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