When the concentration of alkyl halide is tripled and concentration of OH ions is reduced to half of its
originals concentration what will be the change in rate of SN2 reaction ?
Answers
Answered by
0
Answer:1.5 times
Explanation:
Answered by
0
Answer:
New Rate = 1.5 x ( original rate)
Explanation:
SN2 reactions are those that involve the substitution of bimolecular Nucleophilic.
Now this reaction usually is a single step process and its rate of reaction can be written as;
Rate = K [R- X] [OH]
Now as per given information, when we increase concentration of alkyl halide 3 times and concentration of OH ions by half, then overall equation would be
New Rate = K ( 3 x [R- X] ) . ( 1/2 x [OH] )
New Rate = 1.5 x ( original rate)
Similar questions