The equilibrium constant K of a reversible reaction is 10. The rate constant for the reverse reaction is 2.8. What is the rate constant for the forward reaction

Options

A. 0.28
B. 28
C. 0.028
D. 280

Answer

To solve the problem, we need to find the rate constant for the forward reaction (kf) using the information provided about the equilibrium constant (K) and the rate constant for the reverse reaction (kb).

### Step-by-Step Solution:

1. **Understand the relationship between K, kf, and kb**:
The equilibrium constant (K) for a reversible reaction is defined as the ratio of the rate constant for the forward reaction (kf) to the rate constant for the reverse reaction (kb). This relationship can be expressed mathematically as:
\[ K = \frac{k_f}{k_b} \]

2. **Substitute the known values**:
From the problem, we know:
- K = 10 (equilibrium constant)
- kb = 2.8 (rate constant for the reverse reaction)

We can substitute these values into the equation:
\[ 10 = \frac{k_f}{2.8} \]

3. **Rearrange the equation to solve for kf**:
To find kf, we can rearrange the equation:
\[ k_f = K \times k_b \]

4. **Calculate kf**:
Now we substitute the values of K and kb into the rearranged equation:
\[ k_f = 10 \times 2.8 \]
\[ k_f = 28 \]

5. **Conclusion**:
The rate constant for the forward reaction (kf) is 28.

### Final Answer:
The rate constant for the forward reaction is **28**.

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