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. `2xx10^(-3)`
B. `5xx10^(-4)`
C. `1.12xx10^(-3)`
D. `9.0xx10^(-4)`

Explanation

To find the rate constant for the forward reaction, we can use the relationship between the equilibrium constant (K) and the rate constants for the forward (Rf) and backward (Rb) reactions. The equilibrium constant is defined as:

\[ K = \frac{R_f}{R_b} \]

Where:
- \( K \) is the equilibrium constant,
- \( R_f \) is the rate constant for the forward reaction,
- \( R_b \) is the rate constant for the backward reaction.

Given:
- \( R_b = 7.5 \times 10^{-4} \)
- \( K = 1.5 \)

We need to find \( R_f \). Rearranging the equation gives us:

\[ R_f = K \times R_b \]

Now, substituting the values we have:

\[ R_f = 1.5 \times (7.5 \times 10^{-4}) \]

Calculating this:

1. First, multiply the numbers:
\[ 1.5 \times 7.5 = 11.25 \]

2. Now, we keep the power of ten:
\[ R_f = 11.25 \times 10^{-4} \]

3. To express this in standard scientific notation, we can convert it:
\[ R_f = 1.125 \times 10^{-3} \]

Thus, the rate constant for the forward reaction is:

\[ R_f = 1.125 \times 10^{-3} \]

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