the following reaction is a single step bimolecular reaction when the concentrat
ID: 1005423 • Letter: T
Question
the following reaction is a single step bimolecular reaction when the concentrations of CH3Br and NaOH are both 0.200 M, the rate of the reaction is 0.0040 M/s. The following reaction is a single-step, bimolecular reaction: CH,Br+ NaOH CH3OH + NaBr When the concentrations of CH3Br and NaOH are both 0.200 M, the rate of the reaction is 0.0040 M/s. (a) What is the rate of the reaction if the concentration of CHaBr is doubled Number1 M/ s (b) What is the rate of the reaction if the concentration of NaOH is halved? Number M/ s increased (e) What is the rate of the reaction if the concentrations of CH,Br and NaOH are both by a factor of three? Number M/Explanation / Answer
Sine this is a bimolecular reacton, it proceeds via SN2 mechanism. The order of the reaction is 1st order with respect to each of the CH3Br and NaOH. Hence
Rate = k x [CH3Br] x [NaOH]
(a): Since rate is proportional to [CH3Br], when we double the concentration of CB3Br, the rate of reaction will also double. Hence
new rate = 2 x 0.0040 M/s = 0.0080 M/s (answer)
(b): Since rate is proportional to [NaOH], when the concentration of NaOH is halved, the rate of reaction will also be halved. Hence
new rate = (1/2)x 0.0040 M/s = 0.0020 M/s (answer)
(c): Since rate is proportional to both [NaOH] and [CH3Br], when the concentration of both NaOH and CH3Br are increased by a factor of 3, the rate of reaction will increase by 3x3 = 9 times. Hence
new rate = (9)x 0.0040 M/s = 0.0360 M/s (answer)
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