Question

The rate law for a reaction between the substances $P$ and $Q$ is given by rate $=k[P{]}^a[Q{]}^b$. On doubling the concentration of $P$ and reducing the concentration of $Q$ to one-half, the ratio of the new rate to the earlier rate of the reaction will be

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

Answer 1

Answer: (B)

Rate $=k[P{]}^a[Q{]}^b$
New rate $=k[2P{]}^a{\left[\frac{Q}{2}\right]}^b=k\cdot 2^a[P{]}^a\cdot \frac{[Q{]}^b}{2^b}$
$=2^{(a-b)}\cdot k[P{]}^a[Q{]}^b=2^{(a-b)}\cdot$ rate
$\therefore \frac{\text{ New rate }}{\text{ Rate }}=2^{(a-b)}$