At a given temperature the equilibrium constant for the reaction of
`PCl_(5)hArrPCl_(3)+Cl_(2)` is `2.4xx10^(-3)`. At the same temperature, the equilibrium constant for the reaction
`PCl_(3)(g)+Cl_(2)(g)hArrPCl_(5)(g)` is :

Options

A. `2.4xx10^(-3)`
B. `-.24xx10^(-3)`
C. `4.2xx10^(2)`
D. `4.8xx10^(-2)`

Explanation

To solve the problem, we need to determine the equilibrium constant for the reverse reaction given the equilibrium constant for the forward reaction.

### Given:
1. The equilibrium constant for the reaction:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]
is \( K_c = 2.4 \times 10^{-3} \).

### Step 1: Write the expression for the equilibrium constant.
For the forward reaction, the equilibrium constant \( K_c \) is expressed as:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 2: Determine the equilibrium constant for the reverse reaction.
The reverse reaction is:
\[ PCl_3 + Cl_2 \rightleftharpoons PCl_5 \]
For this reaction, the equilibrium constant \( K'_c \) can be expressed as:
\[ K'_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} \]

### Step 3: Relate the equilibrium constants.
The relationship between the equilibrium constants of a reaction and its reverse is given by:
\[ K'_c = \frac{1}{K_c} \]

### Step 4: Substitute the value of \( K_c \).
Now, substituting the value of \( K_c \):
\[ K'_c = \frac{1}{2.4 \times 10^{-3}} \]

### Step 5: Calculate \( K'_c \).
Calculating \( K'_c \):
\[ K'_c = \frac{1}{2.4 \times 10^{-3}} = \frac{1}{0.0024} \approx 416.67 \]

### Conclusion:
The equilibrium constant for the reaction \( PCl_3 + Cl_2 \rightleftharpoons PCl_5 \) is approximately \( 416.67 \).

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