false

Class Notes
(835,065)

Canada
(508,908)

University of Alberta
(13,379)

Chemistry
(644)

CHEM102
(133)

Sai Yiu
(57)

Lecture

Unlock Document

Description

UNIVERSITY OF ALBERTA
INTRODUCTORY UNIVERSITY CHEMISTRY II (CHEM 102 / 105)
Chemical kinetics
You may find the following mathematical formulas useful in this course:
If a = N, x = log a
a0 = 1, log 1 = 0
log M = log M – log N log MN = log M + log N
N log xy= y log x
1
∫ dx = ln x + constant
x
xn+1
n
∫ x dx = + constant
n+1
Introduction
Thermodynamics
Energy Reactants
Thermodynamics
Kinetics
Products
Extent of a chemical reaction
Reaction coordinate of a chemical reaction
Thermodynamics gives the energy change and the extent (chemical equilibrium)
of a chemical reaction
Chemical kinetics concerns about the reaction rates.
The two are totally independent on each other.
The area of chemistry that concerns reaction rates is called chemical kinetics.
1 Rates of reaction
For a chemical reaction such as:
2NO (g) 2NO (g) + O (g) at 300 C o
2 2
A graph of the concentration of NO 2, NO and O p2otted against time:
5
[NO]
4 [NO 2
3
[O 2
2
0.5 x 10 M
1 70 s
0 300
100 200
Time (s)
From the graph, we can see that:
(i) concentration of NO d2creases with time (Not linear).
(ii) concentration of O a2d NO increase with time. (Again they are not linear).
(iii) concentration of NO increases more (in fact doubles that of the O2.)
Change in amount (or concentration ) of substance A
Reaction rate =
Time taken
Concentration of A at t 2 concentration of A at t 1
Reaction rate =
t2– t1
Δ[A]
Reaction rate =
Δt
2 Δ [NO ]2
For NO ,2the rate of reaction = −
Δt
The negative sign indicates concentration decreases with time.
Unit for rate of reaction:
mol L s or M s -1
From the graph; for the first 100 s,
-2 -2
Δ [NO 2 2.2 x 10 M –5.0 x 10 M
- = - = - (-2.8 x 104M s )1
Δt 100. s – 0 s
-4 -1
Therefore, rate of disappearance for NO i2 2.8 x 10 M s
This is the average rate of the reaction within the first 100 seconds.
For NO,
-2
Δ [NO] 2.8 x 10 M – 0 M
Rate of appearance = = = 2.8 x 10 M s -1
Δt 100. s – 0 s
For O 2
Δ [O 2 1.4 x 10 M – 0 M
-4 -1
Rrate of appearance = = = 1.4 x 10 M s
Δt 100. s – 0 s
For this reaction:
Δ[ NO ]2 Δ[NO] Δ[O 2
Rate of reaction = − = =
2 Δt 2 Δt Δt
Generally for a reaction aA + bB cC + dD
Δ[A] Δ[B] Δ [C] Δ [D]
Rate of reaction = − = - = =
a Δt b Δt c Δt d Δt
3 Write the rate expressions for the following reaction in terms of the disappearance
of the reactants and the appearance of the products:
3O 2 (g) 2O 3g)
If the rate of disappearance of O 2is 0.025 mol / L s, what is the rate of appearance
of O 3
Since concentration change is not linear, reaction rate is not constant (it changes
with time)
To measure reaction rate, a tangent is drawn and the gradient or slope gives the
instantaneous rate at that specific time.
From the graph, the instantaneous rate of disappearance for NO 2 when t = 170 s is:
d [NO ] - 0.50 x 10 M2
2 -5 -1
- = - = 7.1 x 10 M s
dt 70. s
The rate laws (effect of concentration on rate)
For a reaction such as the decomposition of nitrogen dioxide
2NO (2) 2NO (g) + O (g2
the reaction rate is dependent on the concentration of the reactant, NO by 2
certain mathematical relationship. We may therefore write
n
Rate of reaction α [NO ] 2
n
Rate of reaction = k[NO ] 2
k = rate constant of the reaction.
• It is a constant for a specific reaction and is dependent on the temperature.
• The faster is a reaction, the higher is the value k.
• The unit of k is dependent on the overall order of reaction.
4 n = the order of the reaction (or reaction order) with respect to NO 2
• It can have a value of 0 (zero order) or any integers such as 1(first order), 2
(second order) or 3 (third order).
• It can have negative value such as n = -1
