`K_(p)` is always equal to `K_(c) " if"Delta n_(g)` is ………..

Answer

To solve the question, we need to understand the relationship between \( K_p \) and \( K_c \) in terms of the change in the number of moles of gas, represented as \( \Delta n_g \).

### Step-by-Step Solution:

1. **Understand the Definitions**:
- \( K_p \) is the equilibrium constant for a reaction expressed in terms of partial pressures of the gases involved.
- \( K_c \) is the equilibrium constant for a reaction expressed in terms of the concentrations of the reactants and products.

2. **Recall the Relationship**:
- The relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c (RT)^{\Delta n_g} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n_g \) is the change in the number of moles of gas, calculated as the moles of gaseous products minus the moles of gaseous reactants.

3. **Analyze the Condition for Equality**:
- For \( K_p \) to be equal to \( K_c \), the term \( (RT)^{\Delta n_g} \) must equal 1.
- This occurs when \( \Delta n_g = 0 \) because any number raised to the power of 0 is 1.

4. **Conclusion**:
- Therefore, \( K_p \) is equal to \( K_c \) if \( \Delta n_g = 0 \).

### Final Answer:
\( K_p \) is always equal to \( K_c \) if \( \Delta n_g = 0 \).