For the reaction: 2 N2O5 (g) → 4 NO2 (g) + O2 (g)

Question

The experimental data for this reaction is :

t = 0.0 s [N2O5] = 0.80 M [NO2] = 0.0 M [O2] = 0.0 M

t = 20.0 s [N2O5] = 0.60 M [NO2] = 0.40 M [O2] = 0.10 M

How much N2O5 was consumed during the 20.0 s time interval, or in other words, what was the change in concentration of N2O5?

Answer 1

delta N2O5 = 0.80M - 0.60M = 0.20 M = the amount of N2O5 consumed
So the AVERAGE change in that 20 s is 0.20/20 = -0.01 M (negative to show decrease).
The rate of the reaction is -1/2(-0.20/20) = + 0.005 M/s.
Notice that the RATE of the reaction is the same 0.005 if we use the NO2 data. O2 data gives 0.005 M/s also.
rate = 1/4[delta (NO2)/delta t] = 1/4(0.40 - 0)/(20 s)] = 1/4(0.4/20) = 1/4(0.02) = 0.005 M/s