Final answer:
The rate constant (k) for the reaction CH₃Br (aq) + OH⁻ (aq) → CH₃OH (aq) + Br⁻ (aq) is calculated to be 172.8 M⁻¹s⁻¹. Tripling the concentration of OH⁻ will triple the reaction rate, and tripling the concentrations of both reactants will increase the reaction rate by a factor of 9.
Step-by-step explanation:
The given reaction CH₃Br (aq) + OH⁻ (aq) → CH₃OH (aq) + Br⁻ (aq) has a rate law that is first order in CH₃Br and first order in OH⁻, making it overall second order. Using the provided reaction rate and concentrations, we can calculate the rate constant (k).
Rate law: Rate = k * [CH₃Br] * [OH⁻]
Given data:
Rate = 0.0432 M/s
[CH₃Br] = 5.0 x 10⁻³ M
[OH⁻] = 0.050 M
Substitute the given data into the rate law:
0.0432 M/s = k * (5.0 x 10⁻³ M) * (0.050 M)
Solve for k:
k = 0.0432 M/s / ((5.0 x 10⁻³ M) * (0.050 M)) = 172.8 M⁻¹s⁻¹
a) The value of the rate constant (k) is 172.8 M⁻¹s⁻¹.
b) The units of the rate constant are M⁻¹s⁻¹.
c) If the concentration of OH⁻ were tripled, the rate of the reaction would also be tripled, since the reaction is first order in OH⁻.
d) If the concentrations of both reactants were tripled, the reaction rate would increase by a factor of 9 (3²), as it is first