To determine the half-life of a reaction given that after 79.0 minutes, 39.0% of a compound has decomposed, and assuming first-order kinetics, we will follow these steps:
### Step-by-Step Solution
1. Understand the Decomposed Fraction:
- The problem states that 39.0% of the compound has decomposed after 79.0 minutes.
- Therefore, the remaining fraction of the compound is \(\( 1 - 0.39 = 0.61 \)\).
2. First-Order Kinetics Equations:
- For a first-order reaction, the relationship between the initial quantity \(\( N_0 \)\), the quantity after time \(\( t \)\) \(\( (N) \)\), and the rate constant \(\( k \)\) is given by:
\(\[
\ln\left(\frac{N_0}{N}\right) = kt
\]\)
- Considering \(\( N_0 \)\) as the initial amount and \(\( N \)\) as the amount remaining after time \(\( t \)\), we have:
\(\[
\ln\left(\frac{N_0}{N}\right) = \ln\left(\frac{1}{0.61}\right) = kt
\]\)
3. Calculate the Rate Constant \(\( k \)\):
- Rearrange the equation to solve for \(\( k \)\):
\(\[
k = \frac{\ln\left(\frac{1}{0.61}\right)}{t}
\]\)
- Substituting \(\( t = 79.0 \)\) minutes:
\(\[
k = \frac{\ln\left(\frac{1}{0.61}\right)}{79.0}
\]\)
- Simplifying, we find \(\( k \)\):
\(\[
k \approx 0.006256915466009875 \, \text{min}^{-1}
\]\)
4. Determine the Half-Life:
- The half-life (\(\( t_{1/2} \)\)) for a first-order reaction is related to the rate constant \(\( k \)\) by the formula:
\(\[
t_{1/2} = \frac{0.693}{k}
\]\)
- Substituting the rate constant \(\( k \)\) we calculated:
\(\[
t_{1/2} = \frac{0.693}{0.006256915466009875}
\]\)
- Simplifying, we find the half-life:
\(\[
t_{1/2} \approx 110.75744969940213 \, \text{minutes}
\]\)
### Conclusion
- The rate constant \(\( k \)\) for the decomposition is approximately \(\( 0.006256915466009875 \, \text{min}^{-1} \)\).
- The half-life of the reaction is approximately \(\( 110.76 \)\) minutes.
