For the reaction 2A + 3B $\to$ products, A is in excess and on changing the concentration of B from 0.1 M to 0.4 M, rate becomes doubled. Thus, rate law is

\frac{d x}{d t} = k [A]^{2} [B]^{2}

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\frac{d x}{d t} = k [A] [B]

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\frac{d x}{d t} = k [A] 0 [B]^{2}

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

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Solution

The correct option is D $\frac{d x}{d t} = k [B]^{1 / 2}$
A is in excess, so rate does not depend on A.

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For the reaction 2A + 3B → products, A is in excess and on changing the concentration of B from 0.1 M to 0.4 M, rate becomes doubled. Thus, rate law is

The correct option is D dxdt=k[B]1/2A is in excess, so rate does not depend on A.

For the reaction 2A + 3B $\to$ products, A is in excess and on changing the concentration of B from 0.1 M to 0.4 M, rate becomes doubled. Thus, rate law is

The correct option is D $\frac{d x}{d t} = k [B]^{1 / 2}$
A is in excess, so rate does not depend on A.