$H_{2} + {B r}_{2} \to 2 H B r$ .
The rate law is $\frac{d x}{d t} = k [H_{2}]^{1 / 2} {[{B r}_{2}]}^{1 / 2}$ :
On increasing the concentration of ${B r}_{2}$ four times, by how much times the rate of reaction will change?

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Solution

The correct option is A 2
$H_{2} + {B r}_{2} \to 2 H B r$ .
The rate law is $\frac{d x}{d t} = k [H_{2}]^{1 / 2} {[{B r}_{2}]}^{1 / 2}$
So, order of reaction $= \frac{1}{2} + \frac{1}{2} = 1$
Molecularity $= 1 + 1 = 2$
$\frac{d x}{d t} \propto {[B r]}^{1 / 2}$
So, reaction rate will be doubled if concentration of ${B r}_{2}$ is increased by $4$ times.

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H_{2} + {B r}_{2} \to 2 H B r

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