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

Problem 13.85.

Hydroxide ion is involved in the mechanism but not consumed in this reaction in aqueous solution.
 OCl–(aq) + I–(aq) OI–(aq) + Cl–(aq)

(a) From the data in the table, determine the order of reaction with respect to OCl, I, and OH, and the overall order.
 [OCl–], M [I–], M [OH–], M Rate of formation of OI–, mol L–1 s–1 0.0040 0.0020 1.00 4.8 × 10–4 0.0020 0.0040 1.00 5.0 × 10–4 0.0020 0.0020 1.00 2.4 × 10–4 0.0020 0.0020 0.50 4.6 × 10–4 0.0020 0.0020 0.25 9.4 × 10–4

(b) Write the rate law, and determine the value of the rate constant, k.

(c) Show that the following mechanism is consistent with the net equation and with the rate law. Which is the rate–determining step?
 OCl –(aq) + H2O(l)HOCl(aq) + OH –(aq) I –(aq) + HOCl(aq) HOI(aq) + Cl –(aq) HOI(aq) + OH –(aq) H2O(l) + OI –(aq)

(d) Is it appropriate to refer to OH as a catalyst in this reaction? Explain.

(a) Use the method of initial rates for a rate law of the general form Rate = k[OCl ]m[ I ]n[ OH ]p. Experiments 1 and 3 give the order in hypochlorite ion (iodide and hydroxide are constant): doubling the concentration of hypochlorite doubles the initial rate, meaning that the reaction is 1st order in hypochlorite ion, m = 1. Experiments 2 and 3 can be used to find the order in iodide ion (hypochlorite and hydroxide are unchanged): doubling the iodide concentration doubles the rate so the reaction is also first order in iodide ion, n = 1. Experiments 3 and 4 can be used to find the order in hydroxide (hypochlorite and iodide are unchanged): doubling the concentration halves the rate, so the order in hydroxide must be –1, p = –1. Mathematically:

The overall order is the sum of each individual order = 1 + 1 + –1 = 1

(b) Based on the orders found above, the rate law is:

The rate constant can be found by using the rate law and the data:
 [OCl –], M [I –], M [OH –], M Rate of formation of OI–, mol L–1 s–1 0.0040 0.0020 1.00 4.8 × 10–4 60.0 s–1 0.0020 0.0040 1.00 5.0 × 10–4 62.5 s–1 0.0020 0.0020 1.00 2.4 × 10–4 60.0 s–1 0.0020 0.0020 0.50 4.6 × 10–4 57.5 s–1 0.0020 0.0020 0.25 9.4 × 10–4 58.75 s–1

The average rate constant (to the correct number of significant figures) = 60. s–1

(c) First, add all the equations in the mechanism together to see if the stoichiometry of the mechanism is correct:

OCl (aq) + H2O(l) + I (aq) + HOCl(aq) + HOI(aq) + OH (aq)
HOCl(aq) + OH (aq) + HOI(aq) + Cl (aq) + H2O(l) + OI (aq)

After eliminating common species:

OCl (aq) + I (aq) OI (aq) + Cl (aq)

which is the correct stoichiometry.

To find the rate–determining step, consider the rate law for each elementary reaction:

Step 1, forward reaction: Rate = kforward[OCl ][H2O]
Step 1, reverse reaction: Rate = kreverse[HOCl][ OH ]
Step 2: Rate = k2[I ][HOCl]
Step 3: Rate = k3[HOI][ OH ]

Step 1, forward does not have enough terms in the rate law. Step 1, reverse and Step 3 both have hydroxide in the numerator but the experimental rate law requires it to be in the denominator. This leaves Step 2 as the likely slow step, but this has an intermediate that must be eliminated by using Step 1, assuming it to be a fast step:

For Step 1, Rateforward = Ratereverse so kforward[OCl ][H2O] = kreverse[HOCl][OH ]

Solving for the intermediate, [HOCl] gives:

Substituting this into the rate law for Step 2, the proposed slow step, gives:

Since water is the solvent, its concentration does not change and should be grouped with the constants:

This now matches the experimental rate law.

(d) Since hydroxide is found in the denominator of the rate law, an increase in concentration slows the reaction down so OH is an inhibitor, not a catalyst. Compare experiments 3, 4, and 5 to see how decreasing the hydroxide increases the reaction rate.