Equilibrium constant for an equilibrium reaction is 100. Its forward reaction rate constant `K_f= 10^(5).` Its backward reaction rate constant `K_b` is

Options

A. `10^(2)`
B. 10
C. `10^(-3)`
D. `10^(3)`

Answer

To find the backward reaction rate constant \( K_b \), we can use the relationship between the equilibrium constant \( K \), the forward reaction rate constant \( K_f \), and the backward reaction rate constant \( K_b \). The relationship is given by the formula:

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

### Step-by-step Solution:

1. **Identify the given values:**
- Equilibrium constant \( K = 100 \)
- Forward reaction rate constant \( K_f = 10^5 \)

2. **Set up the equation using the relationship:**
\[ K = \frac{K_f}{K_b} \]
Plugging in the known values:
\[ 100 = \frac{10^5}{K_b} \]

3. **Rearranging the equation to solve for \( K_b \):**
Multiply both sides by \( K_b \):
\[ 100 K_b = 10^5 \]
Now, divide both sides by 100:
\[ K_b = \frac{10^5}{100} \]

4. **Simplifying the right side:**
We can express 100 as \( 10^2 \):
\[ K_b = \frac{10^5}{10^2} = 10^{5-2} = 10^3 \]

5. **Final answer:**
Therefore, the backward reaction rate constant \( K_b \) is:
\[ K_b = 10^3 \]

### Summary:
The backward reaction rate constant \( K_b \) is \( 10^3 \).