Post on 02-May-2023
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Chapter 13
Chemical Kinetics
Why Study Kinetics?
• We can control the reaction conditions to obtain product as quickly and economically as possible
• To understand reaction mechanisms – a study of kinetics sheds light on how a reaction proceeds
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Factors that control reaction kinetics -
• Chemical makeup of reactants and products
• Concentration of reactants• Temperature• Catalysts• Surface area
Reaction Rates
If driving a car, rate = change in positionchange in time
Rate of a chemicalreaction
change in concentrationchange in time
=
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Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
time
393 nmlight
Detector
∆[Br2] α ∆Absorption
393
nm
Br2 (aq)
Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
average rate = -∆[Br2]∆t
= -[Br2]final – [Br2]initial
tfinal - tinitial
slope oftangent
slope oftangent slope of
tangent
instantaneous rate = rate for specific instance in time
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Reaction Rates and Stoichiometry
2A B
Two moles of A disappear for each mole of B that is formed, so the rate of disappearance of A is twice the rate of appearance of B.
Mathematically:
aA + bB cC + dD
In general -
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Example 13.1 – Write the rate expression for the following reactions in terms of the disappearance of the reactants and the appearance of the products:
(a) I-(aq) + OCl- → Cl-(aq) + OI-(aq)
(b) 3O2(g) → 2O3(g)
(c) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
The Rate LawWe observe in chemical reactions that rates decrease as the reactant concentrations decrease. The rate is found to be proportional to the reactant concentrations raised to some experimentally determined power -
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Reaction OrderThe sum of the powers that each reactant concentration is raised to in the Rate Law.
Rate = k[A]
Rate = k[A][B]
Rate = k[A]2[B]
The Rate Law for a reaction can be determined using the Method of Initial Rates -
Time →
Con
cent
ratio
n →
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F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
Rate doubles
x = 1
Quadruple [ClO2] with [F2] constant
Rate quadruples
y = 1
rate = k [F2][ClO2]
F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2][ClO2]
Rate Laws
• Rate laws are always determined experimentally.
• Reaction order is always defined in terms of reactant (not product) concentrations.
• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
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Example 13.3 – The reaction of nitric oxide with hydrogen at 1280 oC is
2NO(g) + 2H2(g) → N2(g) + 2H2(g)
From the following data collected at this temperature, determine (a) the rate law, (b) the rate constant, and (c) the rate of the reaction when [NO] = 12.0 x 10-3 M and [H2] = 6.0 x 10-3 M.
Experiment [NO] [H2] Initial Rate (M/s)
1 5.0 x 10-3 2.0 x 10-3 1.3 x 10-5
2 10.0 x 10-3 2.0 x 10-3 5.0 x 10-5
3 10.0 x 10-3 4.0 x 10-3 10.0 x 10-5
Relation Between Reactant Concentration and Time:Integrated Rate Laws
I. First Order Kinetics: rate = k1[A]
II. Second Order Kinetics: rate = k2[A]2
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First Order Kinetics
rate = - ∆[A]
∆t= k1[A]
A (+ other reactants) → products
ln[A]t
[A]o= -k1t
First Order Integrated Rate Equation
[A]t = concentration of A at some time = t
[A]o = concentration of A at time t = 0 (initial concentration)
k1 = first order rate constant (units = s-1, min-1, h-1, etc)
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Some mathematics:
Example 13.4 – The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7 x 10-4
s-1 at 500 oC. (a) If the initial concentration of cyclopropane is 0.25 M, what is the concentration after 8.8 min? (b) How long will ittake for the concentration of cyclopropane to decrease from 0.25 to 0.15 M? (c) How long will it take to convert 74 % of the starting material?
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Example 13.4 – solution
Graphical Methods for Determining Reaction Order and the Rate Constant
First Order Kinetics
ln[A]t
[A]o= -k1t
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[A] = [A]0exp(-kt) ln[A] = ln[A]0 - kt
For the reaction 2 N2O5 → 4 NO2(g) + O2(g) graphically determine the order and rate constant for the reaction given thefollowing data obtained at 45 oC:
t (s) [N2O5] ln [N2O5]
0 0.91
300 0.75
600 0.64
1200 0.44
3000 0.16
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Half Life and First Order Reactions
half life - the time it takes for half of the substance to react
[A]t = half life = ½[A]o ln[A]t
[A]o= -k1t
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Example 13.6 - The decomposition of ethane (C2H6) to methyl radicals is a first-order reaction with a rate constant of 5.36 x 10-4
s-1 at 700 oC:
C2H6(g) → 2 CH3(g)
Calculate the half-life of the reaction in minutes.
Some typical half lives for First Order processes -
Radioactive decay of 238U 4.51 X 109 yr 4.87 X 10-18
Radioactive decay of 14C 5.73 X 103 yr 3.83 X 10-12
C12H22O11(aq) + H2O → 8.4 h 2.3 X 10-6
C6H12O6(aq) + C6H12O6(aq)(glucose) (fructose)HC2H3O2 + H2O → 8.9 X 10-7 s 7.8 X 105
H3O+ + C2H3O2-
Process Half life, t1/2 Rate constant, k (s-1)