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Q.

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 reducing the concentration of B to half, the ratio of the new rate to the earlier rate of the reaction will be as

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a

(nm)

b

1/2(m+n)

c

2(nm)

d

(m+n)

answer is A.

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Detailed Solution

r1=k[A]n[B]m and  r2=k[2A]n[B]2m=2(nm)k[A]n[B]m

Hence, r2r1=2(nm)

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