The rate law expresison is given for a typical reaction, `n_(1)A + n_(2) B rarrP` as `r = k[A]^(n)[B]^(n2)`. The reaction completes only in one step and `A` and `B` are present in the solution. If the reaction occurs in more than one step, then the rate law is expressed by consdering the slowest step, i.e., for `S_(N)l` reaction `r = k[RX]`. If the eraction occurs in more than one step and the rates of the steps involved are comparable, then steady state approximation is conisdered, i.e., the rate of formation of intermediate is always equal to the rate of decompoistion of the intermediate. Conisder the reaction:
`[[I_(2) underset(k_(1))overset(k_(2))hArr2I("rapid equilibrium")],[H_(2)+2I overset(k_(3))rarr 2HI("slow")]]`
Which of the following expresison is correct?

Options

A. `(d[HI])/(dt) = (k_(3)k_(1))/(k_(2))[H_(2)][I_(2)]`
B. `(d[HI])/(dt) = (k_(1)k_(3))/(k_(2))2[H_(2)][I_(2)]`
C. `(d[HI])/(dt) = k_(2)[H_(2)][I_(2)]^(2)`
D. None of these

Explanation

To solve the problem, we need to analyze the given reaction and derive the correct rate law expression based on the information provided.

### Step-by-Step Solution:

1. **Identify the Reaction Steps**:
The reaction is given as:
\[ \text{I}_2 \underset{k_1}{\rightleftharpoons} 2\text{I} \quad (\text{rapid equilibrium}) \]
\[ \text{H}_2 + 2\text{I} \overset{k_3}{\rightarrow} 2\text{HI} \quad (\text{slow step}) \]

2. **Determine the Rate-Determining Step**:
The slow step in the reaction mechanism is the second step:
\[ \text{H}_2 + 2\text{I} \rightarrow 2\text{HI} \]
This means that the rate of the overall reaction will be determined by this step.

3. **Write the Rate Law for the Slow Step**:
The rate law for the slow step can be expressed as:
\[ r = k_3[\text{H}_2][\text{I}]^2 \]
where \( k_3 \) is the rate constant for the slow step.

4. **Express [I] in Terms of [I2] Using Equilibrium**:
Since the first step is an equilibrium, we can express the concentration of iodine (\([I]\)) in terms of the concentration of iodine gas (\([I_2]\)):
The equilibrium constant \( K_{eq} \) for the first step is given by:
\[ K_{eq} = \frac{[\text{I}]^2}{[\text{I}_2]} = \frac{k_2}{k_1} \]
Rearranging gives:
\[ [\text{I}]^2 = K_{eq}[\text{I}_2] = \frac{k_2}{k_1}[\text{I}_2] \]
Thus,
\[ [\text{I}] = \sqrt{\frac{k_2}{k_1}[\text{I}_2]} \]

5. **Substitute [I] into the Rate Law**:
Now substitute \([I]\) into the rate law:
\[ r = k_3[\text{H}_2]\left(\sqrt{\frac{k_2}{k_1}[\text{I}_2]}\right)^2 \]
Simplifying gives:
\[ r = k_3 \frac{k_2}{k_1}[\text{H}_2][\text{I}_2] \]

6. **Final Rate Expression**:
The final expression for the rate of formation of HI can be written as:
\[ r = k'[\text{H}_2][\text{I}_2] \]
where \( k' = k_3 \frac{k_2}{k_1} \).

### Conclusion:
Based on the above steps, the correct expression for the rate of the reaction is:
\[ r = k_3 \frac{k_2}{k_1}[\text{H}_2][\text{I}_2] \]
Since none of the options provided in the question matched this expression, the answer is "none of these".