For the reversible reaction,`A+BhArrC`, the specific reaction rates for forward and reverse reactions are `1.25xx10^(3) "and" 2.75xx10^(4)` respectively. The equilibrium constant for the reaction is:

Options

A. `0.0454`
B. `0.022`
C. `2.20`
D. `0.4545`

Answer

To find the equilibrium constant (Kc) for the reversible reaction \( A + B \rightleftharpoons C \), we can use the relationship between the rate constants of the forward and reverse reactions.

### Step-by-step Solution:

1. **Identify the given data:**
- Rate constant for the forward reaction (\( k_f \)) = \( 1.25 \times 10^3 \)
- Rate constant for the reverse reaction (\( k_r \)) = \( 2.75 \times 10^4 \)

2. **Write the formula for the equilibrium constant:**
The equilibrium constant (\( K_c \)) is defined as the ratio of the rate constant of the forward reaction to the rate constant of the reverse reaction:
\[ K_c = \frac{k_f}{k_r} \]

3. **Substitute the values into the formula:**
\[ K_c = \frac{1.25 \times 10^3}{2.75 \times 10^4} \]

4. **Perform the division:**
To calculate this, we can first divide the coefficients and then handle the powers of ten:
\[ K_c = \frac{1.25}{2.75} \times \frac{10^3}{10^4} = \frac{1.25}{2.75} \times 10^{-1} \]

5. **Calculate the numerical value:**
\[ \frac{1.25}{2.75} \approx 0.4545 \]
Now, multiply by \( 10^{-1} \):
\[ K_c \approx 0.4545 \times 0.1 = 0.04545 \]

6. **Final result:**
Rounding off, we get:
\[ K_c \approx 0.0454 \]

### Conclusion:
The equilibrium constant \( K_c \) for the reaction is approximately \( 0.0454 \).