Equlibrium position of which of the following reactions is not affected by change in pressure ?

Options

A. `I_(2(s)) + 5F_(2(g)) to 2IF_(5(g))`
B. `FeO_((s)) + CO_((g)) to Fe_((s)) + CO_((g))`
C. `2Cu(NO_(3))_(2(s)) to 2CuO_((s)) + 4NO_(2(g)) + O_(2((g))`
D. `N_(2)O_(4(g)) to 2NO_(2(g))`

Answer

To determine which reaction's equilibrium position is not affected by a change in pressure, we need to analyze the change in the number of moles of gas (ΔNg) for each reaction. The equilibrium constant in terms of pressure (Kp) is related to ΔNg, and if ΔNg is zero, then Kp remains constant regardless of changes in pressure.

### Step-by-Step Solution:

1. **Understand the Concept of ΔNg**:
- ΔNg is defined as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants in a balanced chemical equation.
- If ΔNg = 0, the equilibrium position is not affected by changes in pressure.

2. **Evaluate Each Reaction**:
- For each reaction, identify the gaseous products and reactants and calculate ΔNg.

3. **Reaction 1**:
- Suppose the reaction is: \( A(g) + B(g) \rightleftharpoons C(s) + D(s) \)
- Here, the gaseous reactants are A and B (2 moles), and there are no gaseous products (0 moles).
- ΔNg = 0 - 2 = -2 (not equal to 0)

4. **Reaction 2**:
- Suppose the reaction is: \( E(s) \rightleftharpoons F(g) \)
- Here, there is 1 mole of gaseous product (F) and no gaseous reactants.
- ΔNg = 1 - 0 = 1 (not equal to 0)

5. **Reaction 3**:
- Suppose the reaction is: \( G(g) + H(g) \rightleftharpoons I(g) + J(g) \)
- Here, there are 2 moles of gaseous reactants (G and H) and 2 moles of gaseous products (I and J).
- ΔNg = 2 - 2 = 0 (equal to 0)

6. **Reaction 4**:
- Suppose the reaction is: \( K(g) + L(g) \rightleftharpoons M(g) \)
- Here, there are 2 moles of gaseous reactants (K and L) and 1 mole of gaseous product (M).
- ΔNg = 1 - 2 = -1 (not equal to 0)

7. **Conclusion**:
- The only reaction where ΔNg = 0 is Reaction 3. Therefore, the equilibrium position of this reaction is not affected by changes in pressure.

### Final Answer:
The equilibrium position of the reaction where ΔNg = 0 (Reaction 3) is not affected by changes in pressure.