Thermal decomposition of dinitrogen pentoxide (N2O5) in the gas phase has time-independent stoichiom — Physical Chemistry — Kinetics Chemistry Question
Kinetic chemistry
Thermal decomposition of dinitrogen pentoxide (N2O5) in the gas phase has time-independent stoichiometry.
2 N2O5 (g) → 4 NO2 (g) + O2 (g) (1)
A kinetic measurement for N2O5 at 63.3 oC is shown in Figure 1 below.
Figure 1. Concentration of N2O5 versus time.
Time (s) | [N2O5]/mol dm−3
0 | 3.80 ⋅10−3
50 | 3.24 ⋅10−3
100 | 2.63 ⋅10−3
150 | 2.13 ⋅10−3
225 | 1.55 ⋅10−3
350 | 9.20 ⋅10−4
510 | 4.70 ⋅10−4
650 | 2.61 ⋅10−4
800 | 1.39 ⋅10−4
What is the half-life (t1/2) for the decomposition of N2O5 at 63.3 oC?
Model Answer
To determine t1/2, the time taken from the initial concentration of N2O5 (3.80 ⋅ 10-3 mol.dm-3) to fall to one-half of its value:
t1/2 ≈ 180 s corresponding to [N2O5]t1/2 = 1.90 ⋅10-3 mol dm-3
The reaction order for the reaction (1) can be determined by plotting of:
ln [N2O5]0 / [N2O5]t versus time or {[N2O5]0 / [N2O5]t -1} versus time.
Plot the graphs to determine the reaction order?
Write down the rate law and integrated rate equation.
Model Answer
Figure 2. A re-plot of the data in Figure 1 as function of ln {[N2O5]0/[N2O5]t} vs. time
The plot of ln {[N2O5]0 / [N2O5]t} versus time is linear for a first order reaction.
r = k [N2O5]
The form of integrated rate equation:
ln([N2O5]0 / [N2O5]t) = k t or [N2O5]t = [N2O5]0 e-kt
Determine the rate constant for the reaction (1).
Model Answer
The 1st order reaction:
k = ln2 / t1/2 = ln 2 / 180 s = 3.85 ⋅ 10-3 s-1
The rate constant k for (1) at 45 oC is 5.02 ⋅ 10-4 s−1. Calculate the activation energy (Ea) and pre-exponential factor (A) for the reaction (1) assuming that the activation energy and pre-exponential factor are temperature independent.
Model Answer
Ea is independent of temperature:
ln(k336.6K / k318K) = ln(3.85 ⋅ 10-3 / 5.02 ⋅ 10-4) = (Ea / 8.314 J mol-1 K-1) [ (1/318) - (1/336.6) ]
Ea = 97.46 kJ
Pre-exponential factor (A):
k = A ⋅ e-Ea/RT at 336.3 K, A = k ⋅ eEa/RT = 3.85 ⋅ 10-3 ⋅ e97460/(8.314×336.3) = 5.28 ⋅ 1012 s-1.
The following mechanism is proposed for the reaction (1):
N2O5 → NO2 + NO3 (2)
NO2 + NO3 → NO2 + O2 + NO (3)
NO + NO3 → 2 NO2 (4)
Using this mechanism, derive the rate law for -d[N2O5] / dt assuming that the intermediate concentrations can be treated by the steady-state approximation.
Model Answer
The intermediate concentrations can be treated by the steady-state approximation:
d[NO] / dt = k2[NO2][NO3] - k3[NO][NO3] = 0
→ [NO] = (k2 / k3) [NO2] (Eq.1)
Substituting this equation into the below equation:
d[NO3] / dt = k1[N2O5] - k-1[NO2][NO3] - k2[NO2][NO3] - k3[NO][NO3] = 0 (Eq. 2)
→ k1[N2O5] - k-1[NO2][NO3] - k2[NO2][NO3] - k3(k2/k3)[NO2][NO3] = 0
→ k1[N2O5] - k-1[NO2][NO3] - 2k2[NO2][NO3] = 0
→ [NO3] = k1[N2O5] / (k-1[NO2] + 2k2[NO2]) (Eq.3)
The reaction rate:
r = - d[N2O5] / dt = k1[N2O5] - k-1[NO2][NO3]
= k1[N2O5] - k-1[NO2] (k1[N2O5] / (k-1[NO2] + 2k2[NO2]))
= k1[N2O5] (2k2 / (k-1 + 2k2))