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Question

# The rate law for the decomposition of gaseous N2O5, N2O5→2NO2+12O2, is observed to be:−d[N2O5]dt=K[N2O5]A reaction mechanism which has been suggested to be consistent with this is:N2O5k⇌NO2+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 :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=kfKbRate 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]=0kf[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]

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