Among the following, the equilibrium which is not affected by an increase in pressure is

Options

A. `2SO_3(g)hArr2SO_(2(g))+O_(2)(g)`
B. `H_2(g)+I_2(s)hArr2HI(g)`
C. `C(s)+H_2O(g)hArrCO(g)+H_2(g)`
D. `3Fe(s) + 4H_(2) O(g) hArr Fe_(3)O_(4)(s) +4H_(2)(g)`

Explanation

To determine which equilibrium is not affected by an increase in pressure, we need to analyze the change in the number of moles of gas on both sides of the equilibrium reactions. The principle of Le Chatelier states that if an external change is applied to a system at equilibrium, the system will adjust to counteract that change.

### Step-by-Step Solution:

1. **Identify the Equilibria**: List the given equilibria and their corresponding states (gases, liquids, solids).

2. **Count Moles of Gases**: For each equilibrium, count the number of moles of gaseous reactants and products.
- If the number of moles of gas on the reactant side is different from the product side, the equilibrium will shift in response to a change in pressure.
- If the number of moles of gas is the same on both sides, the equilibrium will not be affected by pressure changes.

3. **Analyze Each Equilibrium**:
- **Equilibrium 1**: If it has 2 moles of gas on one side and 1 on the other, an increase in pressure will shift the equilibrium towards the side with fewer moles of gas.
- **Equilibrium 2**: If it has equal moles of gas on both sides (e.g., 2 moles on each side), an increase in pressure will not affect the equilibrium.
- **Equilibrium 3**: Similar analysis as above.
- **Equilibrium 4**: If it has 4 moles of gas on both sides, it will also remain unaffected by pressure changes.

4. **Conclusion**: The equilibrium that has the same number of moles of gas on both sides (delta ng = 0) will not be affected by an increase in pressure.

### Final Answer:
The equilibrium that is not affected by an increase in pressure is the one where the number of moles of gas on both sides is equal.