Question

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

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

Answer 1

Answer: (B)

$2^{(n-m)}$

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

A doubled, B halved

New rate $=K{\left[2A\right]}^n{\left[B/2\right]}^m$

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

ratio $=2^{n-m}$