For the equilibrium in a closed vessel`" "PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`,
`K_(p)` is found to be double of `K_(e)` . This is attained when `:`

Options

A. `T=2K`
B. `T=12.18K`
C. `T=24.36K`
D. `T=27.3K`

Explanation

To solve the problem, we need to establish the relationship between \( K_p \) and \( K_c \) for the equilibrium reaction:

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

### Step 1: Understand the relationship between \( K_p \) and \( K_c \)

The relationship between \( K_p \) and \( K_c \) is given by the equation:

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

where:
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n_g \) is the change in the number of moles of gas during the reaction.

### Step 2: Calculate \( \Delta n_g \)

In the reaction, we have:
- Reactants: 1 mole of \( \text{PCl}_5 \)
- Products: 1 mole of \( \text{PCl}_3 \) + 1 mole of \( \text{Cl}_2 \) = 2 moles

Thus, the change in the number of moles of gas is:

\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} = 2 - 1 = 1 \]

### Step 3: Substitute \( \Delta n_g \) into the equation

Now, substituting \( \Delta n_g \) into the relationship gives:

\[ K_p = K_c (RT)^{1} \implies K_p = K_c RT \]

### Step 4: Use the given condition

According to the problem, we know that:

\[ K_p = 2 K_c \]

### Step 5: Set the equations equal

From the previous step, we have:

\[ K_c RT = 2 K_c \]

Assuming \( K_c \neq 0 \), we can divide both sides by \( K_c \):

\[ RT = 2 \]

### Step 6: Solve for \( T \)

Now, we can solve for \( T \):

\[ T = \frac{2}{R} \]

Given that \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \):

\[ T = \frac{2}{0.0821} \approx 24.36 \, \text{K} \]

### Conclusion

The temperature at which \( K_p \) is double \( K_c \) is approximately \( 24.36 \, \text{K} \).