Question

The rate law of reaction between the substance $A$ and $B$ is given by $rate=K{\left[A\right]}^n{\left[B\right]}^m$. On doubling the concentration of $A$ and making the volume of $B$ half the ratio of new rate to the earlier rate of reaction will be:

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

Answer 1

The correct option is D $2^{n-m}$

Rate is given by,

$rate=K\left[A{]}^n\right[B{]}^m$

On Doubling the concentration of $A$ and making Volume of $B$ half, the new rate becomes $2^n\times {\frac{1}{2}}^m$ = $2^{n-m}$

So Option $D$ is correct.