Answer:
Step-by-step explanation:
a) For a first-order reaction, the rate of the reaction is proportional to the concentration of the reactant, i.e., rate = k[A]. The integrated rate law for a first-order reaction is given by:
ln([A]t/[A]0) = -kt
where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration, k is the rate constant, and t is time.
If a reaction is 50% complete in 30.0 min, it means that [A]t/[A]0 = 0.5. Substituting these values into the equation above, we can solve for the rate constant:
ln(0.5) = -k(30.0 min)
k = 0.0231 min^-1
Now, if the reaction is 75% complete, it means that [A]t/[A]0 = 0.25 (since 50% is half of the initial concentration and 75% is a quarter of the initial concentration). Substituting this value and the rate constant into the equation above, we can solve for the time t:
ln(0.25) = -0.0231 min^-1 * t
t = 61.3 min
Therefore, for a first-order reaction, the reaction will be 75% complete after 61.3 min.
b) For a zero-order reaction, the rate of the reaction is independent of the concentration of the reactant, i.e., rate = k. The integrated rate law for a zero-order reaction is given by:
[A]t = -kt + [A]0
where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration, k is the rate constant, and t is time.
If a reaction is 50% complete in 30.0 min, it means that [A]t = 0.5[A]0. Substituting these values into the equation above, we can solve for the rate constant:
0.5[A]0 = -k(30.0 min) + [A]0
k = 0.0167 M/min
Now, if the reaction is 75% complete, it means that [A]t = 0.25[A]0. Substituting this value and the rate constant into the equation above, we can solve for the time t:
0.25[A]0 = -0.0167 M/min * t + [A]0
t = 45.0 min
Therefore, for a zero-order reaction, the reaction will be 75% complete after 45.0 min.