For which of the following reactions at equilibrium at constant temperature, doubling the volume will cause a shift to the right?

Options

A. `N_(2)O_(4)(g) hArr 2NO_(2)(g)`
B. `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)`
C. `2CO(g)+O_(2)(g) hArr 2CO_(2)(g)`
D. `N_(2)(g)+O_(2)(g) hArr 2NO(g)`

Answer

To determine which reaction will shift to the right when the volume is doubled at constant temperature, we can apply Le Chatelier's principle. This principle states that if a system at equilibrium is subjected to a change in conditions (such as volume), the system will adjust to counteract that change and restore a new equilibrium.

### Step-by-Step Solution:

1. **Identify the Reactions**: We have four reactions to analyze:
- **Option 1**: \( N_2O_4(g) \rightleftharpoons 2 NO_2(g) \)
- **Option 2**: \( CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) \)
- **Option 3**: \( 2 CO(g) + O_2(g) \rightleftharpoons 2 CO_2(g) \)
- **Option 4**: \( N_2(g) + O_2(g) \rightleftharpoons 2 NO(g) \)

2. **Count Moles of Gases**: For each reaction, count the number of moles of gaseous reactants and products:
- **Option 1**: 1 mole of \( N_2O_4 \) (reactants) and 2 moles of \( NO_2 \) (products) → Total: 1 mole (reactants) vs. 2 moles (products)
- **Option 2**: 0 moles of gas (reactants) and 1 mole of \( CO_2 \) (products) → Total: 0 moles (reactants) vs. 1 mole (products)
- **Option 3**: 3 moles of gas (2 moles of \( CO \) + 1 mole of \( O_2 \)) (reactants) and 2 moles of \( CO_2 \) (products) → Total: 3 moles (reactants) vs. 2 moles (products)
- **Option 4**: 2 moles of gas (1 mole of \( N_2 \) + 1 mole of \( O_2 \)) (reactants) and 2 moles of \( NO \) (products) → Total: 2 moles (reactants) vs. 2 moles (products)

3. **Analyze the Effect of Doubling the Volume**: When the volume is doubled, the pressure decreases. According to Le Chatelier's principle, the equilibrium will shift toward the side with more moles of gas to counteract the decrease in pressure.
- **Option 1**: Shift to the right (2 moles of products > 1 mole of reactants)
- **Option 2**: Shift to the right (1 mole of products > 0 moles of reactants)
- **Option 3**: Shift to the left (3 moles of reactants > 2 moles of products)
- **Option 4**: No shift (2 moles of reactants = 2 moles of products)

4. **Conclusion**: The reactions that will shift to the right when the volume is doubled are Options 1 and 2. However, since the question asks for a single answer, we consider the reaction with the most significant shift, which is **Option 1**: \( N_2O_4(g) \rightleftharpoons 2 NO_2(g) \).

### Final Answer:
The reaction that will shift to the right when the volume is doubled is **Option 1**: \( N_2O_4(g) \rightleftharpoons 2 NO_2(g) \).