Consider the collision-induced dissociation of N_2O_5(g) via the following mecha
ID: 1070150 • Letter: C
Question
Consider the collision-induced dissociation of N_2O_5(g) via the following mechanism: N_2O_5(g) + N_2O_5(g) N_2O_5(g)* + N_2O_5(g) N_2O_5(g)* NO_2(g) + NO_3(g) The asterisk in the first reaction indicates that the reactant is activated through collision. Experimentally it is found that the reaction can be either first or second order in N_2O_5(g) depending on the concentration of this species. Derive a rate law expression for this reaction consistent with this observation. R = k_2k_1[N_2O_5]^2/k_-1[N_2O_5]+k_1 R = k_2k_1[N_2O_5]^2/k_-1[N_2O_5]+k_2 R = k_2k_1[N_2O_5]/k_-1[N_2O_5]+k_1 R = k_2k_1[N_2O_5]/k_-3[N_2O_5]+k_2Explanation / Answer
By Steady state approximation,
d[N2O5*]/dt = 0 = k1[N2O5]2 - k-1 [N2O5]* [N2O5] - k2[N2O5]*
k1[N2O5]2 - k-1 [N2O5]* [N2O5] - k2[N2O5]* = 0
k1[N2O5]2 = k-1 [N2O5]* [N2O5] + k2[N2O5]*
k1[N2O5]2 = [N2O5]* { k-1 [N2O5] + k2 }
[N2O5]* = k1[N2O5]2 / {k-1 [N2O5] + k2}
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Rate = k2 [N2O5]*
= k2 k1[N2O5]2 / {k-1 [N2O5] + k2}
Therefore,
R = k2 k1[N2O5]2 / {k-1 [N2O5] + k2}