Hoa dai cuong_chuong 15

Philip Dutton
University of Windsor, Canada
N9B 3P4

Prentice-Hall © 2002


General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 15: Chemical Kinetics
Contents
15-1 The Rate of a Chemical Reaction
15-2 Measuring Reaction Rates
15-3 Effect of Concentration on Reaction Rates: The Rate Law
15-4 Zero-Order Reactions
15-5 First-Order Reactions
15-6 Second-Order Reactions
15-7 Reaction Kinetics: A Summary
Contents
15-8 Theoretical Models for Chemical Kinetics
15-9 The Effect of Temperature on Reaction Rates
15-10 Reaction Mechanisms
15-11 Catalysis
Focus On Combustion and Explosions
15-1 The Rate of a Chemical Reaction
Rate of change of concentration with time.
2 Fe3+(aq) + Sn2+ → 2 Fe2+(aq) + Sn4+(aq)
t = 38.5 s [Fe2+] = 0.0010 M
Δt = 38.5 s Δ[Fe2+] = (0.0010 – 0) M
Rates of Chemical Reaction
2 Fe3+(aq) + Sn2+ → 2 Fe2+(aq) + Sn4+(aq)
General Rate of Reaction
a A + b B → c C + d D
Rate of reaction = rate of disappearance of reactants
= rate of appearance of products
15-2 Measuring Reaction Rates
H2O2(aq) → H2O(l) + ½ O2(g)
2 MnO4-(aq) + 5 H2O2(aq) + 6 H+ →
2 Mn2+ + 8 H2O(l) + 5 O2(g)
Example 15-2
-(-1.7 M / 2600 s) =
6  10-4 M s-1
-(-2.32 M / 1360 s) = 1.7  10-3 M s-1
Determining and Using an Initial Rate of Reaction.
Example 15-2
= 2.17 M
What is the concentration at 100s?
[H2O2]i = 2.32 M
15-3 Effect of Concentration on Reaction Rates: The Rate Law
a A + b B …. → g G + h H ….
Rate of reaction = k [A]m[B]n ….
Rate constant = k
Overall order of reaction = m + n + ….
Example 15-3 Method of Initial Rates
Establishing the Order of a reaction by the Method of Initial Rates.
Use the data provided establish the order of the reaction with respect to HgCl2 and C2O22- and also the overall order of the reaction.
Example 15-3
Notice that concentration changes between reactions are by a factor of 2.
Write and take ratios of rate laws taking this into account.
Example 15-3
R2 = k[HgCl2]2m[C2O42-]2n
R3 = k[HgCl2]3m[C2O42-]3n
2m = 2.0 therefore m = 1.0
= k(2[HgCl2]3)m[C2O42-]3n
Example 15-3
R2 = k[HgCl2]21[C2O42-]2n = k(0.105)(0.30)n
R1 = k[HgCl2]11[C2O42-]1n = k(0.105)(0.15)n
2n = 3.98 therefore n = 2.0
+ = Third Order
R2 = k[HgCl2]2 [C2O42-]2
First order
Example 15-3
1
Second order
2
15-4 Zero-Order Reactions
A → products
Rrxn = k [A]0
Rrxn = k
[k] = mol L-1 s-1
Integrated Rate Law
-[A]t + [A]0 = kt
[A]t = [A]0 - kt
Δt
-Δ[A]
= k
And integrate from 0 to time t
15-5 First-Order Reactions
H2O2(aq) → H2O(l) + ½ O2(g)
= -k [H2O2]
d[H2O2 ]
dt
[k] = s-1
First-Order Reactions
Half-Life
t½ is the time taken for one-half of a reactant to be consumed.
- ln 2 = -kt½
Half-Life
ButOOBut(g) → 2 CH3CO(g) + C2H4(g)
Some Typical First-Order Processes
15-6 Second-Order Reactions
Rate law where sum of exponents m + n +… = 2.

A → products
Second-Order Reaction
Pseudo First-Order Reactions
Simplify the kinetics of complex reactions
Rate laws become easier to work with.

