Equilibrium constant for the reaction `H_(2)O_((g))+CO_((g)) Leftrightarrow H_(2(g))+CO_(2(g))` is 81. If the rate constant of the forward reaction is 162 lit `mol^(-1), sec^(-1)`, what is the velocity constnat `("in lit." mole"^(-1) sec^(-1))` for the backward reaction?

Options

A. 131
B. 2
C. 261
D. 243

Answer

To find the velocity constant for the backward reaction, we can use the relationship between the equilibrium constant (K) and the rate constants of the forward (Kf) and backward (Kb) reactions. The relationship is given by the equation:

\[ K = \frac{K_f}{K_b} \]

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Equilibrium constant \( K = 81 \)
- Rate constant for the forward reaction \( K_f = 162 \, \text{lit} \, \text{mol}^{-1} \, \text{sec}^{-1} \)

2. **Set Up the Equation**:
- From the relationship \( K = \frac{K_f}{K_b} \), we can rearrange it to solve for \( K_b \):
\[ K_b = \frac{K_f}{K} \]

3. **Substitute the Known Values**:
- Substitute \( K_f \) and \( K \) into the equation:
\[ K_b = \frac{162}{81} \]

4. **Calculate \( K_b \)**:
- Perform the division:
\[ K_b = 2 \, \text{lit} \, \text{mol}^{-1} \, \text{sec}^{-1} \]

5. **Final Answer**:
- The velocity constant for the backward reaction is:
\[ K_b = 2 \, \text{lit} \, \text{mol}^{-1} \, \text{sec}^{-1} \]