Question

The rate law for a reaction between the substance $ 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 new rate to the earlier rate of the reaction will be as

• A. $\frac{1}{2^{(m+n)}}$
• B. $ (m + n)$
• C. $(n -m )$
• D. $2^{(n-m)} $

Answer 1

Answer: (D)

$ Rate_1 = k[A]^n [B]^m $
On doubling the concentration of $ A$ and halving the concentration of $B, $
$Rate_2 = k[2A]^n [B/2]^m $
Ratio between new and earlier rate
$ \frac {k[2A]^n[B/2]^m}{k[A]^n[B]^m} = 2^n \times \left(\frac{1}{2}\right)^{m} =2 ^{(n-m)} $