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Answer :
To solve this problem, we need to determine the half-life of a first-order reaction given some data. Here's how you can approach it step by step:
1. Understand the problem: We are given that 22% of substance A remains after 38.2 minutes. This is a first-order reaction, which means the rate at which A converts to B is proportional to its concentration.
2. Recall the first-order reaction formula:
For a first-order reaction, the relationship between the concentration of A at any time and its initial concentration can be expressed using the formula:
[tex]\[
\ln\left(\frac{N}{N_0}\right) = -kt
\][/tex]
where:
- [tex]\( N_0 \)[/tex] is the initial concentration of A.
- [tex]\( N \)[/tex] is the remaining concentration of A after time [tex]\( t \)[/tex].
- [tex]\( k \)[/tex] is the rate constant.
- [tex]\( t \)[/tex] is the time elapsed.
3. Use the given information:
- Initially, the concentration [tex]\( N_0 \)[/tex] is 100% (or simply 1 if considering it as a fraction).
- After 38.2 minutes, [tex]\( N \)[/tex] is 22% of the initial concentration, or 0.22 when expressed as a fraction.
- Thus, we have:
[tex]\[
\ln\left(\frac{0.22}{1}\right) = -k \times 38.2
\][/tex]
4. Solve for the rate constant [tex]\( k \)[/tex]:
- Calculate [tex]\(\ln(0.22)\)[/tex].
- Rearrange the formula to solve for [tex]\( k \)[/tex]:
[tex]\[
k = -\frac{\ln(0.22)}{38.2}
\][/tex]
5. Calculate the half-life:
- For a first-order reaction, the half-life ([tex]\( t_{1/2} \)[/tex]) is related to the rate constant by:
[tex]\[
t_{1/2} = \frac{\ln(2)}{k}
\][/tex]
- Using [tex]\( \ln(2) \approx 0.693 \)[/tex], substitute the value of [tex]\( k \)[/tex] into the equation to find [tex]\( t_{1/2} \)[/tex].
6. Result:
- After performing the calculations, the half-life of the reaction is approximately [tex]\( 17.49 \)[/tex] minutes.
Therefore, the half-life of the reaction is [tex]\( t_{1/2} = 17.49 \)[/tex] minutes.
1. Understand the problem: We are given that 22% of substance A remains after 38.2 minutes. This is a first-order reaction, which means the rate at which A converts to B is proportional to its concentration.
2. Recall the first-order reaction formula:
For a first-order reaction, the relationship between the concentration of A at any time and its initial concentration can be expressed using the formula:
[tex]\[
\ln\left(\frac{N}{N_0}\right) = -kt
\][/tex]
where:
- [tex]\( N_0 \)[/tex] is the initial concentration of A.
- [tex]\( N \)[/tex] is the remaining concentration of A after time [tex]\( t \)[/tex].
- [tex]\( k \)[/tex] is the rate constant.
- [tex]\( t \)[/tex] is the time elapsed.
3. Use the given information:
- Initially, the concentration [tex]\( N_0 \)[/tex] is 100% (or simply 1 if considering it as a fraction).
- After 38.2 minutes, [tex]\( N \)[/tex] is 22% of the initial concentration, or 0.22 when expressed as a fraction.
- Thus, we have:
[tex]\[
\ln\left(\frac{0.22}{1}\right) = -k \times 38.2
\][/tex]
4. Solve for the rate constant [tex]\( k \)[/tex]:
- Calculate [tex]\(\ln(0.22)\)[/tex].
- Rearrange the formula to solve for [tex]\( k \)[/tex]:
[tex]\[
k = -\frac{\ln(0.22)}{38.2}
\][/tex]
5. Calculate the half-life:
- For a first-order reaction, the half-life ([tex]\( t_{1/2} \)[/tex]) is related to the rate constant by:
[tex]\[
t_{1/2} = \frac{\ln(2)}{k}
\][/tex]
- Using [tex]\( \ln(2) \approx 0.693 \)[/tex], substitute the value of [tex]\( k \)[/tex] into the equation to find [tex]\( t_{1/2} \)[/tex].
6. Result:
- After performing the calculations, the half-life of the reaction is approximately [tex]\( 17.49 \)[/tex] minutes.
Therefore, the half-life of the reaction is [tex]\( t_{1/2} = 17.49 \)[/tex] minutes.
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Rewritten by : Jeany
