For the reaction, $H_{2} (g) + B r_{2} (g) = 2 H B r (g)$ , the 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 is

s e c^{- 1}

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Molecularity of the reaction is 2

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Solution

The correct option is C Molecularity of the reaction is 2
Molecularity of the reaction is 2 as one molecules of hydrogen and one molecules of bromine participates in the reaction. The order of reaction is 3/2 .

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Similar questions

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