Consider the following data for the decomposition of nitrous oxide (N_2O) to N_2

Question

Consider the following data for the decomposition of nitrous oxide (N_2O) to N_2 and O_2 gas at 333 K: Mechanism 1 N_2O leftharpoonoverrightharpoon N_2 + O fast equilibrium N_2O + O rightarrow N_2 + O_2 slow Mechanism 2 N_2O leftharpoonoverrightharpoon N_4O_ fast equilibrium N_4O_2 rightarrow 2 N_2 + O_2 slow Mechanism 3 2 N_2O leftharpoonoverrightharpoon N_4O_4 fast equilibrium 1/2 N_4O_2 rightarrow N_2 + O_2 slow Mechanism 4 N_2O + N_2O rightarrow 2 N_2 + O + O slowO + O leftharpoonoverrightharpoon O_2 fast equilibrium Mechanism 3 Mechanism 2 Mechanism 1 Mechanism 4

Explanation / Answer

from experiment 2 and 3,
[N2O] doubles, rate becomes 4 times

order of N2O is 2

mechanism 1:
rate = k [N2O][O]

Kc= [N2][O]/[N2O]
[O] = Kc[N2O]/[N2]

rate = kkc [N2O][N2O]/[N2]

not valid

mechanism 2:
rate= k [N4O2]

Kc= [N4O2] / [N2O]^2
[NHO2] = Kc [N2O]^2

rate = kkc*[N2O]^2
this is valid

mechanism 3:
rate= k [N4O2]^1/2

Kc= [N4O2] / [N2O]^2
[NHO2] = Kc [N2O]^2

rate = kkc*[N2O]
this is not valid

mechanism 4:
this is valid

mechanism 2 and mechanism 4 both are valid

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