moles of Products**: 2 (from \( 2H_2 \)) + 1 (from \( O_2 \)) = 3 moles of gas
- **Conclusion**: Since the number of gaseous moles increases from 2 to 3, decreasing pressure will favor the formation of products.

### Final Conclusion
The reactions where the formation of products is favored by a decrease in pressure are:
- Reaction 1: \( CO_2(g) + C(s) \rightleftharpoons 2CO(g) \)
- Reaction 4: \( 2H_2O(g) \rightleftharpoons 2H_2(g) + O_2(g) \)

Thus, the correct options are **1 and 4**.

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