1. Factors that Influence Reaction Rate Expressing the Reaction Rate
Average, Instantaneous, Initial Reaction rates
Rate & Concentration
The Rate Law and its Components
Determining the Initial Rate
Reaction Order Terminology
Determining Reaction Orders
Determining the Rate Constant
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2. Integrated Rate Laws: Concentration Changes Over Time
First, Second, and Zero-Order Reactions
Reaction Order
Reaction Half-Life
The Effect of Temperature on Reaction Rate
Explaining the Effects of Concentration and Temperature
Collision Theory
Transition State Theory
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3. Reaction Mechanisms: Steps in the Overall reaction
Elementary Reactions
The Rate-Determining Step
The Mechanism and the Rate Law
Catalysis: Speeding up a Chemical Reaction
Homogeneous Catalysis
Heterogeneous Catalysis
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4. Chemical Kinetics Chemical Kinetics is the study of:
Chemical Reaction Rates
The changes in chemical concentration of reactants as a function of time
Chemical reactions range from:
Very Fast to Very Slow
Under a given set of conditions each reaction has its own rate
Factors that influence reaction rate:
Concentration
Physical state (surface area)
Temperature (frequency & energy of particle collisions)
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5. Rate  Collision Frequency  Concentration
Chemical Kinetics
Factors That Influence Reaction Rate
Concentration
Molecules must Collide to React
Reaction rate is proportional to the concentration of the reactants
Rate  Collision Frequency  Concentration
Physical State
Molecules must Mix to Collide
The more finely divided a solid or liquid reactant:
The greater its surface area per unit volume
The more contact it makes with the other reactants
The faster the reaction occurs
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6. Rate  Collision Energy  Temperature
Chemical Kinetics
Temperature
Molecules must collide with enough energy to react
At a higher temperature, more collisions occur in a given time
Raising the temperature increases the reaction rate by increasing the number and energy of the collisions
Rate  Collision Energy  Temperature
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7. “how fast does the reaction occur?”
Chemical Kinetics
A fundamental question addressed in chemical reactions is
“how fast does the reaction occur?”
Kinetics is the study of the rate of chemical reactions
rate is a time dependent process
Rate units are concentration over time
Consider the reaction A  B
Reactant concentrations [A] decrease while product concentrations [B] increase
Note: Reaction rate is positive, but the concentration of A at t2 (A2) is always less than the concentration of A at t1 (A1), thus, the change in concentration (final – initial) of reactant A is always negative
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8. Chemical Kinetics Consider the reaction: A + B  C
Concentrations of both reactants ([A] & [B]) decrease at the same rate
d Indicates “Change in” (final - initial)
Brackets [ ] indicate concentration
Note minus sign in front of term reflecting the decrease in concentration with time
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9. Chemical Kinetics Reaction - Butyl Chloride (C4H9Cl) and Water (H2O)
CH3(CH2)2CH2-Cl(l) + H2O(l)  CH3(CH2)2CH2-OH(l) + HCl
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10. Chemical Kinetics Butyl Chloride (C4H9Cl)
When plotting Concentration versus Time for a chemical reaction, the tangent at any point on the curve (drawn through the concentration points) defines the instantaneous rate of the reaction
The average rate of a reaction over some time interval is determined through triangulation of concentration plot (slope of hypotenuse of right triangle)
The rate of the reaction decreases over time as the reactants are consumed
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11. Chemical Kinetics Rate of reaction of the Products
The rate of reaction for the formation of the products is the same as for the reactants, but opposite, that is the concentrations are increasing
