Determine the rate law for this reaction.

Question

The rate of the reaction

NO2​(g) + CO(g)NO(g) + CO2​(g)

was determined in three experiments at 225°C. The results are given in the following table:

Experiment	[NO2]0 (M)	[CO]0 (M)	Initial Rate -ΔΔ[NO2]/ΔΔt (M/s)
1	0.263	0.826	1.44x10-5
2	0.263	0.413	1.44x10-5
3	0.526	0.413	5.76x10-5

a) Determine the rate law for this reaction.

b) Calculate the rate of appearance of CO₂ when [NO₂] = [CO] = 0.518 M.

Answer 1

Compare experiment 2 to 1. You see [CO] doubles while [NO2] remains constant. The rate does NOT change, telling us the reaction is zero order in CO, ie independent of CO.

Next, compare experiments 2 and 3. We see the [NO2] doubles while [CO] remains constant. The rate increases by 4 times telling us the reaction is 2nd order in NO2.

We can now write the rate law for the reaction:

Rate = k[NO2]²

(b). Determine k:

1.44x10^-5 M/s = k (0.263 M)²

k = 2.08x10^-4 M^-1s^-1

Rate of appearance of CO2 when NO2 = CO = 0.528 M is…

rate = 2.08x10^-4 M^-1s^-1(0.528)²

rate = 5.80x10^-5 M/s