The following reaction is a single-step, bimolecular reaction:

CH3Br + NaOH ---> CH3OH + NaBr

When the concentrations of CH3Br and NaOH are both 0.150 M, the rate of the reaction is 0.0090 M/s.

a) what is the rate of the reaction if the concentration of CH3Br is doubled?
______ M/s

b) What is the rate of the reaction if the concentration of NaOH is halved?
______ M/s

c) What is the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of five?
______ M/s

Please help! I have no idea how to start this problem off!

3 answers

According to the problem, the rate constant, k, has units of M/s which is moles/L*s and that is for a zero order equation. The equation then is r = k and the concns don't enter into the equation at all. You can read more about it here.
http://www.chm.davidson.edu/vce/kinetics/DifferentialRateLaws.html

jjh

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