The rate of change of Ethane (C2H4) and Ozone (O3) is the same, but exactly opposite for Acetaldehyde (C2H4O) and Oxygen (O2)
Product concentration increases at the same rate that the reactant concentrations decrease
The curves have the same shape, but are inverted
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12. Chemical Kinetics
The Rate expression must be consistent with stoichiometry
When the stoichiometric molar ratios are not 1:1, the reactants still disappear and the products distill appear, but at different rates
For every molecule of H2 that disappears, one molecule of I2 disappears and 2 molecules of HI appear
The rate of H2 decrease is the same as the rate of I2 decrease, but both are only half the rate of HI increase
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13. Chemical Kinetics Summary equation for any reaction
The rate of a reaction is dependent on the concentration of reactants
The average reaction rate is the change in reactant (or) product concentration over a change in time, t
The instantaneous rate at a time, t, is obtained from the slope of the tangent to a concentration vs. time curve at a given time, t
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14. Chemical Kinetics
As reactant concentrations decrease, the reaction rates decrease with time
Product concentrations increase at the same rate as the reactants relative to the stoichiometric ratios
The rate of a reaction depends on the following variables:
reactant concentration
temperature
presence and concentration of a catalyst
surface area of solids, liquids or catalysts
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15. Sample Problem
Write an expression defining equivalent rates for the loss of NO2 and the formation of NO in the following reaction with respect to the rate of formation of O2 2 NO2(g)  2 NO(g) + O2(g)
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16. Chemical Kinetics –The Rate Law
The dependence of reaction rate on concentrations is expressed mathematically by the rate law
The rate law expresses the rate as a function of reactant concentrations, product concentrations, and temperature
In the following development:
Only the Reactants Appear in the Rate Law
For a general reaction at a fixed temperature:
aA + bB + …  cC dD …
the rate law has the form:
Rate = k[A]m[B]n ...
Note: The Stoichiometric Coefficients – a, b, c – are not used in the rate law equation; they are not related to the reaction order terms – m, n, p, etc
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17. Rate Law Components of the rate law aA + bB + … products
Rate = -[A]/t = k[A]m[B]n[C]p
[A] & [B] = concentrations of reactants (M)
[C] = concentration of catalyst (M, if used)
k = rate constant
m, n, & p = Reaction Orders
Note: Reaction Orders are not related to the Stoichiometric coefficients in the chemical equation
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18. Chemical Kinetics The Rate Law
The rate constant “k” is a proportionality constant
“k” changes with temperature; thus it determines how temperature affects the rate of the reaction
The exponents (m, n, p, etc.) are called reaction orders, which must be determined experimentally
Reaction orders define how the rate is affected by the reactant concentration
If the rate doubles when [A] doubles, the rate depends on [A] raised to the first power, i.e.,
m =1 (a 1st order reaction)
If the rate Quadruples when [B] doubles, the rate depends on [B] raised to the second power, i.e.,
n = 2 (a 2nd order reaction)
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19. Rate Constant - Units
Units of the Rate Constant k change depending on the overall Reaction Order
Overall Reaction Order
Units of k (t in seconds)
mol/Ls (or mol L-1 s-1) 1
1/s (or s-1)
L/mols (or L mol -1 s-1) 2
L2 / mol2 s (or L2 mol-2 s-1) 3
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20. The Rate Law The Rate Law
If the rate does not change even though [A] doubles, the rate does not depend on the concentration of A and m = 0
The Stoichiometric coefficients, a, b, c, etc. in the general balanced equation are not necessarily related in any way to the reaction orders m, n, etc.
