In which one of the following equilibria, the increase of pressure over the equilibrium will favour the backward reaction

Options

A. Decomposition equilibrium of HI
B. Formation equilibrium of `SO_3`
C. Decomposition equilibrium of `NH_3`
D. Formation equilibrium of `PCl_5`

Explanation

To determine in which equilibrium the increase of pressure will favor the backward reaction, we can analyze the given equilibria based on the number of moles of gas on both sides of the reaction. According to Le Chatelier's principle, if the pressure is increased, the equilibrium will shift towards the side with fewer moles of gas.

### Step-by-Step Solution:

1. **Identify the Equilibria**:
- The equilibria we need to consider are:
1. \( 2 HI(g) \rightleftharpoons H_2(g) + I_2(g) \)
2. \( SO_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons SO_3(g) \)
3. \( NH_3(g) \rightleftharpoons N_2(g) + \frac{3}{2} H_2(g) \)
4. \( PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \)

2. **Count the Moles of Gas**:
- For each reaction, count the total number of moles of gas on both sides:
1. \( 2 HI \) (2 moles) \( \rightleftharpoons H_2 + I_2 \) (2 moles)
- Total moles: 2 (reactants) = 2 (products)
2. \( SO_2 + \frac{1}{2} O_2 \) (1.5 moles) \( \rightleftharpoons SO_3 \) (1 mole)
- Total moles: 1.5 (reactants) > 1 (products)
3. \( NH_3 \) (1 mole) \( \rightleftharpoons N_2 + \frac{3}{2} H_2 \) (2.5 moles)
- Total moles: 1 (reactants) < 2.5 (products)
4. \( PCl_3 + Cl_2 \) (2 moles) \( \rightleftharpoons PCl_5 \) (1 mole)
- Total moles: 2 (reactants) > 1 (products)

3. **Analyze the Effect of Pressure**:
- According to Le Chatelier's principle:
- If the total number of moles of gas decreases on the product side compared to the reactant side, an increase in pressure will favor the forward reaction.
- Conversely, if the total number of moles of gas increases on the product side compared to the reactant side, an increase in pressure will favor the backward reaction.

4. **Determine the Correct Equilibrium**:
- For the equilibrium \( NH_3 \rightleftharpoons N_2 + \frac{3}{2} H_2 \):
- Reactants: 1 mole (NH3)
- Products: 2.5 moles (N2 + H2)
- Increase in pressure will favor the backward reaction (towards NH3).
- For the equilibrium \( PCl_3 + Cl_2 \rightleftharpoons PCl_5 \):
- Reactants: 2 moles (PCl3 + Cl2)
- Products: 1 mole (PCl5)
- Increase in pressure will favor the forward reaction (towards PCl5).

5. **Conclusion**:
- The equilibrium that will favor the backward reaction upon an increase in pressure is the decomposition of ammonia:
\[ NH_3(g) \rightleftharpoons N_2(g) + \frac{3}{2} H_2(g) \]

### Final Answer:
The increase of pressure will favor the backward reaction in the equilibrium of \( NH_3(g) \rightleftharpoons N_2(g) + \frac{3}{2} H_2(g) \).