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 halving the
concentration of B, the ratio of the new rate to the earlier rate
of the reaction will be as
(a) (m + n) (b) (n – m) (c) 2⁽ⁿ⁻ᵐ⁾ (d) 1/2⁽ᵐ⁺ⁿ⁾

Answer 1

Answer:

ANSWER

rate = k{[A]}^{n}{[B]}^{m}rate=k[A]

n

[B]

m

On doubling the concentration of A and making the volume of B half, the new rate becomes {2}^{n}\times{\dfrac{1}{2}}^{m}2

n

×

2

1

m

= {2}^{n - m}2

n−m

times the earlier rate.

So Ratio is {2}^{n - m}2

n−m

Answer 2

Hello Friend..!!

the answer of ur question is..!!

Option.D

Thank you