The components of the Rate Law – rate, reaction orders, rate constant – must be determined experimentally; they cannot be deduced or inferred from the balanced stoichiometric equation
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21. NO(g) + O3(g)  NO2(g) + O2(g)
Rate Law
The Rate Law - Examples
NO(g) + O3(g)  NO2(g) O2(g)
Rate = k[NO]1[O3]1
Reaction is 1st order with respect to NO, m=1
Rate depends on [NO] raised to 1st power
Reaction is 1st order with respect to O3, n=1
The overall reaction is 2nd order:
m + n = = 2
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22. Rate Law
The Rate Law - Examples 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) Rate = k[NO]2[H2]1 Reaction is 2nd order in NO and 1st order in H2 Overall reaction is = 3rd order Note: [NO] coefficient (2) is not related to the [NO] reaction order (2)
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23. (CH3)3C–Br(l) + H2O(l)  (CH3)C–OH(l) + H+(aq) + Br-(aq)
Rate Law
The rate law – Examples
(CH3)3C–Br(l) + H2O(l)  (CH3)C–OH(l) + H+(aq) + Br-(aq)
Rate = k [(CH3)3CBr]1[H2O]0 or
Rate = k [(CH3)3CBr]1
Reaction is first order in 2-bromo-2-methyl propane
Reaction is zero order (n=0) in water [H2O]0
Note: zero order reaction order terms, ex. [H2O]0 can be eliminated from the overall rate equation, i.e. any term raised to the “0” power is equal to 1
[H2O]0 = 1
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24. CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g)
Rate Law
The Rate Law - Examples
CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g)
Rate = k[CHCL3][Cl2]1/2
The reaction order means that the rate depends on the square root of the Chlorine (CL2) concentration
If the initial Cl2 concentration is increased by a factor of 4, while the initial concentration of CHCl3 is kept the same, the rate increases by a factor of 2, the square root of the change in Cl2
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25. Rate Law The Rate Law - Examples
Negative reaction orders are used when the law includes the product(s)
If the O2 concentration doubles, the reaction proceeds at one half (1/2) the rate
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26. Rate Law – Reaction Order
Overall reaction order
Sum of Exponents in Rate Equation
Order of Rxn Possible Expression of Rate Law
1 k[A]
2 k[A]2
2 k[A][B]
3 k[A]2[B]
3 k[A][B][C]
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27. CH3CHO(g)  CH4(g) + CO(g)
Practice Problem
What is the reaction order of Acetaldehyde and the overall order in the following reaction
CH3CHO(g)  CH4(g) + CO(g)
Rate = k [CH3CHO]3/2
Ans:
3/2 order in CH3CHO
Overall order: 3/2
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28. Practice Problem Experiments are performed for the reaction A  B + C
and the rate law has the been determined to be of the form
Rate = k [A]x
Determine the value of the exponent “x” for each of the following:
[A] is tripled and you observe no rate change
Ans: x = 0 k[3A]0
[A] is doubled and the rate doubles
Ans: x = 1 k[2A]1
[A] is tripled and the rate increases by a factor of 27
Ans: x = 3 k[3A]3
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29. Experimental Rate Law
Concentration Exponents (reaction orders) must be determined experimentally because the stoichiometric balanced equation with its reaction coefficients, does not indicate the mechanism of the reaction
Experimentally, the reaction is run with varying concentrations of the reactants, while observing the change in rate over time
The initial rate of reaction is observed, where the rate is linear with time (instantaneous rate = average rate); usually just when the reaction begins
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30. Initial Reactant Concentration (mol/L)
Experimental Rate Law
Initial rates of reaction from experiments on the reaction:
O2(g) + 2NO(g)  2 NO2(g)
Rate = k[O2]m[NO]n
Determine “m” & “n” from experimental data
Initial Reactant Concentration (mol/L)
Experiment O2 NO
Initial Rate
Mol/Ls 1
1.10x10-2
1.30x10-2
3.21x10-3 2
2.20x10-2
6.40x10-3 3
2.60x10-2
12.8x10-3 4
3.30x10-2
9.60x10-3 5
3.90x10-2
28.8x10-3
Rate equations from two applicable experiments are combined,
depending on the reactant order to be determined
cont’d
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31. Experimental Rate Law
Select the 1st two experiments where the effect of doubling the concentration of O2 is observed at constant temperature
K is constant and [NO] does not change
Substitute rate values and concentration values
Reaction is 1st order in O2:
When [O2] doubles, the rate doubles
cont’d
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32. Experimental Rate Law Determining the Rate Constant (k)
The rate data from any one of the experiments in the previous table can be used to compute the rate constant
Using the first experiment:
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33. “What are the concentrations of ‘A’ after ‘y’ minutes of the reaction”
Integrated Rate Laws
Concentration changes over time
Previous notes assume that time is not a variable and the rate or concentration for a reaction is at a given instant in time
By using time as a factor in the reaction, the rate law can be integrated
“How long will it takefor x moles per liter of reactant ‘A’ to be used up?”
