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?

Question

(b) how long would it take for 80% of the original N2O5 to react?
(c) Calculate the half-life of the above reaction.

Answer 1

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.