Transcript Chemical Kinetics Chapter 15

Goals: Dependence of rate on conc.
Distinguish different rates
How T affects rate
How catalysts affect rate
Determine simple mechanisms
Suggested Problems: 1, 7, 9, 11,
13, 15, 19, 21, 23, 25, 27, 29, 31,
33, 35, 37, 41, 45, 47, 49, 53, 55,
57, 59, 61, 65
Relative Rates of
Change
Your Turn: Show the relative rates
of change for the reaction:
N2O5(solvent) --> 2 NO2(solvent)+ ½ O2(g)
Hint: First, convert to whole number
stoichiometric coefficients.
Not
Given
Relative Rates of
Change
Problem: Show the relative rates of change
for the reaction:
N2O5(solvent) --> 2 NO2(solvent)+ ½ O2(g)
2 N2O5(solvent) --> 4 NO2(solvent)+ O2(g)
1 N2O5
1 NO 2 
O2 
rrt  


2
t
4
t
t
C/t Deriving Rate Laws
Your Turn: Derive rate law and k for
CH3CHO(g) ---> CH4(g) + CO(g)
from experimental data for rate of disappearance
of CH3CHO: Rate of rxn = - k [CH3CHO]?
Expt.
[CH3CHO]
(mol/L)
Disappear of CH3CHO
RATE (mol/L•sec)
1
0.10
0.020
2
0.20
0.081
3
0.30
0.182
4
0.40
0.318
C/t Deriving Rate Laws
Not
Given
Rate of rxn = k [CH3CHO]2
Here the rate goes up by FOUR when initial conc.
doubles. Therefore, we say this reaction is
Second order.
Now determine the value of k. Use expt. #3 data—
0.182 mol/L•s = - k (0.30 mol/L)2
k = - 2.0 (L / mol•s)
Using k you can calc. rate at other values of
[CH3CHO] at same T.
C/t Deriving Rate Laws
Note: In this example the rate of
removal depends on the conc. in soln.
Increase conc. a increase removal
Go back to the ethanol problem, rate is
independent of the presence of ethanol.
So what will happen?
Your Turn: Determine the rate order
expression for the following data set:
Tim e (m in)
0.00
3.35
6.87
10.60
15.80
20.50
30.60
[C]
50.00
13.90
7.91
5.43
3.78
2.96
2.02
ln[C]
1/[C]
C/t Concentration/Time Relations
Tylenol dose is 650 mg
Human body has 5L of
blood. Thus, Initial
[Tylenol] = Co = 8.61 x
10-4 M (confirm this!)
Acetaminophen
(Tylenol)
C8H9O2N
Rate of removal of
Tylenol is k = 0.21 hr-1
What is conc. when you
need to take another
dose in 4 hours?
How long will it take to drop the conc. by 90% of Co
(10% is left or 8.61 x 10-5 M)?
Relationships
C/t Concentration/Time
Rate of removal of Tylenol = k [Tylenol], k = 0.21 hr . If
-1
initial [Tylenol] = 8.61 x 10-4 M: What is the blood conc.
when you take another dose in 4 hrs; How long will it take
to drop the blood conc. by 90% of Co or to 8.61 x 10-5 M?
Use the first order integrated rate law
ln [A]4.0hr = ln[A]o - kt
-4
-1
ln[A]4.0hr = ln[8.61 x 10 ] - 0.21 hr * 4 hr
ln[A]4.0hr = - 7.9
e
ln[A]4.0hr
= [A]4.0hr = e
-7.9
= 3.72 x 10
-4
M
Relationships
C/t Concentration/Time
Rate of removal of Tylenol = k [Tylenol], k = 0.21 hr . If
-1
initial [Tylenol] = 8.61 x 10-4 M: What is the blood conc.
when you take another dose in 4 hrs; How long will it take
to drop the blood conc. by 90% of Co or to 8.61 x 10-5 M?
Use the first order integrated rate law
[A]t
ln
= - kt
[A]o
ln
8.61 x 10
-5
-4
8.61 x 10
t = 10.96 hr
= ln(0.10) = - 2.30 = -0.21 hr
-1
t
Your Turn
• If k = 0.1 min-1 what is half-life?
• If half-live = 1000 years, what is k?
• THERE IS NO ANSWER TO THIS PROBLEM
ON THE WEB
Pharmacology /
Forensic Science
Your Turn: You are working as a technician for
the State crime lab and you receive a urine sample for
cocaine analysis. The sample is from a driver who fled the
scene of an accident and was apprehended 3 hours later.
It has been 5 hours after the accident when you obtain the
sample and you find a concentration of 6.04 x 10-7 M. The
average removal rate of cocaine from the human body is
0.8 hr-1. The court decides that a concentration above
3.3 x 10-6 M will impair the driver. What was the cocaine
concentration in the driver at the time of the accident?
Your Turn: You are working as a technician for
the State crime lab and you receive a urine sample for
cocaine analysis. The sample is from a driver who fled the
scene of an accident and was apprehended 3 hours later.
It has been 5 hours after the accident when you obtain the
sample and you find a concentration of 6.04 x 10-7 M. The
average removal rate of cocaine from the human body is
0.8 hr-1. The court decides that a concentration above
3.3 x 10-6 M will impair the driver. What was the cocaine
concentration in the driver at the time of the accident?
ln[A]t = ln[A]o - kt
ln[A]o = ln[A]t + kt
ln[A]o = ln[6.04 x 10-7 ] + 0.8 hr -1 * 5 hr
ln[A]o = -14.32 + 4 = -10.32
[A]o = e -10.32 = 3.3 x 10-5 M
Natural Formation of
Radioactive Elements: 146C
N2 in the atmosphere is bombarded by
cosmic radiation. One type of radiation
is neutrons.
14 N + 1 n ===> 14 C + 1 H
7
0
6
1
14
6C
formed in this reaction, goes on to
form radioactive CO2…which is taken
up by plants….which is eaten by
animals….which are eaten by more
animals. With time 146C decays.
14
6C ===>
0
-1b +
14
7N
Half-Life Problem
You are working as an anthropologist, and
you find a well-preserved burial ground
containing wooden artifacts. You have
some of these dated using a 14C dating
technique and the results show that 35%
of the original 14C remains. Given that
the half-live of 14C is 5730 years, what
is the age of the burial site?
ln([A]t/[A]o)=-kt
Hint: calc. k using t½, then calc. age
Half-Life Problem
Soln:
ln([A]t/[A]o)=-kt
ln(0.5)=-k (5735yr)
1.21 x 10-4 /yr= k
ln(0.35) = - (1.21 x 10-4 /yr)t
age = t = 8686 years