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Question

The rate law for the decomposition of gaseous N2O5, N2O52NO2+12O2, is observed to be:
d[N2O5]dt=K[N2O5]
A reaction mechanism which has been suggested to be consistent with this is:
N2O5kNO2+NO3 (fast)
NO2+NO3k1−−−−NO2+NO+O2 (slow)
NO+NO3k2−−−−2NO2 (fast)
Show that the mechanism is consistent with the observed law.

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Solution

Given reaction :
N2O5kNO2+NO3 (fast)...(i)
NO2+NO3K1NO2+NO+O2 (slow)...(ii)
NO+NO3K22NO2 (fast)...(iii)
To Prove
d[N2O5]dt=k[N2O5]
Let us consider kf and kb are the rate constant
for forward and backward reaction
Assume k=kfKb
Rate of the slow step = rate of the reaction
R=K1[NO2][NO3]...(iv)
[NO3] is an intermediate Hence we apply the
steady state approximation
d[NO3]dt=kf[N2O5]+Kb[NO2][NO3]k1[NO2][NO3]K2[NO][NO3]=0
kf[N2O5]=kb[NO2][NO3]+k1[NO2][NO3]+K2[NO][NO3]
[NO3]=kf[N2O5]kb[NO2]+k1[NO2]+k2[NO2]
Putting the value in eqn (iv)
R=k1[NO2]kf[N2O5]kb[NO2+k1[NO2]+k2[NO]
Concentration of [NO] and [NO2]k1 can be neglected because
they are generated in slow step
dRdt=k1[NO2]kf[N2O5]kb[NO2]=k1[N2O5]kfkb=k1k[N2O5]=k[N2O5]

1144819_661286_ans_8eadd6108932449f8650d54739c947ec.jpg

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