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Question

The rate law for a reaction between the substances A and B is given by rate=k[A]n[B]m. On doubling the concentration of A and having the concentration of B halved, the ratio of the new rate to the earlier rate of the reaction will be as:

A
12m+n
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B
(m+n)
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C
(nm)
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D
2(nm)
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Solution

The correct option is D 2(nm)
Given that
R=K[A]n[B]m
after doubling the concentration of A and concentration of B is halfed
R1=K[2A]n[B2]m
R1=(2)nmR
R1R=2nm1


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