The equilibrium constant for the reaction is `H_(2)O_((l))+CO_((g)) Leftrightarrow H_(2(g))+CO_(2(g))` is 64. If the rate constant for the forward reaction is 160, the rate constant for the backward reaction is

Options

A. 0.4
B. 2.5
C. 6.2
D. `10.24 xx 10^(3)`

Answer

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

### Step-by-Step Solution:

1. **Write down the equilibrium expression**:
The equilibrium constant (K) for the reaction can be expressed in terms of the rate constants for the forward (kf) and backward (kb) reactions. The reaction is:
\[ H_2O_{(l)} + CO_{(g)} \leftrightarrow H_2_{(g)} + CO_2_{(g)} \]

2. **Recall the relationship between K and rate constants**:
The relationship is given by:
\[ K = \frac{k_f}{k_b} \]
where:
- \( K \) is the equilibrium constant,
- \( k_f \) is the rate constant for the forward reaction,
- \( k_b \) is the rate constant for the backward reaction.

3. **Substitute the known values**:
From the problem, we know:
- \( K = 64 \)
- \( k_f = 160 \)

We can substitute these values into the equation:
\[ 64 = \frac{160}{k_b} \]

4. **Rearrange the equation to solve for \( k_b \)**:
To find \( k_b \), we can rearrange the equation:
\[ k_b = \frac{160}{64} \]

5. **Calculate \( k_b \)**:
Now, perform the division:
\[ k_b = \frac{160}{64} = 2.5 \]

6. **Conclusion**:
The rate constant for the backward reaction \( k_b \) is 2.5.

### Final Answer:
The rate constant for the backward reaction is \( k_b = 2.5 \).

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