“What are the concentrations of ‘A’ after ‘y’ minutes of the reaction”
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34. Integration of Rate Equation
First Order Reaction
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35. Integration of Rate Equation
Second Order Reaction
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36. Integration of Rate Equation
Zero Order Reaction
Any number raised to the “zero” power is equal to 1
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37. Integrated Rate Law
Integrated Rate Law – Straight Line Plot
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38. Integrated Rate Law
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39. First-Order Concentration vs. Time Graphs
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40. Integrated Rate Law – Half-LIfe
Zero Order Reaction Half-Life
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41. Integrated Rate Law 1st Order Reaction Half-Life
The half-life of a reaction is the time required for the reactant concentration to reach ½ its initial value
At fixed conditions, the half-life of a 1st order reaction is a constant, independent of reactant concentration
[Recall Radioactivity half-life (Chap 24)]
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42. Integrated Rate Law – Half-LIfe
2nd Order Reaction Half-Life
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43. Practice Problem
A reaction is first order with respect to A. The first-order rate constant is 2.61 /min. How long will it take the concentration of A to decrease from M to M? What is the half-life of the reaction? How long will it take for the concentration of A to decrease by 85%?
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44. Practice Problem
A reaction is second order with respect to B. The second-order rate constant is 1.5 L/molmin How long will it take the concentration of B to decrease from M to M?
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45. Temperature Dependence of Reaction Rate
An increase in Temperature (T) generally increases the reaction rate
A 10oC increase in Temperature usually doubles the rate
Temperature affects the rate constant (K) of the rate equation
Temperature effect process is described by Collision Theory
Can calculate the effect of T on rate of a reaction using the Arrhenius Equation
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46. Effects of Concentration & Temperature
Two major models explain the observed effects of Concentration & Temperature on reaction rate
Collision Theory
Views the reaction rate as a result of particles colliding with a certain frequency and minimum energy
Transition State Theory
Close-up view of how the energy of a collision converts reactant to product
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47. Collision Theory Why concentrations are “Multiplied” in the Rate Law
Consider 2 particles of “A” & 2 particles of “B”
Total A-B collisions = 4 (2 x 2)
A1B A1B A2B A2B2
Add additional Particle of “A”
Total A-B collisions = 6 (3 x 2)
A1B1 A1B2 A2B1 A2B2 A3B1 A3B2
It is the product of the number of different particles, not the sum (6 vs 5), that determines the number of collisions (reactions) possible
The number of particles of reactant A (concentration) must be multiplied by the number of particles of Reactant B to account for the total number of collisions (reactions) that occur.
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48. Activation Energy (Ea)
Collision Theory
Increasing the temperature of a reaction increases the average speed of particles; thus, the frequency of collision
Most collisions do not result in a “reaction”
Collision Theory assumes that, for a reaction to occur, reactant molecules must collide with an energy greater than some minimum value and with proper orientation
Activation Energy (Ea)
The rate constant, k, for a reaction is a function of collision related factors:
Z collision frequency
f fraction of collisions => activation energy
p fraction of collisions in proper orientation
k = Zpf
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49. Collision Theory
At a given temperature, the fraction of molecular collisions, f, with energy greater than or equal to the activation energy, Ea, is related to activation energy by the expression:
An equation (Arrhenius) expressing the dependence of the rate constant, k, on temperature can be obtained by combining the relationship between the rate constant and fraction of collisions, f, that are >= to the “activation energy”, Ea
Since
Then
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50. Arrhenius Equation k = Ae -Ea/RT Arrhenius Equation
