At 473 K,`K_(c)` for the reaction
`PCl_(5(g))rArrPCl_(3(g))Cl_(2(g))` is `8.3xx10^(-3)`. What will be the value of `K_(c)` for the formation of `PCl_(5)` at the same temperature ?

Options

A. `8.3xx10^(3)`
B. 120.48
C. `8.3xx10^(-3)`
D. 240 .8

Answer

To find the value of \( K_c \) for the formation of \( PCl_5 \) at 473 K, we need to consider the relationship between the equilibrium constants of a reaction and its reverse reaction.

### Step-by-Step Solution:

1. **Write the given reaction and its equilibrium constant**:
The reaction is:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]
The equilibrium constant \( K_c \) for this reaction is given as:
\[ K_c = 8.3 \times 10^{-3} \]

2. **Identify the reverse reaction**:
The reverse reaction for the formation of \( PCl_5 \) is:
\[ PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \]

3. **Relate the equilibrium constants**:
For a reaction and its reverse, the equilibrium constants are related by:
\[ K_{c, \text{reverse}} = \frac{1}{K_{c, \text{forward}}} \]
Therefore, for our reaction:
\[ K_{c, \text{formation of } PCl_5} = \frac{1}{K_c} \]

4. **Calculate \( K_c \) for the formation of \( PCl_5 \)**:
Substituting the value of \( K_c \):
\[ K_{c, \text{formation of } PCl_5} = \frac{1}{8.3 \times 10^{-3}} \]

5. **Perform the calculation**:
\[ K_{c, \text{formation of } PCl_5} = \frac{1}{8.3 \times 10^{-3}} \approx 120.48 \]

6. **Final answer**:
Thus, the value of \( K_c \) for the formation of \( PCl_5 \) at 473 K is approximately:
\[ K_c \approx 120.48 \]