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

Answer 1

Answer: (D)

Rate becomes $x^{y}$ times if concentration is made

$x$ times of a reactant giving $y^{\text {th }}$ order reaction

Rate $=k[A]^{n}[B]^{m}$

Concentration of $A$ is doubled hence $x=2, y=n$ and rate becomes $=2^{n}$ times

Concentration of $B$ is halved, hence $x=1 / 2$ and

y=m and rate becomes $=\left(\frac{1}{2}\right)^{m} $ times

Net rate becomes $=(2)^{n}\left(\frac{1}{2}\right)^{m} $ times

$=(2)^{n-m} $ times