Question

The breakdown of dopamine is catalyzed by the enzyme monoamine oxidase (MAO). What is the final concentration of product if the starting dopamine concentration is 0.050 M and the reaction runs for 5 seconds. (Assume the rate constant for the reaction is 0.249 s^-1.)
A) 0.050 M
B) 0.014 M
C) 0.018 M
D) 1.2 M
E) 0.025 M

Answer 1

Answer:

The answer is: B) 0.014 M

Explanation:

The reaction is the first order, then:

$-\frac{dC}{dt} =KC\\Integrating\\C_{t} =C_{o} e^{-Kt}$

Where

Ct = concentration in any time

Co = initial concentration = 0.05 M

K = 0.249 s⁻¹

t = 5 s

Replacing:

$C_{t} =0.05e^{-0.249*5} =0.014M$

Answer 2

Answer:

final concentration: Ca = 0.014 M

Explanation:

Velocity of reaction:

• ra = K(Ca)∧α = δCa/δt

∴ α: order of reaction, assuming α = 1

∴ K = 0.249 s-1.......rate constant

∴ Cao = 0.050 M......initial concentration

∴ t = 5 s.......reaction time

⇒ δCa/δt = K*Ca

⇒ ∫δCa/Ca = K*∫δt

⇒ Ln(Cao/Ca) = K*t = (0.249s-1)(5 s) = 1.245

⇒ Cao/Ca = 3.473

⇒ Ca = 0.050/3.473

⇒ Ca = 0.014 M