For the reaction 2N2O5 -> 4NO2 + O2, if the rate of disappearance of N2O5 is 0.020 M/s, what is the rate of appearance of NO2?
Q1. For the reaction 2N2O5 -> 4NO2 + O2, if the rate of disappearance of N2O5 is 0.020 M/s, what is the rate of appearance of NO2?
Answer: 0.040 M/s
Explanation: Rate = -1/2(d[N2O5]/dt) = 1/4(d[NO2]/dt). Thus, d[NO2]/dt = 2 * 0.020 = 0.040 M/s.