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Question

1. Given the following experimental data, find the rate law and the rate constant for the reaction

NO (g) + NO2 (g) + O2 (g)  N2O5 (g)

Run [NO]o, M [NO2]o, M [O2]o, M Initial Rate, Ms-1

1 0.10 M 0.10 M 0.10 M 2.1 x 10-2

2 0.20 M 0.10 M 0.10 M 4.2 x 10-2

3 0.20 M 0.30 M 0.20 M 1.26 x 10-1

4 0.10 M 0.10 M 0.20 M 2.1 x 10-2

Answer 1

Run 1 vs. 2: [NO] doubles, [NO2] is constant, [O2] is constant. Rate doubles, means 1st order in NO.

Run 1 vs. 4: [O2] doubles, [NO] is constant], [NO2] is constant. Rate stays same, means zero order in O2

Run 2 vs 3: [NO2] triples, [NO] is constant, [O2] doubles. Rate increases 3x, means 1st order in NO2 (the reason for this is because it's zero order in O2, so O2 doubling doesn't play any role in the calculation).

Rate law: Rate = k[NO][NO2]

2.1x10^-2 M/s = k[0.10 M][0.10 M] = 0.01 M²*k

k = 2.1 M^-1s^-1