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Question: The rate law for a reaction between the substances A and B is given by rate = k [A]n [B]m. On doubli...

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

12m+n\frac{1}{2^{m + n}}

B

(m + n)(\text{m + n})

C

(n - m)(\text{n - m})

D

2(n - m)2(\text{n - m})

Answer

2(n - m)2(\text{n - m})

Explanation

Solution

Rate1 = k [A]n [B]m\text{1 = k }\lbrack A\rbrack^{n}\ \lbrack B\rbrack^{m}

On doubling the concentration of A and halving the

concentration of B

Rat 2 = k [2A]n [B/2]m\text{2 = k }\lbrack\text{2A}\rbrack^{n}\ \lbrack\text{B/2}\rbrack^{m}

Ratio between new and earlier rate.

k[2A]n[B/2]mk[A]n[B]m=2n×(12)m=2n−m\frac{k\lbrack 2A\rbrack^{n}\lbrack B/2\rbrack^{m}}{k\lbrack A\rbrack^{n}\lbrack B\rbrack^{m}} = 2^{n} \times \left( \frac{1}{2} \right)^{m} = 2^{n - m}