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Question
Answer the following in one or two sentences.
For the reaction,
\[\ce{CH3Br_{(aq)} + OH^-_{ (aq)} -> CH3OH^-_{ (aq)} + Br^-_{ (aq)}}\], rate law is rate = k`["CH"_3"Br"]["OH"^-]`
What is the change in rate if concentrations of both reactants are doubled?
Solution
For the reaction
\[\ce{CH3Br_{(aq)} + Br^-_{ (aq)} -> CH3OH_{ (aq)} + Br^-_{ (aq)}}\]
Rate = k`["CH"_3"Br"]["OH"^-]`
If concentrations of `"CH"_3"Br"` and `"OH"^-` are doubled, rate will increase by a factor of 4.
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