For the reaction $H_{2} + B r_{2} \to 2 H B r$ overall order is found to be 3/2. The rate of reaction can be expressed as

[H_{2}] [B r_{2}]^{1 / 2}

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[H_{2}]^{1 / 2} [B r_{2}]

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[H_{2}]^{3 / 2} [B r_{2}]^{0}

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Solution

The correct option is D $A l l o f t h e s e$
Sum of exponents should be 3/2

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Similar questions

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

\frac{3}{2}

H_{2} + {B r}_{2} ⟶ 2 H B r

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

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

\frac{d x}{d t} = k [H_{2}] {[{B r}_{2}]}^{1 / 2}

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

\frac{d x}{d t} = k [H_{2}]^{1 / 2} {[{B r}_{2}]}^{1 / 2}

{B r}_{2}

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

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