what is the rate of the reaction if the concentration of The following reaction

Question

what is the rate of the reaction if the concentration of The following reaction is a single-step, bimolecular reaction: CH_3Br + NaOH rightarrow CH_3OH + NaBr When the concentrations of CH_3Br and NaOH are both 0.100 M, the rate of the reaction is 0.0060 M/s. What is the rate of the reaction if the concentration of CH_3Br is doubled? What is the rate of the reaction if the concentration of NaOH is halved? What is the rate of the reaction if the concentrations of CH_3Br and NaOH are both increased by a factor of three?

Explanation / Answer

a)

if bimolecular, assume it is RAte = k[A][B]

initial rate = 0.006 M/s

so

if we double, then

the rate should be doubled, since it is directly propoertional to CH3Br

so

0.006*2 = 0.012 M/s

b)

if we halve NaOH, then

the rate will be halved

so

0.006 /2 M/s = 0.003 M/s

c)

if we increase both by a factor of 3

this will be:

3x3 = 9 factor

so

0.006*9

0.054 M/s

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