Reaction: O3(g) +NO(g) = O2(g)+NO2 has at 310K the rate law

Question

Rate of reaction =k[O3][NO]

k= 3.0x10^6

Given that [O3]= 5.0x10^-4M and [NO]=6.0x10^-5M at t=0

calculate that rate of the reaction at t=0

Answer 1

O₃(g) + NO(g) <==> O₂(g) + NO₂(g)

rate = k[O₃][NO] ... k = 3.0x10⁶ M^-1s^-1 (2nd order reaction, k has units of M^-1s^-1)

rate = 3.0x10⁶ M^-1s^-1 [5.0x10^-4 M][6.0x10^-5 M]

rate = (3.0x10⁶ M^-1s^-1) (3.0x10^-8 M²)

rate = 0.09 Ms^-1