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

Options

A. `T=6.09 K`
B. `T=12.18K`
C. `T=36.54K`
D. `T=27.3K`

Explanation

To solve the problem, we need to find the temperature at which the equilibrium constant \( K_p \) is half of \( K_c \) for the reaction:

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

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

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

\[ K_p = K_c \cdot R T^{\Delta n} \]

where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas (moles of products - moles of reactants).

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

For the reaction:

- Products: \( PCl_3(g) + Cl_2(g) \) → 2 moles of gas
- Reactants: \( PCl_5(g) \) → 1 mole of gas

Thus,

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

### Step 3: Set up the equation for the given condition

We are given that \( K_p = \frac{1}{2} K_c \). Substituting this into the relationship gives:

\[ \frac{1}{2} K_c = K_c \cdot R T^{\Delta n} \]

### Step 4: Simplify the equation

Cancelling \( K_c \) from both sides (assuming \( K_c \neq 0 \)):

\[ \frac{1}{2} = R T^{\Delta n} \]

Substituting \( \Delta n = 1 \):

\[ \frac{1}{2} = R T \]

### Step 5: Solve for \( T \)

Now, substituting the value of \( R \):

\[ \frac{1}{2} = 0.0821 \cdot T \]

Rearranging gives:

\[ T = \frac{1}{2 \cdot 0.0821} = \frac{1}{0.1642} \]

Calculating \( T \):

\[ T \approx 6.09 \text{ K} \]

### Conclusion

The temperature at which \( K_p \) is half of \( K_c \) is approximately:

\[ \boxed{6.09 \text{ K}} \]