The rate law for the Br^- - catalyzed reaction H^+ + HNO_2 + C_6H_5NH_2 C_6H_5N_

Question

The rate law for the Br^- - catalyzed reaction H^+ + HNO_2 + C_6H_5NH_2 C_6H_5N_2^+ + H_2O is observed to be R = k[H^+] [HNO_2] [Br^-]. A proposed mechanism is H^+ + HNO_2 H_2NO_2^+ (fast equilibrium) H_2NO_2^+ + Br' k_2 ONBr + H_2O (slow) ONBr + C_6H_5NH_2 rightarrow C_6H_5N_2^+ + H_2 O + Br^- (fast) Apply steady-state approximation to the intermediates H_2NO_2^+ and ONBr and determine the rate law and compare with that observed experimentally Under what conditions would agreement with experiment be found?

Explanation / Answer

From Eq.3 of proposed mechanism

Rate , R = K3[ONBr] [C6H5NH2]

RONBr = K2 [H3NO2+] [Br-] is the slowest step and rate determining step

Rate = K3K2[H2NO2+] [C6H5NH2] [Br-]

But K1[H+] [HNO2] = K-1[H2NO2+]

[H2NO2+] = K1[H+] [HNO2]/K-1

Rate = K3K2[H+] [HNO2] [Br-] [C6H5NH2]/K-1 =

Rate= K’ [C6H5NH2] [H+] [HNO2] [Br-] where K’ = K3K2/K-1 (1)

Rate = K[ H+] [HNO2] [Br-]

When excess of C6H5NH2 is taken, its concentration can be assumed to remain the same through out the reaction. Then Eq.1 becomes

Rate= K [H+] [HNO2] [Br-] where K = K’[C6H5NH2]

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