Question

Part A: The rate constant for a certain reaction is k = 5.00×10−3 s−1 . If the initial reactant concentration was 0.250 M, what will the concentration be after 2.00 minutes?

Part B: A zero-order reaction has a constant rate of 3.40×10−4 M/s. If after 80.0 seconds the concentration has dropped to 6.00×10−2 M, what was the initial concentration?

Answer 1

Part A

he has given the rate constant k = 5.00×10−3 s⁻¹

units of the rate constant = s^-1 that means it is first order

take th first order rate constant

k = 2.303 / t log[A0/A]

A0 is initial concentration

A = concentration after 2 minitus

t he has given in minitues convert in to sec 2 min = 2 x 60 = 120 sec

now all the values are ther

5.00 x 10^-3 = 2.303 / 120 log[0.250/A]

log[0.250/A] = 0.26

0.25 / A = 1.82

A = 0.140 M

paer B

rate equation for the zero order reaction

A = [A0]/kt

k = rate constant = 3.40×10−4 M/s

t = time = 80.0 sec

A0 = intal concentration need to find out

A = concentration after 80 sec time = 6.00×10−2 M

put all these values in above equations

6.00×10−2 M = [A0] - 3.40×10−4 x 80

[A0] = 0.06 + 0.0272

[A0] = 8.72 x 10^-2 M