The half life of a first order reaction is 100 seconds. What is the time required for 90% completion of the reaction?

Options

A. 100 sec
B. 200 sec
C. 333 sec
D. 500 sec

Answer

To solve the problem, we need to find the time required for 90% completion of a first-order reaction given that its half-life is 100 seconds.

### Step-by-step Solution:

1. **Understand the Half-life of a First-Order Reaction**:
The half-life (\(t_{1/2}\)) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \(k\) is the rate constant.

2. **Calculate the Rate Constant \(k\)**:
Given that the half-life is 100 seconds, we can rearrange the formula to find \(k\):
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{100 \, \text{s}} = 0.00693 \, \text{s}^{-1} \]

3. **Determine the Time for 90% Completion**:
For a first-order reaction, the relationship between the initial concentration (\(A_0\)), the concentration at time \(t\) (\(A\)), and the rate constant \(k\) is given by:
\[ \ln\left(\frac{A_0}{A}\right) = kt \]
In this case, we want to find the time when 90% of the reaction is complete. This means that 10% of the reactant remains. If we assume \(A_0 = 100\), then \(A = 10\).

4. **Substituting Values into the Equation**:
Now, substituting the values into the equation:
\[ \ln\left(\frac{100}{10}\right) = kt \]
This simplifies to:
\[ \ln(10) = kt \]
We know that \(\ln(10) \approx 2.303\).

5. **Rearranging to Solve for Time \(t\)**:
Now we can substitute \(k\) into the equation:
\[ 2.303 = (0.00693)t \]
Rearranging gives:
\[ t = \frac{2.303}{0.00693} \]

6. **Calculating the Time**:
Performing the calculation:
\[ t \approx 333.2 \, \text{s} \]

### Conclusion:
The time required for 90% completion of the reaction is approximately **333 seconds**.