Hi guys, my question is:
Bromoethane is converted to methanol in an alkaline solution. This reaction is first order with respect to each reactant.
CH3Br(aq) + NaOH(aq) --> CH3OH(aq) + NaBr(aq)
a) What is the rate law for this reaction?
b) Predict how the reaction rate will change when the concentration of NaOH is decreased by a factor of 5.
c) Predict the change in reaction rate when the concentrations of both reactant are doubled?

A) The rate law can only be on the reactant side and you can only determine it after you get the net ionic equation because of spectators cancelling out. So in this case the rate law is k=[CH3Br]^1 [OH-]^1. The powers are there because the rxn is first order.

B) Since the rxn is first order anything you do to it will be the exact same "counter rxn" per say so since you are decreasing the OH- by 5 the rate will decease by 5

C) The rate will increase by 4 since you are doubling both you have to multiply them both.

andromache andromache · Answer 2 · 2022-03-01T15:24:35+01:00

a. The rate law for the reaction should be [CH3Br]^1 [OH-]^1

b. The reaction rate should be decrease by 5.

c. The change in reaction rate should be increased by 4.

Rate law, reaction rate:

a. The rate law should be on the reactant side only and it can only measured when the net ionic equation should be considered due to this, spectators cancelling out. So in the given situation, the rate law is k=[CH3Br]^1 [OH-]^1. The powers should be there since the rxn is first order.

B) Since the rxn is first order so it should be the same "counter rxn" so since it decreased the OH- by 5 due to this, the rate will decease by 5

C) The rate will increase by 4 because it is doubling both you have to multiply them both.

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