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 for this reaction can be expressed based on the order with respect to the reactants. Since the reaction is first-order with respect to each reactant (bromoethane, CH3Br, and sodium hydroxide, NaOH), the rate law will reflect this. The rate law is given by:

Rate = k [CH3Br][NaOH]

where: - Rate is the rate of the reaction. - k is the rate constant for the reaction. - [CH3Br] is the molar concentration of bromoethane. - [NaOH] is the molar concentration of sodium hydroxide.

b) If the concentration of NaOH is decreased by a factor of 5, the reaction rate will also decrease by a factor of 5. This is because the rate is directly proportional to the concentration of NaOH, as seen in the rate law. So if [NaOH] is reduced by 5 times, the rate becomes 1/5 of the original rate.

c) When the concentrations of both reactants are doubled, the rate of the reaction will change by a factor determined by the individual rate changes in each reactant. Since both reactants are first-order:

- Doubling [CH3Br] will increase the reaction rate by a factor of 2. - Doubling [NaOH] will also increase the reaction rate by a factor of 2.

Therefore, doubling both [CH3Br] and [NaOH] will increase the reaction rate by a factor of 2 * 2, which is 4. So the reaction rate will be four times faster when the concentrations of both reactants are doubled.