The rate constant for forward reaction `A(g) hArr 2B(g)` is `1.5 xx 10^(-3) s^(-1)` at `100 K`. If `10^(-5)` moles of `A` and `100 "moles"` of `B` are present in a `10-L` vessel at equilibrium, then the rate constant for the backward reaction at this temperature is

Options

A. `A. 1.50 xx 10^(-4) L mol s^(-1)`
B. `B. 1.50 xx 10^(-11) L mol^(-1) s^(-1)`
C. `C. 1.50 xx 10^(-10) L mol^(-1) s^(-1)`
D. D. None of these

Explanation

To solve the problem, we need to find the rate constant for the backward reaction given the rate constant for the forward reaction and the equilibrium concentrations of the reactants and products.

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction is given as:
\[ A(g) \rightleftharpoons 2B(g) \]

2. **Identify Given Data**:
- Rate constant for the forward reaction, \( k_f = 1.5 \times 10^{-3} \, s^{-1} \)
- Moles of \( A = 10^{-5} \)
- Moles of \( B = 100 \)
- Volume of the vessel = 10 L

3. **Calculate Concentrations**:
- Concentration of \( A \):
\[ [A] = \frac{\text{moles of } A}{\text{volume}} = \frac{10^{-5} \, \text{moles}}{10 \, \text{L}} = 10^{-6} \, \text{mol/L} \]
- Concentration of \( B \):
\[ [B] = \frac{\text{moles of } B}{\text{volume}} = \frac{100 \, \text{moles}}{10 \, \text{L}} = 10 \, \text{mol/L} \]

4. **Write the Expression for Equilibrium Constant**:
The equilibrium constant \( K_{eq} \) for the reaction can be expressed as:
\[ K_{eq} = \frac{[B]^2}{[A]} = \frac{(10)^2}{10^{-6}} = \frac{100}{10^{-6}} = 10^8 \]

5. **Relate Equilibrium Constant to Rate Constants**:
The relationship between the equilibrium constant and the rate constants for the forward and backward reactions is given by:
\[ K_{eq} = \frac{k_f}{k_b} \]
Rearranging this gives:
\[ k_b = \frac{k_f}{K_{eq}} \]

6. **Substitute the Values**:
Substitute \( k_f \) and \( K_{eq} \) into the equation:
\[ k_b = \frac{1.5 \times 10^{-3}}{10^8} \]

7. **Calculate \( k_b \)**:
\[ k_b = 1.5 \times 10^{-3} \times 10^{-8} = 1.5 \times 10^{-11} \, \text{L}^{-1} \text{mol}^{-1} \text{s}^{-1} \]

### Final Answer:
The rate constant for the backward reaction \( k_b \) is:
\[ k_b = 1.5 \times 10^{-11} \, \text{L}^{-1} \text{mol}^{-1} \text{s}^{-1} \]