Temperature dependence of reaction rate
k = Ae -Ea/RT Arrhenius Equation
k = rate constant
A = frequency factor (pZ)
Ea = activation energy (J)
R = gas constant (8.314 J/molK)
T = temperature (K)
The Relationship between temperature (T) in the e -Ea/RT term and the rate constant (k) means that as the temperature increases, the negative exponent (-Ea/RT) becomes smaller, and the e -Ea/RT term becomes larger, so the value of k becomes larger, which means that the rate of the reaction increases
Higher T  Larger k  Increased Reaction Rate
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51. Arrhenius Equation
The activation energy (Ea) can be calculated from the Arrhenius equation by taking the natural logarithm (lne) of both sides and rearranging the equation into a “straight line (y = b + mx) form
Note: The natural logarithm (lne) is usually presented as: “ln” omitting the subscript “e”
A plot of l nk (y) vs. 1/T (x) gives a
straight line whose slope (m) is -Ea/R and whose y intercept is Ln A (b)
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52. Arrhenius Equation
Ea can be determined graphically from a series of k values at different temperatures
Determine the slope from the plot
Use slope formula = -Ea/R
Alternate Approach - Compute Ea mathematically if the rate constants at two temperatures are known
“ln A” term
drops out
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53. Practice Problem
Find the Activation Energy (Ea) for the decomposition of Hydrogen Iodide (HI)
2HI(g)  H2(g) + I2(g)
The rate constants are:
9.51x10-9 L/mols at 500oK
1.10x10-5 L/mols at 600oK
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54. Rate – Affects of Temperature
Collision Energy
Collision Energy
ACTIVATED STATE
Ea (forward)
Ea (reverse)
REACTANTS
PRODUCTS
Molecules must collide with sufficient energy to reach “activation” status
Minimum collision energy is “energy of activation, Ea”
The forward reaction is Exothermic because the reactants have more energy than the products.
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55. Collision Theory: Proper Orientation (p)
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56. Transition-State Theory
Transition-state theory explains the reaction in terms of the collision of two high energy species – activated complexes
An activated complex (transition state) is an unstable grouping of atoms that can break up to form products
A simple analogy would be the collision of three billiard balls on a billiard table
Suppose two balls are coated with a slightly stick adhesive
We’ll take a third ball covered with an extremely sticky adhesive and collide it with our joined pair
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57. Transition-State Theory
Transition-state theory (cont’)
At the instant of impact, when all three spheres are joined, we have an unstable transition- state complex
The “incoming” billiard ball would likely stick to one of the joined spheres and provide sufficient energy to dislodge the other, resulting in a new “pairing”
If we repeated this scenario several times, some collisions would be successful and others (because of either insufficient energy or improper orientation) would not be successful.
We could compare the energy we provided to the billiard balls to the activation energy, Ea
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58. Transition State Theory
Reaction of Methyl Bromide & OH-
Reaction is Exothermic – reactants are higher in energy than products
Forward activation energy Ea(fwd) is less than reverse Ea(rev)
Difference in activation energies is “Heat of Reaction”
Hrxn = Ea(fwd) - Ea(rev)
Note the partial elongated C-O and C-Br bonds and the trigonal bipyramidal shape of the transition state
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59. Exothermic Reaction Pathway
Transition State
Hrxn = Ea(fwd) - Ea(rev)
Ea(fwd) < Ea(rev)
 Hrxn < 0 Reaction is Exothermic
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60. Endothermic Reaction Pathway
Hrxn = Ea(fwd) - Ea(rev)
Ea(fwd) > Ea(rev)
 Hrxn > 0 Reaction is Endothermic
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61. Reaction Mechanisms
Steps in the overall reaction that detail how reactants change into products
Reaction Mechanism – set of elementary reactions that leads to overall chemical equation
Reaction Intermediate – species produced during a chemical reaction that do not appear in chemical equation
Elementary Reactions – single molecular event resulting in a reaction
Molecularity – number of molecules on the reactant side of elementary reaction
Rate Determining Step (RDS) – slowest step in the reaction mechanism
