Question

The rate law for a reaction between the substances A and B is given by $rate=K\left[A{]}^n\right[B{]}^m$. On doubling the concentration of A and having the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as:

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

Answer 1

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

$r=K\left[A{]}^n\right[B{]}^m$
$r_1=K[2A{]}^n{\left[\frac{B}{2}\right]}^m$
$\therefore \frac{r_1}{r}=2^{n-m}$