`PCI_(5), PCI_(3)` and `CI_(2)` are in equilibrium at 500 K in a closed container and their concentration are `0.8 xx 10^(-3) " mol" L^(-1) " and " 1.2 xx 10^(-3) "mol" L^(-1) " and " 1.2 xx 10^(-3) "mol" L^(-1)` respectively. The value of `K_(c)` for the reaction `PCI_(5) (g) hArr PCI_(3) (g) + CI_(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 follow these steps:

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

The equilibrium constant \( K_c \) is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their coefficients in the balanced equation. For the given reaction, the expression for \( K_c \) is:

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

### Step 2: Substitute the equilibrium concentrations

From the problem, 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, we can substitute 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:

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

To simplify this, we can rewrite the denominator:

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

Calculating \( \frac{1.44}{0.8} \):

\[ \frac{1.44}{0.8} = 1.8 \]

Now, putting it all together:

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

### Final Answer

Thus, the value of \( K_c \) is:

\[ K_c = 1.8 \times 10^{-3} \, \text{mol L}^{-1} \]

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