A first order reaction takes 100 min for completion of 60 % of reaction ,The time eequired for completion of 90% of the reaction is

Options

A. `150 min `
B. `200`min
C. `220.9`min
D. `246.6` min

Explanation

To solve the problem of determining the time required for the completion of 90% of a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction Order**:
The reaction is first-order, which means the rate of reaction depends linearly on the concentration of one reactant.

2. **Use the First-Order Kinetics Equation**:
The equation for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \]
where:
- \( t \) = time taken for the reaction
- \( k \) = rate constant
- \( [A_0] \) = initial concentration of reactant
- \( [A] \) = concentration of reactant at time \( t \)

3. **Determine the Concentration Values**:
- Given that 60% of the reaction is complete in 100 minutes, this means 40% of the reactant remains.
- Thus, if we assume the initial concentration \( [A_0] = 100 \) (arbitrary units), then at 60% completion, \( [A] = 40 \).

4. **Calculate the Rate Constant \( k \)**:
Using the time for 60% completion:
\[ 100 = \frac{2.303}{k} \log \left( \frac{100}{40} \right) \]
Calculate \( \log \left( \frac{100}{40} \right) = \log(2.5) \approx 0.3979 \).
Substitute this value into the equation:
\[ 100 = \frac{2.303}{k} \times 0.3979 \]
Rearranging gives:
\[ k = \frac{2.303 \times 0.3979}{100} \]

5. **Calculate Time for 90% Completion**:
For 90% completion, 10% of the reactant remains. Thus, \( [A] = 10 \).
Using the same first-order kinetics equation:
\[ t_{90} = \frac{2.303}{k} \log \left( \frac{100}{10} \right) \]
Calculate \( \log \left( \frac{100}{10} \right) = \log(10) = 1 \).
Substitute \( k \) from the previous calculation:
\[ t_{90} = \frac{2.303}{k} \times 1 \]

6. **Final Calculation**:
Substitute the value of \( k \) into the equation for \( t_{90} \):
\[ t_{90} = 2.303 \times \frac{100}{2.303 \times 0.3979} \]
Simplifying this gives:
\[ t_{90} \approx 246.6 \text{ minutes} \]

### Conclusion:
The time required for the completion of 90% of the reaction is approximately **246.6 minutes**.

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