The rate constans of forward and backward reactions are `8.5 xx 10^(-5)` and `2.38 xx 10^(-4)` respectively. The equilibrium constant if

Options

A. 0.35
B. 0.42
C. 12.92
D. 0.292

Explanation

To find the equilibrium constant (K) given the rate constants of the forward and backward reactions, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Rate constant for the forward reaction (k_f) = \( 8.5 \times 10^{-5} \)
- Rate constant for the backward reaction (k_b) = \( 2.38 \times 10^{-4} \)

2. **Write the formula for the equilibrium constant**:
\[ K = \frac{k_f}{k_b} \]

3. **Substitute the values into the formula**:
\[ K = \frac{8.5 \times 10^{-5}}{2.38 \times 10^{-4}} \]

4. **Simplify the expression**:
- First, divide the coefficients:
\[ \frac{8.5}{2.38} \approx 3.57 \]
- Next, handle the powers of ten:
\[ 10^{-5} \div 10^{-4} = 10^{-5 + 4} = 10^{-1} \]

5. **Combine the results**:
\[ K = 3.57 \times 10^{-1} = 0.357 \]

6. **Determine significant figures**:
- The values \( 8.5 \) and \( 2.38 \) have 2 and 3 significant figures respectively. The result should be reported with the least number of significant figures, which is 2.
- Therefore, rounding \( 0.357 \) to 2 significant figures gives us \( 0.36 \).

### Final Answer:
The equilibrium constant \( K \) is \( 0.36 \).