Which of the following expression is true regarding formation of `PCl_(5)` by equation given
`PCl_(3)(g)+Cl_(2)(g)hArr PCl_(5)(g)`

Options

A. `(K_(p))/(K_(c))lt1`
B. `(K_(p))/(K_(c))gt1`
C. `(K_(p))/(K_(c))=1`
D. None of these

Answer

To determine the correct expression regarding the formation of \( PCl_5 \) from the reaction:

\[ PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \]

we need to analyze the relationship between the equilibrium constants \( K_p \) and \( K_c \).

### Step-by-Step Solution:

1. **Identify the Reaction and Equilibrium Constants**:
The given reaction is:
\[ PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \]
- \( K_p \) is the equilibrium constant in terms of partial pressures.
- \( K_c \) is the equilibrium constant in terms of concentrations.

2. **Use the Relationship Between \( K_p \) and \( K_c \)**:
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.

3. **Calculate \( \Delta N_g \)**:
To find \( \Delta N_g \):
- Count the moles of gaseous reactants and products.
- On the reactant side, we have 2 moles (\( 1 \, PCl_3 + 1 \, Cl_2 \)).
- On the product side, we have 1 mole (\( 1 \, PCl_5 \)).
- Therefore, \( \Delta N_g = \text{(moles of products)} - \text{(moles of reactants)} = 1 - 2 = -1 \).

4. **Substitute \( \Delta N_g \) into the Equation**:
Substitute \( \Delta N_g \) into the relationship:
\[ K_p = K_c (RT)^{-1} \]
This can be rearranged to:
\[ \frac{K_p}{K_c} = \frac{1}{RT} \]

5. **Analyze the Expression**:
Since \( R \) (the gas constant) and \( T \) (temperature in Kelvin) are both positive values, \( RT \) will also be positive. Therefore:
\[ \frac{K_p}{K_c} < 1 \]
This implies that \( K_p < K_c \).

### Conclusion:
The correct expression regarding the formation of \( PCl_5 \) is:
\[ K_p < K_c \]