`K_(p)andK_(c)` are related by `K_(p)=K_(c)(RT)^(Deltan)`. Under what practical condition/s, `K_(p)=K_(c)` ?

Options

A. RT = 1
B. T = 12K
C. `Deltan = 0`
D. `Rprop1/T`

Answer

To determine under what practical conditions \( K_p = K_c \), we start with the relationship between the two equilibrium constants:

\[ K_p = K_c (RT)^{\Delta n} \]

Where:
- \( K_p \) is the equilibrium constant in terms of partial pressures.
- \( K_c \) is the equilibrium constant in terms of concentrations.
- \( R \) is the universal gas constant.
- \( T \) is the temperature in Kelvin.
- \( \Delta n \) is the change in the number of moles of gas, calculated as the moles of gaseous products minus the moles of gaseous reactants.

### Step-by-step Solution:

1. **Understand the relationship**: The equation \( K_p = K_c (RT)^{\Delta n} \) shows that \( K_p \) and \( K_c \) are related through the term \( (RT)^{\Delta n} \).

2. **Identify \( \Delta n \)**: \( \Delta n \) represents the difference in the number of moles of gaseous products and reactants. It is calculated as:
\[ \Delta n = n_{\text{products}} - n_{\text{reactants}} \]

3. **Set \( \Delta n \) to zero**: For \( K_p \) to equal \( K_c \), we need the term \( (RT)^{\Delta n} \) to equal 1. This occurs when:
\[ \Delta n = 0 \]

4. **Conclusion**: When \( \Delta n = 0 \), the equation simplifies to:
\[ K_p = K_c \cdot (RT)^0 = K_c \cdot 1 = K_c \]
Therefore, the practical condition under which \( K_p = K_c \) is when the number of moles of gaseous products equals the number of moles of gaseous reactants, leading to \( \Delta n = 0 \).

### Final Answer:
\( K_p = K_c \) when \( \Delta n = 0 \) (i.e., the number of moles of gaseous products equals the number of moles of gaseous reactants).