At a given temperature the equilibrium constant for the reaction `PCl_(5) (g) Leftrightarrow PCl_(3)(g)+Cl_(2)(g)` is `2.4 xx 10^(-3)`. At the same temperature the equilibrium constant for the reaction `PCl_(3) (g)+Cl_(2)(g) Leftrightarrow PCl_(5) (g)`

Options

A. `2.4 x 10^(-3)`
B. `-2.4 xx 10^(-3)`
C. `4.2 xx 10^(2)`
D. `4.8 xx 10^(-2)`

Answer

To solve the problem, we need to find the equilibrium constant for the reverse reaction of the given equilibrium. The steps are as follows:

### Step-by-Step Solution:

1. **Identify the Given Reaction and Its Equilibrium Constant**:
The first reaction is:
\[ PCl_5 (g) \leftrightarrow PCl_3 (g) + Cl_2 (g) \]
The equilibrium constant for this reaction is given as:
\[ K_1 = 2.4 \times 10^{-3} \]

2. **Write the Reverse Reaction**:
The reverse reaction is:
\[ PCl_3 (g) + Cl_2 (g) \leftrightarrow PCl_5 (g) \]

3. **Use the Property of Equilibrium Constants**:
According to the properties of equilibrium constants, when a reaction is reversed, the equilibrium constant for the reverse reaction (let's call it \( K_2 \)) is the reciprocal of the equilibrium constant for the forward reaction. Therefore:
\[ K_2 = \frac{1}{K_1} \]

4. **Calculate \( K_2 \)**:
Substitute the value of \( K_1 \) into the equation:
\[ K_2 = \frac{1}{2.4 \times 10^{-3}} \]

5. **Perform the Calculation**:
\[ K_2 = \frac{1}{2.4} \times 10^{3} = 0.41667 \times 10^{3} = 4.1667 \times 10^{2} \]

6. **Final Result**:
Thus, the equilibrium constant for the reverse reaction is:
\[ K_2 \approx 4.17 \times 10^{2} \]