A) \[\frac{1}{{{2}^{m+n}}}\]
B) \[(m+n)\]
C) \[(n-m)\]
D) \[{{2}^{(n-m)}}\]
Correct Answer: D
Solution :
Rate becomes \[{{x}^{y}}\] times if concentration is made \[x\] times of a reactant giving \[{{y}^{th}}\] order reaction. Rate\[=k{{[A]}^{n}}{{[B]}^{m}}\] Concentration of \[A\] is doubled, hence,\[x=2\],\[y=n\]and rate becomes\[={{2}^{n}}\]times Concentration of \[B\] is halved, hence \[x=\frac{1}{2}\] and \[y=m\] and rate becomes \[={{\left( \frac{1}{2} \right)}^{m}}\] times Net rate becomes \[={{(2)}^{n}}{{\left( \frac{1}{2} \right)}^{m}}\] times \[={{(2)}^{n-m}}\]
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