In which of the following equilibrium, `(K_c = K_p)` ?

Options

A. `[N_2(g) + O_2(g) ltimplies 2NO(g)]`
B. `[4NH_3(g) + 5O_2(g) ltimplies 4NO(g) + 6H_2O(g)]`
C. `[N_2(g) + 3H_2(g)ltimplies 2NH_3(g)]`
D. `[2NO(g) + O_2(g) ltimplies 2NO_2(g)]`

Answer

To determine in which equilibrium \( K_c = K_p \), we need to analyze the relationship between \( K_c \) and \( K_p \) given by the equation:

\[ K_p = K_c R T^{\Delta N_g} \]

Where:
- \( R \) is the gas constant,
- \( T \) is the temperature,
- \( \Delta N_g \) is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.

### Step-by-Step Solution:

1. **Understand the Relationship**:
The relationship shows that \( K_p \) and \( K_c \) are equal when \( \Delta N_g = 0 \). This means that the number of moles of gaseous products is equal to the number of moles of gaseous reactants.

2. **Calculate \( \Delta N_g \) for Each Reaction**:
We need to find \( \Delta N_g \) for each given equilibrium reaction.

- **First Reaction**:
- Products: 2 moles (gaseous)
- Reactants: 2 moles (gaseous)
- \( \Delta N_g = \text{moles of products} - \text{moles of reactants} = 2 - 2 = 0 \)

- **Second Reaction**:
- Products: 10 moles (gaseous)
- Reactants: 9 moles (gaseous)
- \( \Delta N_g = 10 - 9 = 1 \)

- **Third Reaction**:
- Products: 2 moles (gaseous)
- Reactants: 4 moles (gaseous)
- \( \Delta N_g = 2 - 4 = -2 \)

- **Fourth Reaction**:
- Products: 2 moles (gaseous)
- Reactants: 3 moles (gaseous)
- \( \Delta N_g = 2 - 3 = -1 \)

3. **Determine When \( K_c = K_p \)**:
From our calculations:
- First Reaction: \( \Delta N_g = 0 \)
- Second Reaction: \( \Delta N_g = 1 \)
- Third Reaction: \( \Delta N_g = -2 \)
- Fourth Reaction: \( \Delta N_g = -1 \)

The only reaction where \( \Delta N_g = 0 \) is the **first reaction**.

4. **Conclusion**:
Therefore, the equilibrium in which \( K_c = K_p \) is the **first option**.