Question

The rate law for a reaction between the substances A and B is given by Rate = k [A]^n [B]^m. If concentration of A is doubled and concentration of B is reduced to half, the ratio of the new rate to the earlier rate of reaction will be
(a)2^(n – m)
(b)1/2^m+n
(c)m + n
(d)n – m

Answer 1

r=K[A]

n

[B]

m

r

1

=K[2A]

n

[

2

B

]

m

∴

r

r

1

=2

n−m

Hope it helps....