In a chemical equilibrium, the rate constant for the backward reaction is `7.5xx10^(-4)` and the equilibrium constant is `1.5` the rate constant for the forward reaction is:

Options

A. `2 xx 10^(-3)`
B. `15 xx 10^(-4)`
C. `1.125 xx10^(-3)`
D. `9.0 xx10^(-4)`

Answer

To find the rate constant for the forward reaction in a chemical equilibrium, we can use the relationship between the equilibrium constant (K), the rate constant for the forward reaction (k_f), and the rate constant for the backward reaction (k_b). The relationship is given by:

\[ K = \frac{k_f}{k_b} \]

### Step-by-Step Solution:

1. **Identify Given Values**:
- Rate constant for the backward reaction (k_b) = \( 7.5 \times 10^{-4} \)
- Equilibrium constant (K) = \( 1.5 \)

2. **Write the Equation for Equilibrium Constant**:
- From the definition of the equilibrium constant, we have:
\[ K = \frac{k_f}{k_b} \]

3. **Rearrange the Equation to Solve for k_f**:
- We can rearrange the equation to find the rate constant for the forward reaction (k_f):
\[ k_f = K \times k_b \]

4. **Substitute the Known Values**:
- Substitute the values of K and k_b into the equation:
\[ k_f = 1.5 \times (7.5 \times 10^{-4}) \]

5. **Perform the Calculation**:
- Calculate \( k_f \):
\[ k_f = 1.5 \times 7.5 = 11.25 \]
- Now, considering the powers of ten:
\[ k_f = 11.25 \times 10^{-4} = 1.125 \times 10^{-3} \]

6. **Final Result**:
- The rate constant for the forward reaction (k_f) is:
\[ k_f = 1.125 \times 10^{-3} \]

### Conclusion:
The rate constant for the forward reaction is \( 1.125 \times 10^{-3} \).

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