Question

Reaction rate between two substance $A$ and $B$ is expressed as following:
rate = $k[A]^n[B]^m$
If the concentration of A is doubled and concentration of B is made half of initial concentration, the ratio of the new rate to the earlier rate will be:

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

Answer 1

Answer: (D)

$Rate_{1}= k\left[A\right]^{n} \left[5\right]^{m}$
$Rate_{2}=k\left[2A\right]^{n}\left[\frac{1}{2}B\right]$
$\therefore \frac{Rate_{2}}{Rate_{1}}=\frac{k\left[2A\right]^{n}\left[\frac{1}{2}B\right]^{m}}{k\left[A\right]^{n}\left[B\right]^{m}}=\left(2\right)^{n}\left(\frac{1}{2}\right)^{m}$
$=2^{n}. \left(2\right)^{-m}=2^{n-m}$