What will be the equilibrium constant of the given reaction carried out in a 5 L vessel and having equilibrium amounts of `A_2` and A as 0.5 mol and `2 xx 10^(-6)` mol respectively? The reaction : `A_2 hArr 2A`

Options

A. `0.16xx10^(-11)`
B. `0.25xx10^5`
C. `0.4xx10^(-5)`
D. `0.2xx10^(-11)`

Answer

To find the equilibrium constant \( K_c \) for the reaction \( A_2 \rightleftharpoons 2A \), we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction \( A_2 \rightleftharpoons 2A \) is given by:

\[ K_c = \frac{[A]^2}{[A_2]} \]

### Step 2: Calculate the concentrations of \( A \) and \( A_2 \)
We are given the equilibrium amounts of \( A_2 \) and \( A \) as follows:
- Moles of \( A_2 = 0.5 \, \text{mol} \)
- Moles of \( A = 2 \times 10^{-6} \, \text{mol} \)

Since the reaction is taking place in a 5 L vessel, we can calculate the concentrations:

\[ [A_2] = \frac{0.5 \, \text{mol}}{5 \, \text{L}} = 0.1 \, \text{M} \]

\[ [A] = \frac{2 \times 10^{-6} \, \text{mol}}{5 \, \text{L}} = 4 \times 10^{-7} \, \text{M} \]

### Step 3: Substitute the concentrations into the \( K_c \) expression
Now we can substitute the concentrations we calculated into the \( K_c \) expression:

\[ K_c = \frac{(4 \times 10^{-7})^2}{0.1} \]

### Step 4: Calculate \( K_c \)
Calculating the numerator:

\[ (4 \times 10^{-7})^2 = 16 \times 10^{-14} = 1.6 \times 10^{-13} \]

Now substituting back into the equation for \( K_c \):

\[ K_c = \frac{1.6 \times 10^{-13}}{0.1} = 1.6 \times 10^{-12} \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is:

\[ K_c = 1.6 \times 10^{-12} \]

---