Question
The reaction 2N2O5(g) ---> 2N2O4 (g) + O2(g) obeys the rate law, r = k [N2O5] in which the specific rate constant is 8.40 x 10 ^-3/s at a certain temperature. (a) if 2.50 moles of N2O5 were placed in a 5.00L container at that temperature, how many moles of N2O5 would remain after one minute?
(b) how long would it take for 80% of the original N2O5 to react?
(c) Calculate the half-life of the above reaction.
(b) how long would it take for 80% of the original N2O5 to react?
(c) Calculate the half-life of the above reaction.
Answers
GPT-4o mini
To solve this problem, we need to utilize the information given about the reaction and the rate law provided. The reaction that we are dealing with is first-order with respect to \(N_2O_5\).
### Part (a): Moles of \(N_2O_5\) remaining after 1 minute
The first-order rate equation is given by:
\[
r = -\frac{d[N_2O_5]}{dt} = k [N_2O_5]
\]
Integrating the first-order rate law, we have:
\[
\ln\left(\frac{[N_2O_5]_0}{[N_2O_5]} \right) = kt
\]
Where:
- \([N_2O_5]_0\) is the initial concentration,
- \([N_2O_5]\) is the concentration at time \(t\),
- \(k\) is the rate constant,
- \(t\) is time.
1. **Calculate Initial Concentration:**
- We have 2.50 moles of \(N_2O_5\) in a 5.00 L container:
\[
[N_2O_5]_0 = \frac{2.50 \, \text{moles}}{5.00 \, \text{L}} = 0.500 \, \text{M}
\]
2. **Time Conversion:**
- One minute is 60 seconds:
\[
t = 60 \, \text{s}
\]
3. **Plugging Values into the Integrated Rate Law:**
- The rate constant \(k = 8.40 \times 10^{-3} \, \text{s}^{-1}\):
\[
\ln\left(\frac{0.500}{[N_2O_5]}\right) = (8.40 \times 10^{-3} \, \text{s}^{-1})(60 \, \text{s})
\]
Calculate the right side:
\[
\ln\left(\frac{0.500}{[N_2O_5]}\right) = 0.504
\]
Now, exponentiate both sides:
\[
\frac{0.500}{[N_2O_5]} = e^{0.504} \approx 1.653
\]
Rearranging gives:
\[
[N_2O_5] = \frac{0.500}{1.653} \approx 0.303 \, \text{M}
\]
4. **Calculate Moles Remaining:**
- To find the remaining moles:
\[
\text{Moles remaining} = [N_2O_5] \times \text{Volume} = 0.303 \, \text{M} \times 5.00 \, \text{L} = 1.515 \, \text{moles}
\]
After one minute, approximately **1.515 moles of \(N_2O_5\)** would remain.
### Part (b): Time for 80% of \(N_2O_5\) to React
If 80% of the original \(N_2O_5\) reacts, 20% remains. Thus, we want to find the time when:
\[
[N_2O_5] = 0.20 \times [N_2O_5]_0 = 0.20 \times 0.500 \, \text{M} = 0.100 \, \text{M}
\]
Using the integrated rate law:
\[
\ln\left(\frac{0.500}{0.100}\right) = kt
\]
\[
\ln(5) = (8.40 \times 10^{-3} \, \text{s}^{-1}) t
\]
Calculate \( \ln(5) \):
\[
\ln(5) \approx 1.609
\]
Now solve for \(t\):
\[
1.609 = (8.40 \times 10^{-3}) t
\]
\[
t = \frac{1.609}{8.40 \times 10^{-3}} \approx 191.5 \, \text{s}
\]
It will take approximately **191.5 seconds** for 80% of the original \(N_2O_5\) to react.
### Part (c): Calculate the Half-Life of the Reaction
For a first-order reaction, the half-life (\(t_{1/2}\)) is given by the formula:
\[
t_{1/2} = \frac{0.693}{k}
\]
Substituting the rate constant into the half-life formula:
\[
t_{1/2} = \frac{0.693}{8.40 \times 10^{-3}} \approx 82.5 \, \text{s}
\]
The half-life of the reaction is approximately **82.5 seconds**.
### Part (a): Moles of \(N_2O_5\) remaining after 1 minute
The first-order rate equation is given by:
\[
r = -\frac{d[N_2O_5]}{dt} = k [N_2O_5]
\]
Integrating the first-order rate law, we have:
\[
\ln\left(\frac{[N_2O_5]_0}{[N_2O_5]} \right) = kt
\]
Where:
- \([N_2O_5]_0\) is the initial concentration,
- \([N_2O_5]\) is the concentration at time \(t\),
- \(k\) is the rate constant,
- \(t\) is time.
1. **Calculate Initial Concentration:**
- We have 2.50 moles of \(N_2O_5\) in a 5.00 L container:
\[
[N_2O_5]_0 = \frac{2.50 \, \text{moles}}{5.00 \, \text{L}} = 0.500 \, \text{M}
\]
2. **Time Conversion:**
- One minute is 60 seconds:
\[
t = 60 \, \text{s}
\]
3. **Plugging Values into the Integrated Rate Law:**
- The rate constant \(k = 8.40 \times 10^{-3} \, \text{s}^{-1}\):
\[
\ln\left(\frac{0.500}{[N_2O_5]}\right) = (8.40 \times 10^{-3} \, \text{s}^{-1})(60 \, \text{s})
\]
Calculate the right side:
\[
\ln\left(\frac{0.500}{[N_2O_5]}\right) = 0.504
\]
Now, exponentiate both sides:
\[
\frac{0.500}{[N_2O_5]} = e^{0.504} \approx 1.653
\]
Rearranging gives:
\[
[N_2O_5] = \frac{0.500}{1.653} \approx 0.303 \, \text{M}
\]
4. **Calculate Moles Remaining:**
- To find the remaining moles:
\[
\text{Moles remaining} = [N_2O_5] \times \text{Volume} = 0.303 \, \text{M} \times 5.00 \, \text{L} = 1.515 \, \text{moles}
\]
After one minute, approximately **1.515 moles of \(N_2O_5\)** would remain.
### Part (b): Time for 80% of \(N_2O_5\) to React
If 80% of the original \(N_2O_5\) reacts, 20% remains. Thus, we want to find the time when:
\[
[N_2O_5] = 0.20 \times [N_2O_5]_0 = 0.20 \times 0.500 \, \text{M} = 0.100 \, \text{M}
\]
Using the integrated rate law:
\[
\ln\left(\frac{0.500}{0.100}\right) = kt
\]
\[
\ln(5) = (8.40 \times 10^{-3} \, \text{s}^{-1}) t
\]
Calculate \( \ln(5) \):
\[
\ln(5) \approx 1.609
\]
Now solve for \(t\):
\[
1.609 = (8.40 \times 10^{-3}) t
\]
\[
t = \frac{1.609}{8.40 \times 10^{-3}} \approx 191.5 \, \text{s}
\]
It will take approximately **191.5 seconds** for 80% of the original \(N_2O_5\) to react.
### Part (c): Calculate the Half-Life of the Reaction
For a first-order reaction, the half-life (\(t_{1/2}\)) is given by the formula:
\[
t_{1/2} = \frac{0.693}{k}
\]
Substituting the rate constant into the half-life formula:
\[
t_{1/2} = \frac{0.693}{8.40 \times 10^{-3}} \approx 82.5 \, \text{s}
\]
The half-life of the reaction is approximately **82.5 seconds**.