This is the reaction used to construct the rate law; it is not necessarily the overall reaction
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62. Reaction Mechanisms Proposed Overall Reaction
2 NO2(g) H2(g)  2 H2O(g) + N2(g)
A mechanism in 3 elementary reactions:
2 NO2  N2O (slow) (RDS)
H2 + N2O2  H2O + N2O (fast)
H2 + N2O  H2O + N (fast)
The Overall Reaction from Elementary Reactions:
N2O2 and N2O are reaction intermediates
Develop “Rate Law” from the “Rate Determining Step” (RDS)
Rate law = Rate = k[NO2]2
Note: The reaction order in an elementary reaction comes from the stoichiometric coefficient, i.e., 2
Adding together the reactions in the mechanism provides the overall chemical equation
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63. Reaction Mechanisms
Elementary Reactions – The individual steps, which together make up a proposed reaction mechanism
Each elementary reaction describes a single molecular event, such as one particle decomposing or two particles colliding and combining
The reaction order in an elementary reaction comes from the stoichiometric coefficient, i.e., 2, unlike the rate law developed from the overall reaction – see slides 17 & 18, where the reaction orders must be determined experimentally
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64. Reaction Mechanisms
An elementary step is characterized by its “Molecularity – the number of reactant particles involved in the step
2O3(g)  3O2(g)
Proposed mechanism – 2 steps
1st step – Unimolecular reaction (decomposition)
O3(g)  O2(g) O(g)
2nd step – Bimolecular reaction (2 particles react)
O3(g) O(g)  2O2(g)
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65. Rate Law & Reaction Mechanisms
Rate law for an elementary reaction can be deduced directly from molecularity of reaction (w/o experimentation)
An elementary reaction occurs in one step
Its rate must be proportional to the product of the reactant concentrations
The stoichiometric coefficients are used as the reaction orders in the rate law for an elementary step
The above statement holds only for an elementary reaction
In an overall reaction the reaction orders must be determined experimentally
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66. Rate Law & Reaction Mechanisms
Steps in determining rate law from reaction mechanism
Identify the rate determining step (RDS) of the mechanism
Write out the preliminary rate law from RDS
Remove expressions for intermediates algebraically
Substitute into preliminary rate law to obtain final rate law expression
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67. Practice Problem
The following two reactions are proposed as elementary steps in the mechanism of an overall reaction:
(1)
NO2Cl(g)
NO2(g) + Cl(g)
(2)
NO2Cl(g) + Cl(g)
NO2(g) + Cl2(g)
Write the overall balanced equation
Determine the molecularity of each step
What are the reaction intermediates
Write the rate law for each step
PLAN:
The overall equation is the sum of the steps
Molecularity is the sum of the reactant particles in the step
SOLUTION:
Step(1) is unimolecular.
Step(2) is bimolecular.
(b)
NO2(g) + Cl (g)
(1)
NO2Cl(g)
(a)
(c) Cl(g) is reaction intermediate
(2)
NO2Cl(g) + Cl (g)
NO2(g) + Cl2(g)
rate1 = k1[NO2Cl]
(d)
2NO2Cl(g)
2NO2(g) + Cl2(g)
rate2 = k2[NO2Cl][Cl]
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68. Correlating Rate Law & Mechanism
Criteria required for proposed reaction mechanism
The elementary steps must add up to the overall balanced equation
The number of reactants and products in the elementary reactions must be consistent with the overall reaction
The elementary steps must be physically reasonable – they should involve one reactant (unimolecular) or at most two reactant particles (bimolecular)
The mechanism must correlate with the “rate law” – The mechanism must support the experimental facts shown by the rate law, not the other way around
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69. Practice Problem
If a slow step precedes a fast step in a two-step mechanism, do the substances in the fast step appear in the rate law? Ans: No, the overall rate law must contain reactants only (no intermediates) and is determined by the slow step If the first step in a reaction mechanism is slow, the rate law for that step is the overall rate law If a fast step precedes a slow step in a two-step mechanism, how is the fast step affected? Ans: If the slow step is not the first one, the faster preceding step produces intermediates that accumulate before being consumed in the slow step How is this effect used to determine the validity of the mechanism? Ans: Substitution of the intermediates into the rate law for the slow step will produce the overall rate law
