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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry, 4th edition.

Problem 13.67.

Show that the proposed mechanism is consistent with the rate law for the following reaction in aqueous solution,

Hg22+(aq) + Tl3+(aq) 2Hg2+(aq) + Tl+(aq)

for which the observed rate law is

Proposed Mechanism:

Fast:

Hg22+(aq) Hg2+(aq) + Hg(s)

Slow:

Hg(s) + Tl3+(aq) Hg2+(aq) + Tl+(aq)






Answer:

First, check to see that the proposed mechanism matches the experimental stoichiometry.

Add the two reactions together to give:

Hg22+(aq) + Hg(s) + Tl3+(aq) Hg2+(aq) + Hg(s) + Hg2+(aq) + Tl+(aq)

The Hg(s) common to both sides of the equation cancel out and the two moles of Hg2+(aq) can be collected to give the observed reaction, so the stoichiometry matches.


The proposed mechanism can be rewritten as

Hg22+(aq) Hg2+(aq) + Hg(s)

Rate = k1[Hg22+] (fast)

Hg2+(aq) + Hg(s) Hg22+(aq)

Rate = k–1[Hg2+][Hg] (fast)

Hg(s) + Tl3+(aq) Hg2+(aq) + Tl+(aq)

Rate = k2[Hg2+][Hg] (slow)

The overall rate is determined by the slow step, but this has an intermediate ([Hg}) that must be eliminated in order to evaluate the mechanism. This can be done using the two fast steps, which must have approximately the same rate, or:

k1[Hg22+] = k–1[Hg2+][Hg]

so

[Hg] = k1[Hg22+]/k–1[Hg2+]

Using this expression in the rate law for the slow step gives

Rate = k2[Hg2+] k1[Hg22+]/ k–1[Hg2+]

or

Rate = (k2k1/k–1)[Hg2+ [Hg22+]/[Hg2+]

which exactly matches the observed rate law when k = k2k1/k–1