A first order reaction is found to have a rate constant `k = 11 × 10^(–14) s–1`. Find the half life of the reaction.

Answer

To find the half-life of a first-order reaction given the rate constant \( k = 11 \times 10^{-14} \, \text{s}^{-1} \), we can use the formula for the half-life of a first-order reaction:

\[ t_{1/2} = \frac{0.693}{k} \]

### Step-by-step Solution:

1. **Identify the formula for half-life**:
The formula for the half-life of a first-order reaction is:
\[ t_{1/2} = \frac{0.693}{k} \]

2. **Substitute the value of k**:
We know that \( k = 11 \times 10^{-14} \, \text{s}^{-1} \). Now substitute this value into the formula:
\[ t_{1/2} = \frac{0.693}{11 \times 10^{-14}} \]

3. **Calculate the denominator**:
First, calculate \( 11 \times 10^{-14} \):
\[ 11 \times 10^{-14} = 1.1 \times 10^{-13} \]

4. **Perform the division**:
Now, divide \( 0.693 \) by \( 1.1 \times 10^{-13} \):
\[ t_{1/2} = \frac{0.693}{1.1 \times 10^{-13}} \approx 0.630 \times 10^{13} \, \text{s} \]

5. **Express in scientific notation**:
To express \( 0.630 \times 10^{13} \) in standard scientific notation:
\[ t_{1/2} \approx 6.30 \times 10^{12} \, \text{s} \]

### Final Answer:
The half-life of the reaction is approximately \( 6.30 \times 10^{12} \, \text{s} \).

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