The reaction:
`OCl^(ɵ) + I^(ɵ)overset(overset(ɵ)(OH))rarr OI^(ɵ)+Cl^(ɵ)`
takes place in the following steps:
(i) `OCl^(ɵ)+H_(2)O underset(k_(2))overset(k_(1))hArr HOCl+overset(ɵ) (OH) ("fast")`
(ii) `I^(ɵ) + HOCloverset(k_(3))rarrHOI + Cl^(ɵ)` (slow)
(iii) `overset(ɵ)(OH) + HOI underset(k_(1)')overset(k_(2)')hArr H_(2)O+OI^(ɵ)" " ("fast")`
The rate of consumption of `I^(ɵ)` in the following equation is

Options

A. `(k_(3)k_(1))/(k_(2))([OCl^(ɵ)][I^(ɵ)])/([overset(ɵ)(OH)]^(2))`
B. `(k_(1)k_(2))/(k_(3))([OCl^(ɵ)][I^(ɵ)])/([overset(ɵ)(OH)])`
C. `(k_(1)k_(3))/(k_(2))([OCl^(ɵ)][I^(ɵ)])/([overset(ɵ)(OH)])`
D. `(k_(2)k_(3))/(k_(1))([OCl^(ɵ)][I^(ɵ)])/([overset(ɵ)(OH)])`

Explanation

To determine the rate of consumption of \( I^- \) in the given reaction, we will analyze the provided steps and identify the rate-determining step.

### Step 1: Identify the Rate-Determining Step
The reaction consists of three steps:
1. \( OCl^- + H_2O \overset{k_1}{\rightarrow} HOCl + OH^- \) (fast)
2. \( I^- + HOCl \overset{k_3}{\rightarrow} HOI + Cl^- \) (slow)
3. \( OH^- + HOI \overset{k_1'}{\rightarrow} H_2O + OI^- \) (fast)

Since the second step is the slow step, it is the rate-determining step. The rate of the overall reaction will be determined by this step.

### Step 2: Write the Rate Law for the Rate-Determining Step
For the slow step:
\[ \text{Rate} = k_3 [I^-][HOCl] \]

### Step 3: Express \( [HOCl] \) in Terms of Other Concentrations
The first step is at equilibrium, so we can express \( [HOCl] \) using the equilibrium constant derived from the first step:
\[ K = \frac{[HOCl][OH^-]}{[OCl^-][H_2O]} \]
Since \( H_2O \) is a liquid, it does not appear in the equilibrium expression. Rearranging gives:
\[ [HOCl] = \frac{K \cdot [OCl^-]}{[OH^-]} \]
Where \( K = \frac{k_1}{k_2} \).

### Step 4: Substitute \( [HOCl] \) Back into the Rate Law
Substituting \( [HOCl] \) into the rate law:
\[ \text{Rate} = k_3 [I^-] \left(\frac{k_1}{k_2} \cdot \frac{[OCl^-]}{[OH^-]}\right) \]

### Step 5: Simplify the Rate Expression
Combining the constants, we get:
\[ \text{Rate} = \frac{k_1 k_3}{k_2} [I^-] [OCl^-] \frac{1}{[OH^-]} \]

### Final Rate Expression
Thus, the final expression for the rate of consumption of \( I^- \) is:
\[ \text{Rate} = \frac{k_1 k_3}{k_2} [I^-] [OCl^-] \frac{1}{[OH^-]} \]

### Conclusion
The correct answer is:
\[ \text{Rate} = \frac{k_1 k_3}{k_2} \cdot \frac{[OCl^-] [I^-]}{[OH^-]} \]

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