• It can have fractions, for example, n = 0.5
The expression: Rate of reaction = k[NO ] 2 n is the rate law or the rate equation.
A rate law is an expression relating the rate of a reaction to the rate constant and
the concentrations of the reactants.
It is important to know that the order n does not have to relate to the
stoichiometry of the chemical equation and must be determined
experimentally.
Measurement of reaction order provides important insight into the molecular
detail (mechanism) of a reaction.
For reaction, aA + bB cC + dD
Rate of reaction = k[A] m[B] n
m + n = overall order of the reaction
For example:
2NO (g) + 2H (g)2 N 2g) + 2H O 2l)
2
Rate law determined is: Rate = k [NO] [H ] 2
nd
The reaction is 2 order with respect to NO
1 order with respect to H 2
rd
The overall order of the reaction is 2 + 1 = 3 (3 order)
Experimental determination of rate laws (The Initial rates)
For example: The decomposition of hydrogen peroxide
2H 2 2aq) 2H O2(l) + O (2)
5
4 [H22 ]
mol/L)
-2
3
Initial rate
2
Concentration (x 10 Reaction rate at 200s
1
0
100 5 200 300
Time (s) For a reaction with more than one reactant, the rate law can be determined by:
Fixing the concentrations of all reactants except one and record the initial rate of
the reaction as a function of the concentration of that reactant. Any changes in
reaction rate must be due to that reactant.
An example is given below:
A + 2B C
The following data were observed:
Experiment Initial [A] (M) Initial [B] (M) Initial rate (M / s)
1 0.100 0.100 5.50 x 10-6
-5
2 0.200 0.100 2.20 x 10
3 0.400 0.100 8.80 x 10-5
-5
4 0.100 0.300 1.65 x 10
5 0.100 0.600 3.30 x 10-5
Let the rate law be:
rate of reaction = k[A] [B]
Substituting into experiment 1
6 **Note that the unit of k is dependent on overall orders
Thus the reaction is First order for B and Second order for A and the overall order of
reaction is 1 + 2 = 3. We say that it is a Third order reaction.
Homework:
o
1 The following is the data of a reaction at 800 C. :
2H (g) + 2NO(g) 2H O(g) + N (g)
2 2 2
Determine the rate law and the rate constant of the reaction.
Experiment Initial [2] (M) Initial [NO] (M) Initial rate (M / min)
1 0.50 1.50 4.2 x 10 -3
-2
2 1.50 1.50 1.3 x 10
3 3.00 3.00 5.2 x 10 -2
2 For a reaction: A + 2B C
If the rate law is
o
Rate = k [A] [B]
What does it mean for a zero order for A? What is the overall order of reaction?
7 Integrated rate laws: Concentration changes over time
There are two types of rate law:
1 The experimental rate law (differential rate law) expresses how the rate
depends on the concentration.
Rate = k [A] x[B] y
2 The integrated rate law expresses how the concentration of the reactants
changes with time.
From the experimental rate law we know the reaction order of the reactants. By
using integral calculus we can derive the integrated rate law for different orders of
reactions
Zero order reactions
Suppose a reaction is zero order with respect to A in the reaction
aA products
Rate = - d[A] / dt = k[A] o
o
-d[A] = k[A] dt
-d[A] = k dt
∫ -d[A] = k∫ dt
-[A] = kt + constant
[A] = - k t + constant
When t = 0, [A] = constant
But when t = 0, [A] = [A] =oinitial concentration. And so constant = [A] o
[A] = - k t + [Ao
8 The slope gives the rate constant, k of a zero order of reaction.
Half–life of zero order reaction
The half-life ‰t ) of a reaction is the time required for a reactant so that half its
original concentration is reacted.
When [A] = ‰ [A] o t = t‰
Substituting into the zero order rate equation gives
‰ [A]o= - k‰t + [Ao
‰ [A]o= k t‰
[A]o
t = _____
‰
2 k
Thus, half-life of zero order reaction is dependent on initial concentration.
9 First order reactions
Rate = - d[A] / dt = k[A]
-d[A] / [A]= k dt
∫ -d[A] / [A] = k ∫ dt
