12.4 Integrated rate law

Question 1

What is the difference between the differential rate law and the integrated rate law?

Answer 1

differential rate law: rate as a function of concentration

integrated rate law: concentration as a function of time

Question 2

What is the integrated rate law of first order reaction

Answer 2

ln [A] = -kt + ln [A]o

Question 3

2 H2O2(aq) → 2 H2O(l) + O2 (g)
The reaction is 1st order in H2O2, the rate constant for consumption of H2O2 at 20 °C is 1.8*10-5 s-1 and [H2O2]0 = 0.30M
a) What is the concentration of H2O2 after 4.00h?
b) How long will it take for the H2O2 concentration to drop to 0.12M? c) How long will it take for 90% of the H2O2 to decompose?

Answer 3

a) [H2O2]t = (0.30M)(0.772) = 0.23 M
b) 5.1 104 s (14hours)
c) [H2O2]t / [H2O2]0 =(0.10)(.30M)/(0.30M) = 0.10
t= - {1/(1.810-5 s-1)} ( Ln0.10) = 1.3*105 s

Question 4

What type of slope is are the reaction? What does the slope say about the order?

Answer 4

all 3 order equation can be compared to y=ax+b. Which ever equation makes a straight line is the order of the reaction

Question 5

How will we calculate the half life of 1st order?

Answer 5

make [A]t=[Ao]/2

THE t1/2 = 0.693/k!!

Question 6

A certain 1st order reaction has 1⁄2 life of 20.0 minutes:

a) Calculate the rate cte for this reaction?
b) How much time is required for this reaction to be 75% complete?

Answer 6

a) t1/2 = 0.693 / k → k=3.4710-2 min-1
b) use Ln { [A]0/[A]} = kt → [A]= 0.25[ A]0
t= Ln (4.0) / (3.4710-2/min)= 40 min

Question 7

What is the integrated rate law of second order reaction

Answer 7

1/ [A]t = kt + 1/[A]o

Question 8

half life equation for 2nd order

Answer 8

t 1⁄2 = 1 / k [A]0

Question 9

Time(s)- [NO2]
0-8.0010-3
50-6.5810-3
100-5.5910-3
150-4.8510-3
200-4.2910-3
300-3.4810-3
400-2.9310-3
500-2.5310-3
a) What is the value of the rate cte(k) ?
b) What is the conc of NO2 at t=20.0 min?
c)What is the 1⁄2 life of the reaction when the initial conc of [NO2] is 6.0010-3 M ?
d) What is the t 1⁄2 when [NO2] = 3.0010-3 M?

Answer 9

a) k= slope= Δy/Δx =( 340M- 1 -150 M-1) / (400s -50s) = 0.54 M-1.s-1
b) 1/[NO2]t = kt + 1/[NO2] 0
t= 20.0 min (1.20103 s) as [NO2]0 = 8.0010-3 M
k= 0.54 M-1.s-1
[NO2]t = 1.310-3 M
c) t 1/2= 1/k[NO2]0 = 1 / (0.54/M.s) (6.0010-3 M) = 3.1 102 s
d) When [NO2]0 = 3.0010-3 M we find from the previous equation t 1⁄2 = 6.2102 s
Twice as long as when [NO2]0 was 6.00*10-3 M because [NO2]0 is smaller by a factor of 2.

Question 10

What is the integrated rate law of 0 order reaction

Answer 10

[A] = - kt + [A]0