Question

Consider the single‑step, bimolecular reaction.

CH₃Br+NaOH⟶CH₃OH+NaBr When the concentrations of CH₃Br and NaOH are both 0.120 M, the rate of the
reaction is 0.0080 M/s. What is the rate of the reaction if the concentration of CH₃Br is doubled?
rate: M/s What is the rate of the reaction if the concentration of NaOH is halved?
rate: M/s What is the rate of the reaction if the concentrations of CH₃Br and NaOH are both increased
by a factor of 5?
rate:___ M/s

Answer 1

The rate of the reaction is doubled when the concentration of CH₃Br is doubled, and the rate is not affected by changing the concentration of NaOH.

Step-by-step explanation:

The given reaction is represented by the equation: CH₃Br + NaOH ⟶ CH₃OH + NaBr

The rate law for this reaction is rate = k[CH₃Br]. Since the exponent for CH₃Br is 1, the reaction is first order in CH₃Br. Therefore, doubling the concentration of CH₃Br will double the reaction rate, and halving the concentration of CH₃Br will halve the reaction rate.

Conversely, the concentration of NaOH does not affect the reaction rate. Thus, increasing or decreasing the concentration of NaOH will not have an impact on the rate of the reaction.