The initial concentration of N2O5(g) in the first-order reaction
2N2O5(g)?4NO2(g)+O2(g)
is 0.372 mol/L. After 30.0 seconds, the concentration of N2O5(g) is 0.0744 mol/L.
What is the rate constant k for the reaction?
Your answer should have 3 significant figures (report your answer in scientific notation).
Answer in units of 1/s
The correct answer and explanation is:
To determine the rate constant kk for the reaction, we use the integrated rate law for a first-order reaction: ln([N2O5]0[N2O5])=kt\ln \left( \frac{[N_2O_5]_0}{[N_2O_5]} \right) = k t
Where:
- [N2O5]0[N_2O_5]_0 is the initial concentration of N2O5N_2O_5,
- [N2O5][N_2O_5] is the concentration of N2O5N_2O_5 at time tt,
- kk is the rate constant,
- tt is the time elapsed.
Step 1: Substitute the given values
[N2O5]0=0.372 mol/L, [N2O5]=0.0744 mol/L, t=30.0 s.[N_2O_5]_0 = 0.372 \, \text{mol/L}, \, [N_2O_5] = 0.0744 \, \text{mol/L}, \, t = 30.0 \, \text{s}. ln([N2O5]0[N2O5])=ln(0.3720.0744)\ln \left( \frac{[N_2O_5]_0}{[N_2O_5]} \right) = \ln \left( \frac{0.372}{0.0744} \right) ln(0.3720.0744)=ln(5.0)≈1.609\ln \left( \frac{0.372}{0.0744} \right) = \ln (5.0) \approx 1.609
Step 2: Solve for kk
Rearranging the equation to isolate kk: k=ln([N2O5]0[N2O5])tk = \frac{\ln \left( \frac{[N_2O_5]_0}{[N_2O_5]} \right)}{t}
Substituting the values: k=1.60930.0≈0.0536 s−1k = \frac{1.609}{30.0} \approx 0.0536 \, \text{s}^{-1}
Final Answer:
k=5.36×10−2 s−1k = 5.36 \times 10^{-2} \, \text{s}^{-1}
Explanation:
The reaction is first-order with respect to N2O5N_2O_5, meaning the rate of decomposition depends only on the concentration of N2O5N_2O_5. The integrated rate law for first-order reactions relates the natural logarithm of the ratio of initial to current concentration to the time and rate constant.
By calculating the ratio of initial to final concentrations and taking the natural logarithm, we determined the total decay over the 30.0-second interval. Dividing this by the time gives the rate constant, kk, which describes how quickly the reaction proceeds. The result is reported in scientific notation with three significant figures as 5.36×10−2 s−15.36 \times 10^{-2} \, \text{s}^{-1}, which aligns with the precision of the given data.