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Sagot :
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 [tex]\( 1 - 0.39 = 0.61 \)[/tex].
2. First-Order Kinetics Equations:
- For a first-order reaction, the relationship between the initial quantity [tex]\( N_0 \)[/tex], the quantity after time [tex]\( t \)[/tex] [tex]\( (N) \)[/tex], and the rate constant [tex]\( k \)[/tex] is given by:
[tex]\[ \ln\left(\frac{N_0}{N}\right) = kt \][/tex]
- Considering [tex]\( N_0 \)[/tex] as the initial amount and [tex]\( N \)[/tex] as the amount remaining after time [tex]\( t \)[/tex], we have:
[tex]\[ \ln\left(\frac{N_0}{N}\right) = \ln\left(\frac{1}{0.61}\right) = kt \][/tex]
3. Calculate the Rate Constant [tex]\( k \)[/tex]:
- Rearrange the equation to solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\ln\left(\frac{1}{0.61}\right)}{t} \][/tex]
- Substituting [tex]\( t = 79.0 \)[/tex] minutes:
[tex]\[ k = \frac{\ln\left(\frac{1}{0.61}\right)}{79.0} \][/tex]
- Simplifying, we find [tex]\( k \)[/tex]:
[tex]\[ k \approx 0.006256915466009875 \, \text{min}^{-1} \][/tex]
4. Determine the Half-Life:
- The half-life ([tex]\( t_{1/2} \)[/tex]) for a first-order reaction is related to the rate constant [tex]\( k \)[/tex] by the formula:
[tex]\[ t_{1/2} = \frac{0.693}{k} \][/tex]
- Substituting the rate constant [tex]\( k \)[/tex] we calculated:
[tex]\[ t_{1/2} = \frac{0.693}{0.006256915466009875} \][/tex]
- Simplifying, we find the half-life:
[tex]\[ t_{1/2} \approx 110.75744969940213 \, \text{minutes} \][/tex]
### Conclusion
- The rate constant [tex]\( k \)[/tex] for the decomposition is approximately [tex]\( 0.006256915466009875 \, \text{min}^{-1} \)[/tex].
- The half-life of the reaction is approximately [tex]\( 110.76 \)[/tex] minutes.
### 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 [tex]\( 1 - 0.39 = 0.61 \)[/tex].
2. First-Order Kinetics Equations:
- For a first-order reaction, the relationship between the initial quantity [tex]\( N_0 \)[/tex], the quantity after time [tex]\( t \)[/tex] [tex]\( (N) \)[/tex], and the rate constant [tex]\( k \)[/tex] is given by:
[tex]\[ \ln\left(\frac{N_0}{N}\right) = kt \][/tex]
- Considering [tex]\( N_0 \)[/tex] as the initial amount and [tex]\( N \)[/tex] as the amount remaining after time [tex]\( t \)[/tex], we have:
[tex]\[ \ln\left(\frac{N_0}{N}\right) = \ln\left(\frac{1}{0.61}\right) = kt \][/tex]
3. Calculate the Rate Constant [tex]\( k \)[/tex]:
- Rearrange the equation to solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\ln\left(\frac{1}{0.61}\right)}{t} \][/tex]
- Substituting [tex]\( t = 79.0 \)[/tex] minutes:
[tex]\[ k = \frac{\ln\left(\frac{1}{0.61}\right)}{79.0} \][/tex]
- Simplifying, we find [tex]\( k \)[/tex]:
[tex]\[ k \approx 0.006256915466009875 \, \text{min}^{-1} \][/tex]
4. Determine the Half-Life:
- The half-life ([tex]\( t_{1/2} \)[/tex]) for a first-order reaction is related to the rate constant [tex]\( k \)[/tex] by the formula:
[tex]\[ t_{1/2} = \frac{0.693}{k} \][/tex]
- Substituting the rate constant [tex]\( k \)[/tex] we calculated:
[tex]\[ t_{1/2} = \frac{0.693}{0.006256915466009875} \][/tex]
- Simplifying, we find the half-life:
[tex]\[ t_{1/2} \approx 110.75744969940213 \, \text{minutes} \][/tex]
### Conclusion
- The rate constant [tex]\( k \)[/tex] for the decomposition is approximately [tex]\( 0.006256915466009875 \, \text{min}^{-1} \)[/tex].
- The half-life of the reaction is approximately [tex]\( 110.76 \)[/tex] minutes.
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