Gas-phase bromine monoxide, BrO, decomposes according to two possible pathways: 2 BrO(g) → O2(g) + 2 — Kinetics Chemistry Question
Problem Context
Gas-phase bromine monoxide, BrO, decomposes according to two possible pathways:
2 BrO(g) → O2(g) + 2 Br(g) (4a)
2 BrO(g) → O2(g) + Br2(g) (4b)
The reaction can be studied by generating BrO(g) by using a laser. If this is carried out in the presence of excess ozone, then any Br(g) that is present reacts with ozone to produce bromine monoxide much more rapidly than either reaction 4a or 4b.
Br(g) + O3(g) → BrO(g) + O2(g) fast
The disappearance of BrO(g) was studied spectrophotometrically at both 267 K and 298 K, either in the absence of excess ozone (open squares) or in the presence of excess ozone (open circles). The reciprocal of the concentration of BrO as a function of time is graphed below for the four experiments.
What is the reaction order in BrO for its decomposition in the absence of O3? Justify your answer based on the experimental data.
Model Answer
Since 1/[BrO] increases linearly with time, the reaction is second-order in BrO.
A sample of BrO in N2 at 298 K with an initial concentration of 4.0 × 1014 molecules cm-3 is prepared. How much time will it take for the BrO concentration to decay to half its original value?
Model Answer
1/[BrO]t = 1/[BrO]0 + kt
1/[2.0 × 1014 molecule cm-3] = 1/[4.0 × 1014 molecule cm-3] + (5.80 × 10-12 cm3 molecule-1)t
t = 4.3 × 10-4 s
Is the observed decomposition of BrO faster or slower in the presence of ozone? Justify your answer based on the experimental data and provide a chemical explanation for the observed difference.
Model Answer
The steeper the slope of the plot, the faster BrO is disappearing. Thus the rates are slower in the presence of ozone than in its absence. The reason is that in the presence of ozone, Br(g) reacts rapidly to form equimolar BrO(g), so reaction 4a does not result in the net disappearance of BrO(g)—it is regenerated from the Br(g) and O3(g). In the presence of ozone, kobs = 2k4b, while in its absence kobs = 2k4a + 2k4b, which must be larger.
[The factor of two is because the reaction consumes two moles of BrO(g); this convention will be used in part (d), but students will not be penalized for answers that omit the factor of 2.]
Calculate the values of k4a and k4b at 298 K.
Model Answer
The slopes of the second-order plots are equal to the kobs for the reaction. From part b, the reaction in the presence of O3 gives
k4b = 0.5(6.24 × 10-13 cm3 molecule-1 s-1) = 3.12 × 10-13 cm3 molecule-1 s-1.
In the absence of O3,
k4a + k4b = 0.5(5.80 × 10-12 cm3 molecule-1 s-1) = 2.90 × 10-12 cm3 molecule-1 s-1
k4a = (2.90 × 10-12 – 3.12 × 10-13) cm3 molecule-1 s-1 = 2.59 × 10-12 cm3 molecule-1 s-1.
Calculate the activation energy for the decomposition of BrO in the presence of ozone.
Model Answer
ln(k2/k1) = (Ea/R)(1/T1 – 1/T2)
ln(6.24 × 10-13/8.64 × 10-13) = (Ea/8.314 J mol-1 K-1)(1/267 K – 1/298 K)
Ea = –6.94 kJ mol-1
The reaction goes slower at higher temperatures, so it has a negative activation energy! This is rare, but sometimes is observed in reactions where there is a lot of bond-forming to make the activated complex. Here, the reaction is hypothesized to involve side-on combination of the two OBr molecules to form an O2Br2 square, which then fragments to give the O2 + Br2 products.