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Question

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

A
dxdt=k[A]2[B]2
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B
dxdt=k[A][B]
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C
dxdt=k[A]0[B]2
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D
dxdt=k[B]1/2
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Solution

The correct option is D dxdt=k[B]1/2
2A + 3B product

Since 'A' is taken in excess, so the rate of reaction doesn't depend on 'A'.

Rate =K[B]N given Rate1=K[0.1]n .....[1]

given Rate2=2×Rate1=K[0.4]n ......[2]

From equation (1) and (2),

n =12

Rate law is given by,

dxdt=K[B]12

Hence, option D is correct.

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