posted by Bryan
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?
b) What is the rate of the reaction if the concentration of NaOH is halved?
c) What is the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of five?
Please help! I have no idea how to start this problem off!
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.