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