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70. Mechanism with a Slow Initial Step
Reaction between Nitrogen Dioxide & Fluorine gas Overall reaction 2NO2(g) + F2(g)  2NO2F(g) Experimental Rate Law Rate = k[NO2][F2] (1st order in NO2 & F2) Proposed Mechanism (1) NO2(g) + F2(g)  NO2F(g) + F(g) [slow, rds] (2) NO2(g) + F(g)  NO2F(g) [fast] Overall: 2NO2(g) + F2(g)  2NO2F(g) Criteria 1: Elementary steps add up to experimental Criteria 2: Both steps “Bimolecular”
Con’t on next Slide
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71. Mechanism with a Slow Initial Step
Criteria 3:
Elementary Reaction Rate Laws
Rate1 = k1[NO2][F2]
Rate2 = k2[NO2][F]
Experimental Rate Law: k[NO2][F2] (from rds)
Rate 1 (k1) from rds is same as overall k
The 2nd [NO2] term (in Rate2) does not appear in the overall rate law
Each step in mechanism has its own transition state
Proposed transition state is shown in step 1
Reactants for 2nd step are the F atom intermediate and the 2nd molecule of NO2
First step is slower – Higher Ea
Overall reaction is exothermic - Hrxn < 0
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72. Mechanism with a Fast Initial Step
Nitric oxide, NO, is believed to react with Chlorine (Cl2) according to the following mechanism
NO Cl2  NOCl2 (Fast, equilibrium)
NOCl2 + NO  2 NOCl (slow, RDS)
1. What is the overall chemical equation for the reaction?
2. Identify the reaction intermediates
3. Propose a viable rate law from the mechanism
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73. Catalysis – Speeding Up Reaction
It is often necessary to “Speed up” a reaction in order to make it useful and in the case of industry, profitable
Approaches
More energy (heat) – could be expensive!!
Catalyst – Stoichiometrically small amount of a substance that increases the rate of a reaction; it is involved in the reaction, but ultimately is not consumed
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74. Catalysis – Speeding Up Reaction
Catalyst:
Causes lower “activation energy”, (Ea)
Lower activation energy is provided by a change in the reaction mechanism
Makes Rate constant larger
Promotes higher reaction rate
Speeds up forward & reverse reactions
Does not improve yield – just makes it faster
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75. Homogeneous Catalysts
Exist in “Solution” with the reactant mixture
All homogenous catalysts are gases, liquids, or “soluble” solids
Speeds up a reaction that occurs in a separate phase
Ex. A solid interacting with gaseous or liquid reactants
The solid would have extremely large surface area for contact
If the rate-determining step occurs on the surface of the catalyst, many reactions are zero order, because once the surface area is covered by the reactant, increasing the concentration has no effect on the rate
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76. Homogeneous Catalysts
Catalytic H+ ion bonds to electron rich carbonyl oxygen
H+ , the catalyst, is a proton supplied by a strong acid
The increased positive charge on the Carbon attracts the partially negative oxygen of the water more strongly, increasing the fraction of effective collisions, speeding up this rate determining step
Acid
The result of the hydrolysis of an ester is the formation of an acid and an alcohol
Alcohol
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77. Heterogeneous Catalysts
Hydrogenation of Ethylene (Ethene) to Ethane catalyzed by Nickel (Ni), Palladium (Pd), or Platinum (Pt)
H2C=CH2(g) + H2(g)  H3C – CH3
Finely divided Group 8B metals catalyze by adsorbing the reactants onto their surface
H2 lands and splits into separate H atoms chemically bound to solid catalyst’s metal atoms
H – H + 1catM(s)  2catM – H
Then C2H4 absorbs and reacts with two H atoms, one at a time, to form H3C–CH3
The H-H split is the rate determining step providing a lower energy of activation
Ni, Pd, Pt
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78. Effect of A Catalyst Comparison of Activation Energies in the Uncatalyzed and Catalyzed Decompositions of Ozone
Catalyst: provides alternative mechanism for a reaction that has a lower activation energy
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79. Practice Problem
Ethyl Chloride, CH3CH2Cl2, used to produce tetraethyllead gasoline additive, decomposes, when heated, to give Ethylene and Hydrogen Chloride. The reaction is first order. In an experiment, the initial concentration of Ethyl Chloride was M. After heating at 500oC for 155 s, this was reduced to M. What was the concentration of Ethyl Chloride after a total of 256 s?
