Question

Consider the following mechanism: (1) ClO−(aq) + H2O(l) HClO(aq) + OH−(aq) [fast] (2) I−(aq) + HClO(aq) HIO(aq) + Cl−(aq) [slow] (3) OH−(aq) + HIO(aq) H2O(l) + IO−(aq) [fast] (a) What is the overall equation? (b) Identify the intermediate(s), if any. (c) What are the molecularity and the rate law for each step? (d) Is the mechanism consistent with the actual rate law: Rate = k[ClO−][I−]?

Answer 1

Explanation:

$ClO^-(aq)+H_2O(l) \rightarrow HClO (aq)+OH^-(aq)\ [fast]$

$I^-(aq)+HClO(aq) \rightarrow HIO(aq)+Cl^-(aq)\ [slow]$

$OH^-(aq)+HIO(aq) \rightarrow H_2O(l)+IO^-(aq)\ [fast]$

(a)

Cancelling common terms, the overall reaction is as follows:

$ClO^-(aq)+I^-(aq) \rightarrow Cl^- (aq)+IO^-(aq)$

(b)

Chemical species which are generated and consumed during the reaction are called intermediates. Therefore, in the given reaction, HClO(aq), OH-(aq) and HIO(aq) are intermediate species.

(c)

two reactant molecules are involved in each steps, therefore, each steps are bimolecular. Since water is in excess, therefore will not be rate law

Rate law for first step:

$rate=k_1[ClO^-]$

Rate law for second step:

$rate=k_2 [I^-][HClO]$

Rate law for third step:

$rate=k_3[OH^-][HIO]$

(d)

Step 2 is slow step, therefore this step will be rate determining step.

So, rate = k [I-] [HClO]

So, it consisted with the actual rate law