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 halving
the concentration of $$B,$$ the ratio of the new rate to the earlier rate of the reaction will be as
A.
$$\left( {m + n} \right)$$
B.
$$\left( {n - m} \right)$$
C.
$${2^{\left( {n + m} \right)}}$$
D.
$$\frac{1}{{{2^{\left( {m + n} \right)}}}}$$
Answer :
$${2^{\left( {n + m} \right)}}$$
Solution :
$$\eqalign{
& Rat{e_1} = k{\left[ A \right]^n}{\left[ B \right]^m};\,Rat{e_2} = k{\left[ {2A} \right]^n}{\left[ {\frac{1}{2}B} \right]^m} \cr
& \therefore \,\,\frac{{Rat{e_2}}}{{Rat{e_1}}} = \frac{{k{{\left[ {2A} \right]}^n}{{\left[ {\frac{1}{2}B} \right]}^m}}}{{k{{\left[ A \right]}^n}{{\left[ B \right]}^m}}} \cr
& = {\left[ 2 \right]^n}{\left[ {\frac{1}{2}} \right]^m} \cr
& = {2^n}{.2^{ - m}} \cr
& = {2^{n - m}} \cr} $$