- ln [A] = kt + constant
ln [A] = - kt + constant
When t = 0, ln [A] = constant
But when t = 0, [A] = [A] o
Therefore ln [A] o = constant
ln [A] = - kt + ln [A]
o
Another form of first order reaction:
-kt
[A] = [A] eo
Slope = rate constant k
A plot of ln [A] versus time
for 1 order reaction
(Fig 16.8B p702)
10 Half–life of first order reaction
The half-life of A from the above graph is 7 seconds.
If we take the initial concentration to be 0.10 M, what is its half-life?
We could pick any initial concentration and find that the half-life is still 7 seconds.
Hence we may say that for first order reactions, the half-life is independent on
initial concentrations.
From the equation for first order reaction
[A] = [A] o e –kt
When [A] = ‰ [A] 0 t = t‰
‰ [A] = [A] e –kt 1/2
o o
e –kt ‰= ‰
k t‰ = ln 2 = 0.693
ln 2 0.693
t‰ = _______ = __________
k k
Hence from the half-life of a first order reaction, we can find the rate constant, k.
Thus, the rate constant for the above reaction is:
Second order reactions
For a reaction aA products
Rate = - d[A] / dt = k[A] 2
2
-d[A] / [A] = k dt
2
-∫ d[A] / [A] = k ∫dt
11 1/ [A] + constant = kt
The integrated rate
When t = 0, constant = 1 / [A] law is:
But when t = 0, [A] = [A] o
1 1
Therefore, constant = 1 / [A] = + kt
o [A] [A]
o
Graph of 1 / [A] Vs time for a
second order reaction gives a
straight line with the slope = k
(Fig 116.7B p701)
The 2 order reaction equation:
1 1
= + kt
[A] [A] o
At half-life, [A] = ‰ [Ao
We can substitute into the equation as:
2 1
= + kt1/2
[A]o [A] o
2 - 1 1
kt1/2 = 1
[A] o [A] o Half-life (t1/2 = ______
1 k [A]
o
t1/2=
k [A] o
12 Thus, half-life is dependent on initial concentration.
Summary of reaction kinetics
Order of Integrated Plot for
Unit for k Slope Half-life
reaction Rate law rate equation straight line
Mol L -1s -1 - k [A]o
0 Rate = k [A]t= - kt + [Ao [A] Vs time 2k
ln 2
1 Rate = k[A] s-1 ln[A]t= - kt + ln[A]o ln[A] Vs time - k
k
1 1 1
2 Rate = k[A] 2 L Mol -1s-1 = kt + Vs time k 1
[A] [A] k[A]
t o [A] o
Work example
-4
The decomposition of H O f2ll2ws first-order kinetics. If the rate constant is 7.30 x 10
s , what percentage of H O 2as2decomposed in the first 500.0 s after the reaction?
Effect of temperature on reaction rates
Rate of reaction is sensitive to change in temperature. In most reactions:
• The change in rate constant, k to the temperature (K) is exponential
• For many reactions, an increase of 10 C in temperature doubles the
reaction rate.
13 Fig 16.10 p705
Why does a small rise in temperature result in such a large increase in reaction
rate?
The Arrhenius equation
Put forward by Arrhenius in 1889, it is in the form of
-Ea/RT
k = Ae
Equation links the rate constant k with activation energy, E aand temperature, T
A = Arrhenius constant and represents collision frequency factor. (How easy do
the molecules collide with each other.)
E a activation energy of the reaction /J mol -1
e = base of natural logarithm and is equal to 2.718
T = absolute temperature (K)
R = gas constant and is equal to 8.314 J K mol-1 -1
Activation energy
The minimum energy the molecules must possess for a chemical reaction to take
place is called the activation energy.
14 Exothermic reaction
Endothermic reaction
Ea
Potential Potential
energy Reactants energy Products
Ea
Products Reactants
Progress of reaction Progress of reaction
Consequently, not all molecular collisions (Collision Theory) lead to a reaction.
They must have sufficient energy to overcome the activation energy barrier before
forming products in a chemical reaction.
Some significant facts derived from the Arrhenius equation:
There are 3 factors affecting the rate constant, k
•

More
Less
Related notes for CHEM102

Join OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.