In the formation of `HBr` form `H_(2)` and `Br_(2)`, following mechanism is observed:
I. `Br_(2) hArr 2Br^(*)` (Equilibrium step)
(II) `H_(2) + Br^(*)rarr HBr+H^(*)` (Slow step)
(III) `H^(*) + Br_(2) rarr HBr + H^(*)` (Fast step)
The rate law for the above reaction is

Options

A. `r=k[H_(2)][Br_(2)]`
B. `r=k[H_(2)][Br_(2)]^(1//2)`
C. `r=k[H_(2)]^(1//2)[Br_(2)]`
D. `r=k[H_(2)]^(1//2)[Br]^(1//2)`

Explanation

To derive the rate law for the formation of HBr from H2 and Br2 based on the given mechanism, we can follow these steps:

### Step 1: Identify the Mechanism
The mechanism consists of three steps:
1. **Equilibrium Step**: \( \text{Br}_2 \rightleftharpoons 2 \text{Br}^* \)
2. **Slow Step**: \( \text{H}_2 + \text{Br}^* \rightarrow \text{HBr} + \text{H}^* \)
3. **Fast Step**: \( \text{H}^* + \text{Br}_2 \rightarrow \text{HBr} + \text{H}^* \)

### Step 2: Determine the Rate-Determining Step
The slow step (Step II) is the rate-determining step, which means that the rate of the overall reaction is determined by this step.

### Step 3: Write the Rate Law for the Slow Step
For the slow step, the rate law can be expressed as:
\[ \text{Rate} = k_1 [\text{H}_2][\text{Br}^*] \]
where \( k_1 \) is the rate constant for the slow step.

### Step 4: Express the Concentration of the Intermediate
Since \( \text{Br}^* \) is a free radical and not a reactant or product in the overall reaction, we need to express its concentration in terms of the reactants. From the equilibrium step, we can write:
\[ K_{eq} = \frac{[\text{Br}^*]^2}{[\text{Br}_2]} \]
Rearranging gives:
\[ [\text{Br}^*]^2 = K_{eq} [\text{Br}_2] \]
Taking the square root of both sides, we find:
\[ [\text{Br}^*] = \sqrt{K_{eq} [\text{Br}_2]} \]

### Step 5: Substitute the Expression for \( [\text{Br}^*] \) into the Rate Law
Substituting the expression for \( [\text{Br}^*] \) into the rate law, we get:
\[ \text{Rate} = k_1 [\text{H}_2] \sqrt{K_{eq} [\text{Br}_2]} \]
This can be simplified to:
\[ \text{Rate} = k_1 \sqrt{K_{eq}} [\text{H}_2] [\text{Br}_2]^{1/2} \]
Let \( k = k_1 \sqrt{K_{eq}} \), which is a new rate constant.

### Step 6: Final Rate Law Expression
Thus, the final rate law for the reaction is:
\[ \text{Rate} = k [\text{H}_2] [\text{Br}_2]^{1/2} \]

### Conclusion
The correct option for the rate law based on the given mechanism is:
\[ \text{Rate} = k [\text{H}_2] [\text{Br}_2]^{1/2} \]