Compounds A and B react with a common reagent with first order kinetics in both cases. If 99% of A must react before 1% of B has reacted, what is the minimum ratio for their respective rate constants ?

Options

A. 916
B. 229
C. 500
D. 458

Explanation

To solve the problem, we need to determine the minimum ratio of the rate constants \( k_A \) and \( k_B \) for the reactions of compounds A and B with a common reagent, given that 99% of A must react before 1% of B has reacted.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- We know that both compounds A and B react with a common reagent and that the reactions follow first-order kinetics.

2. **Defining the Reacted and Remaining Fractions**:
- When 99% of A has reacted, the remaining fraction of A is:
\[ [A] = 1 - 0.99 = 0.01 \]
- At this point, only 1% of B has reacted, so the remaining fraction of B is:
\[ [B] = 1 - 0.01 = 0.99 \]

3. **Using the First-Order Kinetics Formula**:
- For a first-order reaction, the rate constant \( k \) can be expressed in terms of the fraction of the reactant that has reacted:
\[ k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]}\right) \]
- For compound A, since 99% has reacted:
\[ k_A = \frac{1}{t_A} \ln\left(\frac{100}{1}\right) = \frac{1}{t_A} \ln(100) \]

4. **Calculating for Compound B**:
- For compound B, since only 1% has reacted:
\[ k_B = \frac{1}{t_B} \ln\left(\frac{100}{99}\right) \]

5. **Finding the Ratio of Rate Constants**:
- We need to find the ratio \( \frac{k_A}{k_B} \):
\[ \frac{k_A}{k_B} = \frac{\frac{1}{t_A} \ln(100)}{\frac{1}{t_B} \ln\left(\frac{100}{99}\right)} = \frac{t_B \ln(100)}{t_A \ln\left(\frac{100}{99}\right)} \]

6. **Assuming Time Relation**:
- Since we want the minimum ratio, we can assume \( t_A = t_B \) for simplicity, leading to:
\[ \frac{k_A}{k_B} = \frac{\ln(100)}{\ln\left(\frac{100}{99}\right)} \]

7. **Calculating the Logarithms**:
- We know:
\[ \ln(100) = 4.605 \]
\[ \ln\left(\frac{100}{99}\right) = \ln(100) - \ln(99) \approx 4.605 - 4.595 = 0.01005 \]

8. **Final Calculation**:
- Now substituting the values:
\[ \frac{k_A}{k_B} \approx \frac{4.605}{0.01005} \approx 458.7 \]

### Conclusion:
The minimum ratio of the rate constants \( \frac{k_A}{k_B} \) is approximately **458**.