The rate constant for a forward reaction in a reversible reaction `(K_(eq) = 10^(6))` is `10^(2)` . Calculate rate constant for its backward reaction

Options

A. `10^(4)`
B. `10^(-4)`
C. `10^(-6)`
D. `10^(6)`

Answer

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

### Step-by-Step Solution:

1. **Understand the Relationship**:
The relationship between the equilibrium constant (\(K_{eq}\)), the rate constant for the forward reaction (\(k_{forward}\)), and the rate constant for the backward reaction (\(k_{backward}\)) is given by the formula:
\[ K_{eq} = \frac{k_{forward}}{k_{backward}} \]

2. **Identify Given Values**:
From the problem, we have:
- \(K_{eq} = 10^6\)
- \(k_{forward} = 10^2\)

3. **Rearrange the Formula**:
To find \(k_{backward}\), we can rearrange the formula:
\[ k_{backward} = \frac{k_{forward}}{K_{eq}} \]

4. **Substitute the Values**:
Now, substitute the known values into the rearranged formula:
\[ k_{backward} = \frac{10^2}{10^6} \]

5. **Perform the Calculation**:
When dividing powers of 10, we subtract the exponents:
\[ k_{backward} = 10^{2 - 6} = 10^{-4} \]

6. **Final Answer**:
Therefore, the rate constant for the backward reaction is:
\[ k_{backward} = 10^{-4} \]

### Summary:
The rate constant for the backward reaction is \(10^{-4}\).