`PCl_(5), PCl_(3) and Cl_(2)` are at equilibrium at 500 K in a closed container and their concentrations are `0.8xx10^(-3) ` mol `L^(-1), 1.2xx10^(-3) ` mol `L^(-1) and 1.2xx10^(-3) ` mol `L^(-1)` respectively. The value of `K_(c)` for the reaction `PCl_(5)(g) hArrPCl_(3)(g)+Cl_(2)(g)` will be

Options

A. `1.8xx10^(-3) ` mol `L^(_1)`
B. `1.8xx10^(-3)`
C. `1.8xx10^(-3) L "mol"^(-1)`
D. `0.55xx10^(4)`

Explanation

To find the equilibrium constant \( K_c \) for the reaction

\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

we will use the equilibrium concentrations provided.

### Step 1: Write the expression for \( K_c \)

The equilibrium constant \( K_c \) is defined as:

\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]

where \([\text{PCl}_3]\), \([\text{Cl}_2]\), and \([\text{PCl}_5]\) are the equilibrium concentrations of the respective species.

### Step 2: Substitute the given concentrations

From the question, we have the following equilibrium concentrations:

- \([\text{PCl}_5] = 0.8 \times 10^{-3} \, \text{mol L}^{-1}\)
- \([\text{PCl}_3] = 1.2 \times 10^{-3} \, \text{mol L}^{-1}\)
- \([\text{Cl}_2] = 1.2 \times 10^{-3} \, \text{mol L}^{-1}\)

Now, substituting these values into the \( K_c \) expression:

\[ K_c = \frac{(1.2 \times 10^{-3})(1.2 \times 10^{-3})}{0.8 \times 10^{-3}} \]

### Step 3: Calculate the numerator

First, calculate the numerator:

\[ (1.2 \times 10^{-3}) \times (1.2 \times 10^{-3}) = 1.44 \times 10^{-6} \]

### Step 4: Calculate \( K_c \)

Now substitute the numerator back into the \( K_c \) expression:

\[ K_c = \frac{1.44 \times 10^{-6}}{0.8 \times 10^{-3}} \]

### Step 5: Simplify the expression

Now, simplify the expression:

\[ K_c = \frac{1.44 \times 10^{-6}}{0.8 \times 10^{-3}} = \frac{1.44}{0.8} \times 10^{-3} = 1.8 \times 10^{-3} \]

### Final Answer

Thus, the value of \( K_c \) for the reaction at 500 K is:

\[ K_c = 1.8 \times 10^{-3} \]

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