At 473 K,`K_(c)` for the reaction
`PCl_(5(g))hArrPCl_(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. (a) `8.3xx10^(3)`
B. (b) `120.48`
C. (c) `8.3xx10^(-3)`
D. (d) 240.8

Explanation

To find the value of \( K_c \) for the formation of \( PCl_5 \) at 473 K, we will follow these steps:

### Step 1: Write the given reaction and its equilibrium constant
The reaction given is:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]
The equilibrium constant for this reaction at 473 K is given as:
\[ K_c = 8.3 \times 10^{-3} \]

### Step 2: Write the reverse reaction for the formation of \( PCl_5 \)
The formation of \( PCl_5 \) can be represented as the reverse of the given reaction:
\[ PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \]

### Step 3: Write the expression for the equilibrium constant for the reverse reaction
For the reverse reaction, the equilibrium constant \( K_c' \) can be expressed as:
\[ K_c' = \frac{[PCl_5]}{[PCl_3][Cl_2]} \]

### Step 4: Relate \( K_c' \) to \( K_c \)
The relationship between the equilibrium constants of a reaction and its reverse is:
\[ K_c' = \frac{1}{K_c} \]
Thus, substituting the value of \( K_c \):
\[ K_c' = \frac{1}{8.3 \times 10^{-3}} \]

### Step 5: Calculate \( K_c' \)
Now, we calculate \( K_c' \):
\[ K_c' = \frac{1}{8.3 \times 10^{-3}} \approx 120.48 \]

### Conclusion
The value of \( K_c \) for the formation of \( PCl_5 \) at 473 K is approximately:
\[ K_c' \approx 120.48 \]