[13%] The initial rate of decomposition of ozone to molecular oxygen has been examined under a varie — Kinetics Chemistry Question
Problem Context
[13%] The initial rate of decomposition of ozone to molecular oxygen has been examined under a variety of conditions by measuring the change in pressure as the reaction takes place.
2 O3(g) 3 O2(g)
At 90 ºC, in the presence of relatively small amounts of O3 compared to O2 (present in constant amount), the following data were obtained:
If the pressure changes at a rate of 1.21 10–3 (mm Hg) s–1 at 90 ºC, what is the rate of disappearance of O3 in mol L–1 s–1?
Model Answer
With each mol of O3 that reacts of the reaction, 0.5 mol of additional total gas is produced. So if the total pressure is increasing by 1.21 10 -3 (mm Hg) s -1 , then the pressure of O3 is decreasing by twice this, or 2.42 10 -3 (mm Hg) s -1 . To convert to mol L -1 s -1 , one needs to convert mm Hg to mol L -1 at 90 ºC:
PV = nRT
n/V = P/RT
n/V = (2.42 10 -3 mm Hg)/(62.4 L mol -1 K -1 )(363 K)
n/V = 1.07 10 -7 mol L -1
So the rate of disappearance of O3 is 1.07 10 -7 mol L -1 s -1 .
What is the order in O3 under these conditions?
Model Answer
When O3 pressure increases by a factor of 2.24, the rate increases by a factor of 4.79, close to (2.24) 2 = 5.01. So the reaction is second order in O3 under these conditions.
Under slightly different conditions, with the initial pressures of O3 held constant, the initial rates were measured as a function of O2 pressure at 90 ºC and at 100 ºC:
What is the order in O2?
Model Answer
Doubling the O2 pressure results in roughly a factor of two decrease in the rate. Thus the order in O2 is –1.
What is the activation energy for the reaction?
Model Answer
ln(k2/k1) = (Ea/R)(1/T1 – 1/T2)
ln(3.64 10 -3 /1.45 10 -3 ) = (Ea/8.314 J mol -1 K -1 )(1/363 K – 1/373 K)
Ea = 104 kJ mol -1 (using the data from 200 mm Hg O2 gives 90.9 kJ mol -1 )
The following mechanism has been proposed for the reaction:
O3(g) O2(g) + O(g) (rate constants k1 forward, k–1 reverse)
O(g) + O3(g) 2 O2(g) (rate constant k2)
Using the steady-state approximation, derive the rate law predicted by this mechanism. Under what circumstances, if any, is this consistent with the experimental data?
Model Answer
Applying the steady-state approximation to [O] gives:
k1[O3] = k–1[O][O2] + k2[O][O3]
[O] = k1[O3] / (k-1[O2] + k2[O3])
Rate = k2[O3][O] = k1k2[O3]2 / (k-1[O2] + k2[O3])
This is consistent with the experimental data if k–1[O2] >> k2[O3]. This is likely to be fulfilled under these conditions of relatively high O2 pressure and low O3 pressure.