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
(n−m)
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D
2(n−m)
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Solution
The correct option is D2(n−m) Given that
R=K[A]n[B]m
after doubling the concentration of A and concentration of B is halfed