CH3CO2C2H5 + H2O → CH3CO2H + C2H5OH
If the concentration of water does not change appreciably during the reaction.
Rate law appears to be first order.
Typically hold one or more reactants constant by using high concentrations and low concentrations of the reactants under study.
Testing for a Rate Law
Plot [A] vs t.
Plot ln[A] vs t.
Plot 1/[A] vs t.
15-7 Reaction Kinetics: A Summary
Calculate the rate of a reaction from a known rate law using:


Determine the instantaneous rate of the reaction by:



Rate of reaction = k [A]m[B]n ….
Finding the slope of the tangent line of [A] vs t or,
Evaluate –Δ[A]/Δt, with a short Δt interval.
Summary of Kinetics
Determine the order of reaction by:
Using the method of initial rates.
Find the graph that yields a straight line.
Test for the half-life to find first order reactions.
Substitute data into integrated rate laws to find the rate law that gives a consistent value of k.
Summary of Kinetics
Find the rate constant k by:





Find reactant concentrations or times for certain conditions using the integrated rate law after determining k.
Determining the slope of a straight line graph.
Evaluating k with the integrated rate law.
Measuring the half life of first-order reactions.
15-8 Theoretical Models for
Chemical Kinetics
Kinetic-Molecular theory can be used to calculate the collision frequency.
In gases 1030 collisions per second.
If each collision produced a reaction, the rate would be about 106 M s-1.
Actual rates are on the order of 104 M s-1.
Still a very rapid rate.
Only a fraction of collisions yield a reaction.
Collision Theory
Activation Energy
For a reaction to occur there must be a redistribution of energy sufficient to break certain bonds in the reacting molecule(s).

Activation Energy is:
The minimum energy above the average kinetic energy that molecules must bring to their collisions for a chemical reaction to occur.
Activation Energy
Kinetic Energy
Collision Theory
If activation barrier is high, only a few molecules have sufficient kinetic energy and the reaction is slower.

As temperature increases, reaction rate increases.

Orientation of molecules may be important.
Collision Theory
Transition State Theory
The activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state.
15-9 Effect of Temperature on
Reaction Rates
Svante Arrhenius demonstrated that many rate constants vary with temperature according to the equation:
k = Ae-Ea/RT
ln k = + ln A
R
-Ea
T
1
Arrhenius Plot
N2O5(CCl4) → N2O4(CCl4) + ½ O2(g)
-Ea = 1.0102 kJ mol-1
Arrhenius Equation
k = Ae-Ea/RT
15-10 Reaction Mechanisms
A step-by-step description of a chemical reaction.
Each step is called an elementary process.
Any molecular event that significantly alters a molecules energy of geometry or produces a new molecule.
Reaction mechanism must be consistent with:
Stoichiometry for the overall reaction.
The experimentally determined rate law.
Elementary Processes
Unimolecular or bimolecular.
Exponents for concentration terms are the same as the stoichiometric factors for the elementary process.
Elementary processes are reversible.
Intermediates are produced in one elementary process and consumed in another.
One elementary step is usually slower than all the others and is known as the rate determining step.
A Rate Determining Step
Slow Step Followed by a Fast Step
H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)
dt
= k[H2][ICl]
d[P]
Postulate a mechanism:
H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)
slow
fast
Slow Step Followed by a Fast Step
Fast Reversible Step Followed by a Slow Step
2NO(g) + O2(g) → 2 NO2(g)
Postulate a mechanism:
2NO(g) + O2(g) → 2 NO2(g)
The Steady State Approximation
The Steady State Approximation
k1[NO]2 = [N2O2](k2 + k3[O2])
Kinetic Consequences of Assumptions
dt
d[NO2]
k1k3[NO]2[O2]
=
(k2 + k3[O2])
11-5 Catalysis
Alternative reaction pathway of lower energy.
Homogeneous catalysis.
All species in the reaction are in solution.
Heterogeneous catalysis.
The catalyst is in the solid state.
Reactants from gas or solution phase are adsorbed.
Active sites on the catalytic surface are important.
11-5 Catalysis
Catalysis on a Surface
Enzyme Catalysis
Saturation Kinetics
k1[E][S] = (k-1+k2 )[ES]
[E] = [E]0 – [ES]
k1[S]([E]0 –[ES]) = (k-1+k2 )[ES]
Michaelis-Menten
Chapter 15 Questions

Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.

Choose a variety of problems from the text as examples.

Practice good techniques and get coaching from people who have been here before.