For the reaction `Pcl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`

Options

A. `K_(p)=K_(c)`
B. `K_(p)=K_(c)(RT)^(-1)`
C. `K_(p)=K_(c)(RT)`
D. `K_(p)=K_(c)(RT)^(2)`

Explanation

To solve the problem regarding the equilibrium constants \( K_p \) and \( K_c \) for the reaction:

\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

we need to understand the relationship between \( K_p \) and \( K_c \) in terms of the change in the number of moles of gas during the reaction.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The given reaction is:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

2. **Determine the Change in Moles of Gas (\( \Delta N_g \))**:
- On the reactant side, we have 1 mole of \( \text{PCl}_5 \).
- On the product side, we have 1 mole of \( \text{PCl}_3 \) and 1 mole of \( \text{Cl}_2 \), totaling 2 moles.
- Thus, the change in moles of gas is:
\[ \Delta N_g = \text{Moles of products} - \text{Moles of reactants} = 2 - 1 = 1 \]

3. **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.

4. **Substitute \( \Delta N_g \)**:
Since we found that \( \Delta N_g = 1 \), we can substitute this value into the equation:
\[ K_p = K_c (RT)^{1} \]
This simplifies to:
\[ K_p = K_c \cdot RT \]

5. **Conclusion**:
Therefore, the correct relationship is:
\[ K_p = K_c \cdot RT \]
The answer corresponds to option 3.