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Q.

The rate law for a reaction between the substance A and B is given by Rate =k[A]n[B]mOn doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier ratio of the reaction will be

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a

1(2)m+n

b

(m + n)

c

(n - m)

d

2n-m

answer is D.

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Detailed Solution

Rate =kA]n[B]mWhen A' = 2A and B' = 0.58The new rate becomes Rate' = [2A]n, [0.5 B]m  Rate'  Rate =k2n[A]n(0.5)m[B]mk[A]n[B]m=2n12m=2(n−m)
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