For the reaction `H_(2)(g) + Br_(2)(g) rarr 2HBr (g)` the experimental data suggested that `r = k[H_(2)][Br_(2)]^(1//2)`
The molecularity and order of the reaction are respectively:

Options

A. `2,3//2`
B. `3//2,3//2`
C. 1,1
D. `1,1//2`

Answer

To solve the problem, we need to determine the molecularity and order of the reaction given by the equation:

\[ H_2(g) + Br_2(g) \rightarrow 2HBr(g) \]

The rate law provided is:

\[ r = k[H_2][Br_2]^{1/2} \]

### Step 1: Determine the Molecularity
Molecularity is defined as the total number of reactant species (atoms or molecules) that are involved in the rate-determining step of a reaction.

For the reaction:

- There is 1 molecule of \( H_2 \)
- There is 1 molecule of \( Br_2 \)

Thus, the total number of reactant species is:

\[ \text{Molecularity} = 1 + 1 = 2 \]

### Step 2: Determine the Order of the Reaction
The order of a reaction is determined by the sum of the powers of the concentration terms in the rate law expression.

From the rate law:

\[ r = k[H_2][Br_2]^{1/2} \]

We can see that:

- The exponent of \( [H_2] \) is 1.
- The exponent of \( [Br_2] \) is \( \frac{1}{2} \).

Now, we sum these exponents to find the order:

\[ \text{Order} = 1 + \frac{1}{2} = \frac{3}{2} \]

### Conclusion
Thus, the molecularity of the reaction is 2 and the order of the reaction is \( \frac{3}{2} \).

### Final Answer
The molecularity and order of the reaction are respectively: **2, \( \frac{3}{2} \)**.

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