For the reaction, $H_{2} (g) + B r_{2} (g) \to 2 H B r (g)$ , then reaction $r a t e = K [H_{2}] [B r_{2}]^{1 / 2}$ . Which statement is true about this reaction :

The reaction is of second order

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Molecularity of the reaction is

3 / 2

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The unit of

k

{sec}^{- 1}

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Molecularity of the reaction is

2

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Solution

The correct option is A Molecularity of the reaction is $2$
$H_{2} + {B r}_{2} \to 2 H B r$
Rate= $K [H_{2}] {[{B r}_{2}]}_{2}^{\frac{1}{2}}$
Order= $\frac{1}{2} + 1 = \frac{3}{2}$
Molecularity= 2
So, option 'D' is correct.

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Similar questions

H_{2} (g) + B r_{2} (g) = 2 H B r (g)

r a t e = K [H_{2}] [B r_{2}]^{1 / 2}

H_{2} (g) + {B r}_{2} (g) \to 2 H B r (g)

= k [H_{2}] {[{B r}_{2}]}^{1 / 2}

H_{2} (g) + B r_{2} (g) \to 2 H B r (g)

r = k [H_{2}] [B r_{2}]^{1 / 2}

H_{2} (g) + B r_{2} (g) \to 2 H B r (g)

r = k [H_{2}] [B r_{2}]^{1 / 2}

H_{2} + {B r}_{2} ⟶ 2 H B r

= K [H_{2}] {[{B r}_{2}]}^{\frac{1}{2}}

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