In which of the following system, doubling the volume of the container causes a shift to the right

Options

A. `H_(2)(g)+Cl_(2)(g)hArr 2HCl(g)`
B. `2CO(g)+O_(2)(g)hArr2CO_(2)(g)`
C. `N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`
D. `PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)`

Answer

To determine which system will shift to the right when the volume of the container is doubled, we can apply Le Chatelier's principle. This principle states that if an external change is applied to a system at equilibrium, the system will adjust to counteract that change and restore a new equilibrium.

### Step-by-step Solution:

1. **Understand the Effect of Volume Change**:
- When the volume of a container is increased (doubled in this case), the pressure inside the container decreases. According to Le Chatelier's principle, the equilibrium will shift towards the side with more moles of gas to counteract this decrease in pressure.

2. **Analyze Each Reaction**:
- We need to evaluate each given reaction to see which one has more moles of gas on the product side compared to the reactant side.

3. **Evaluate the Reactions**:
- **Reaction 1**: \( H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g) \)
- Reactants: 1 mole \( H_2 \) + 1 mole \( Cl_2 \) = 2 moles
- Products: 2 moles \( HCl \)
- **Moles are equal**: No shift.

- **Reaction 2**: \( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \)
- Reactants: 2 moles \( SO_2 \) + 1 mole \( O_2 \) = 3 moles
- Products: 2 moles \( SO_3 \)
- **More moles on reactant side**: Shift to the left.

- **Reaction 3**: \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \)
- Reactants: 1 mole \( N_2 \) + 3 moles \( H_2 \) = 4 moles
- Products: 2 moles \( NH_3 \)
- **More moles on reactant side**: Shift to the left.

- **Reaction 4**: \( N_2(g) + 2O_2(g) \rightleftharpoons 2NO(g) \)
- Reactants: 1 mole \( N_2 \) + 2 moles \( O_2 \) = 3 moles
- Products: 2 moles \( NO \)
- **More moles on reactant side**: Shift to the left.

4. **Identify the Correct Reaction**:
- The only reaction that has more moles on the product side compared to the reactant side is the fourth reaction. Thus, when the volume is doubled, this reaction will shift to the right.

### Final Answer:
The correct answer is **Reaction 4**: \( N_2(g) + 2O_2(g) \rightleftharpoons 2NO(g) \).