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Answer :
Sure, let's go through the problem step-by-step.
a. Concentration of Nitrous Oxide after 150 minutes:
For a second-order reaction, the half-life is dependent on the initial concentration. We use the half-life formula for a second-order reaction:
[tex]\[ t_{1/2} = \frac{1}{k \cdot [A]_0} \][/tex]
Where [tex]\( t_{1/2} \)[/tex] is the half-life and [tex]\([A]_0\)[/tex] is the initial concentration. Here, we follow the half-life decay pattern to determine how much remains after 150 minutes.
1. The half-life is 75 minutes.
2. Time elapsed is 150 minutes.
Since 150 minutes is double the half-life (150 / 75 = 2), the concentration will halve twice:
- After the first 75 minutes, half of the initial concentration decomposes.
- After the next 75 minutes (totaling 150 minutes), half of what remains decomposes again.
Given these steps, the concentration of nitrous oxide after 150 minutes is calculated to be approximately 0.005 M.
b. Time for 40% Decomposition:
To determine how long it takes for 40% of the sample to decompose, we need 60% of the original concentration remaining, since:
[tex]\[ \text{Remaining concentration} = \text{Initial concentration} \times (1 - \text{Fraction decomposed}) \][/tex]
1. 40% decomposed means 60% remains, so the concentration left is:
[tex]\([A] = 0.60 \times [A]_0\)[/tex]
For a second-order reaction, the integrated rate law is used:
[tex]\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \][/tex]
Rearranging this for time when 60% of the original concentration remains:
1. Solve for [tex]\( t \)[/tex] given the 60% remaining and using the calculated rate constant and initial concentration.
From this calculation, the time it takes for 40% of the sample to decompose is found to be approximately 55.27 minutes.
This detailed explanation breaks down the concepts and calculations used to derive the final values for both parts of the question.
a. Concentration of Nitrous Oxide after 150 minutes:
For a second-order reaction, the half-life is dependent on the initial concentration. We use the half-life formula for a second-order reaction:
[tex]\[ t_{1/2} = \frac{1}{k \cdot [A]_0} \][/tex]
Where [tex]\( t_{1/2} \)[/tex] is the half-life and [tex]\([A]_0\)[/tex] is the initial concentration. Here, we follow the half-life decay pattern to determine how much remains after 150 minutes.
1. The half-life is 75 minutes.
2. Time elapsed is 150 minutes.
Since 150 minutes is double the half-life (150 / 75 = 2), the concentration will halve twice:
- After the first 75 minutes, half of the initial concentration decomposes.
- After the next 75 minutes (totaling 150 minutes), half of what remains decomposes again.
Given these steps, the concentration of nitrous oxide after 150 minutes is calculated to be approximately 0.005 M.
b. Time for 40% Decomposition:
To determine how long it takes for 40% of the sample to decompose, we need 60% of the original concentration remaining, since:
[tex]\[ \text{Remaining concentration} = \text{Initial concentration} \times (1 - \text{Fraction decomposed}) \][/tex]
1. 40% decomposed means 60% remains, so the concentration left is:
[tex]\([A] = 0.60 \times [A]_0\)[/tex]
For a second-order reaction, the integrated rate law is used:
[tex]\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \][/tex]
Rearranging this for time when 60% of the original concentration remains:
1. Solve for [tex]\( t \)[/tex] given the 60% remaining and using the calculated rate constant and initial concentration.
From this calculation, the time it takes for 40% of the sample to decompose is found to be approximately 55.27 minutes.
This detailed explanation breaks down the concepts and calculations used to derive the final values for both parts of the question.
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