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 \[={{(2)}^{n}}{{\left( \frac{1}{2} \right)}^{m}}\text{times}\] \[={{(2)}^{n-m}}\,\text{times}\]
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