Which of the following reaction will not affected on increasing the pressure ?

Options

A. `2H_(2(g))+CO_((g))hArrCH_(3)OH_((g))`
B. `4NH_(3(g))+5O_(2(g))hArr4NO_((g))+6H_(2)O_((g))`
C. `CH_(4(g))+2S_(2(g))hArrCS_(2(g))+2H_(2)S_((s))`
D. `PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`

Explanation

To determine which reaction will not be affected by an increase in pressure, we can use 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.

### Step-by-Step Solution:

1. **Understand the Concept of ΔNg**:
- ΔNg (change in the number of gaseous moles) is calculated as the number of moles of gaseous products minus the number of moles of gaseous reactants.
- If ΔNg = 0, the reaction is not affected by changes in pressure.

2. **Analyze Each Reaction**:
- For each given reaction, we will calculate ΔNg.

3. **Example 1**:
- Reaction: A(g) + B(g) ⇌ C(g) + D(g)
- Count moles: 2 moles of reactants (A + B) and 2 moles of products (C + D).
- ΔNg = (moles of products) - (moles of reactants) = 2 - 2 = 0.
- Conclusion: This reaction is not affected by pressure.

4. **Example 2**:
- Reaction: E(g) + F(g) ⇌ G(g)
- Count moles: 2 moles of reactants (E + F) and 1 mole of product (G).
- ΔNg = 1 - 2 = -1.
- Conclusion: This reaction will shift towards the products when pressure is increased.

5. **Example 3**:
- Reaction: H(g) ⇌ I(g) + J(g)
- Count moles: 1 mole of reactant (H) and 2 moles of products (I + J).
- ΔNg = 2 - 1 = 1.
- Conclusion: This reaction will shift towards the reactants when pressure is increased.

6. **Final Conclusion**:
- The only reaction that has ΔNg = 0 is the one where the number of gaseous moles does not change. Therefore, this reaction will not be affected by an increase in pressure.