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 determine the rate law for the formation of HBr from H2 and Br2, we will analyze the given reaction mechanism step by step.

### Step 1: Identify the Rate-Determining Step
The rate law for a reaction is primarily determined by the slowest step in the mechanism, which is known as the rate-determining step. In this case, the slow step is:
\[ \text{H}_2 + \text{Br}^* \rightarrow \text{HBr} + \text{H}^* \]

### Step 2: Write the Rate Expression for the Slow Step
For the slow step, we can write the rate expression as:
\[ \text{Rate} = k_1 [\text{H}_2][\text{Br}^*] \]
where \( k_1 \) is the rate constant for this step.

### Step 3: Determine the Concentration of the Intermediate (Br*)
The concentration of the bromine radical (\( \text{Br}^* \)) is not directly given, as it is an intermediate. To express it in terms of the reactants, we need to look at the equilibrium step:
\[ \text{Br}_2 \rightleftharpoons 2\text{Br}^* \]
This equilibrium can be described by the equilibrium constant \( K_{eq} \):
\[ K_{eq} = \frac{[\text{Br}^*]^2}{[\text{Br}_2]} \]

From this, we can express \( [\text{Br}^*] \):
\[ [\text{Br}^*] = \sqrt{K_{eq} \cdot [\text{Br}_2]} \]

### Step 4: Substitute the Expression for Br* into the Rate Law
Now, substituting the expression for \( [\text{Br}^*] \) back into the rate law:
\[ \text{Rate} = k_1 [\text{H}_2] \sqrt{K_{eq} \cdot [\text{Br}_2]} \]

### Step 5: Simplify the Rate Law
This can be simplified as:
\[ \text{Rate} = k_1 \sqrt{K_{eq}} [\text{H}_2] [\text{Br}_2]^{1/2} \]

Let \( k = k_1 \sqrt{K_{eq}} \), which is a new rate constant. Thus, the final rate law becomes:
\[ \text{Rate} = k [\text{H}_2] [\text{Br}_2]^{1/2} \]

### Conclusion
The rate law for the formation of HBr from H2 and Br2 is:
\[ \text{Rate} = k [\text{H}_2] [\text{Br}_2]^{1/2} \]