Given reaction :
N2O5k⇌NO2+NO3 (fast)...(i)
NO2+NO3K1→NO2+NO+O2 (slow)...(ii)
NO+NO3K2→2NO2 (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]