The reaction 5 Cl -(aq) + CIO_3 (aq) + 6 H^+(ag) rightarrow 3 Cl_2(g) + 3 H_20(I

Question

The reaction 5 Cl -(aq) + CIO_3 (aq) + 6 H^+(ag) rightarrow 3 Cl_2(g) + 3 H_20(I) is expected to obey the following mechanism. CIO_3^-(ag) + H+(ag) HCIO_3(ag) Fast equilibrium HC1O3_(aq) + H^+-(aq)Fast equilibrium CI-(a1) H_2CIO_3+(ag) rightarrow (CI-CIO_2)(aq) +H_2O(I) slow (CI - CIO_2) (aq) + 4 ^+ (aq) + 4 CI^- (aq) rightarrow products Fast Write the rate law for this reaction. (Enter your answer using the generic rate constant k. Rate expressions take the general form: rate = k.[H_2]. [Cl_2].)

Explanation / Answer

the rate law is determined by the slowest reaction

so

rate = k3 [Cl-] [H2Cl03+]

but

H2Cl03+ is not in the final reaction

so we need to eliminate it

now

consider the equilibrium

HCl03 + H+ ---> H2Cl03+

k2/k-2 = [H2Cl03+] / [HCl03] [H+]

[H2Cl03+] = k2 [HCl03] [H+] / k-2

now

consider the equilibrium

Cl03- + H+ ---> HCl03

k1/k-1 = [HCl03] / [H+] [Cl03-]

[HCl03] = k1 [H+] [Cl03-] / k-1

so

[H2Cl03+] = k2 x (k1 [H+] [Cl03-] / k-1) [H+] / k-2

[H2Cl03+] = ( k2 x k1 / k-1 x k-2) [H+]^2 [Cl03-]

now

rate = k3 [Cl-] [H2Cl03+]

rate = k3 [Cl-] ( k2 x k1 / k-1 x k-2) [H+]^2 [Cl03-]

rate = ( k1 x k2 x k3 / k-1 x k-2) [H+]^2 [Cl-] [Cl03-]

so

the rate law is

rate = k [H+]^2 [Cl-] [Cl03-]

where

k = ( k1 x k2 x k3 / k-1 x k-2)

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