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80. Practice Problem
The rate of a reaction increases by a factor of 2.4 when the temperature is increased from 275 K to 300 K. What is the activation energy of the reaction?
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81. Practice Problem
The rate constant of a reaction at 250 oC is 2.69 x /Ms (L/mols). Given the activation energy for the reaction is 250 kJ, what is the rate constant for the reaction at 100 oC, assuming activation energy is independent of temperature?
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82. Practice Problem
If the half-life of a first-order reaction is 25 min, how long will it take for 20% of the reactant to be consumed?
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83. Practice Problem
For a reaction with the rate law given as rate = k[A]2, [A] decreases from 0.10 to M in 161 min. What is the half-life of the reaction?
(See slide # 42)
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84. Practice Problem
The decomposition of the herbicide Atrazine in the atmosphere by sunlight is first order, with a rate constant of 1.1 x /s. Following field application by spaying it is found that the atmospheric concentration of Atrazine is x 10-6 ppm at mid-day. How long (in hours) will it take for the atmospheric concentration of Atrazine to reach the air quality standard of 1.0 x 10-9 ppm?
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85. [Atr]/t = kph[Atr][OH]
Practice Problem
The indirect photolysis of the pesticide Atrazine (Atr, C8H14ClN5) in air by hydroxyl radical (OH) is shown below
C8H14ClN5 + OH  C8H13ClN5 + H2O
The reaction is second order and follows the rate law:
[Atr]/t = kph[Atr][OH]
The concentration of OH is at steady-state during daylight hours at ~1.0 x M, and kph is 5.0 x /Ms.
How long (in min) will it take for Atr to decrease from 2.5 x M to 1.0 x M (the air quality criteria) following application to a golf course assuming pseudo-first-order kinetics during daylight?
a b c d. 4, e. 119
Solution on next Slide
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86. Practice Problem Photolysis of Atrazine (con’t)
The 2nd order rate law ([Atr]/t = kph[Atr][OH]) is stated in terms of two reactants
This would result in a different integrated form of the 2nd order reaction, i.e., more complicated math
Since the concentration of hydroxyl, [OH], is constant, the rate law can be reduced to a pseudo 1st order reaction by combining the Kph & [OH] terms (both constants) into a new rate constant:
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87. Practice Problem
A convenient rule of thumb is that the rate of a reaction doubles for a 10o C change in temperature.
What is the activation energy for a reaction whose rate doubles from 10.0o C to 20.0o C?
a kJ b kJ c kJ d kJ e kJ
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88. Rate Equations - Summary
Integrated Rate Laws
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89. Rate Equations - Summary
An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions
Zero Order
First Order
Second Order
Rate law
rate = k
rate = k[A]
rate = k[A]2
Units for k
mol/L • s
1/s
L/mol • s
Integrated rate law in straight-line form
[A]t = -kt + [A]0
ln [A]t = -kt + ln [A]0
1/[A]t = kt + 1/[A]0
Plot for straight line
[A]t vs. t
ln [A]t vs. t
1/[A]t = t
Slope, y intercept
k, [A]0
k, 1/[A]0
-k, ln [A]0
Half-life
[A]0/2k
ln 2/k
1/k[A]0
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90. Rate Equations - Summary
Activation Energy (Ea) k = Zpf f = e -Ea/RT k = Zpe -Ea/RT = Ae -Ea/RT k, rate constant Z, collision frequency f, fraction of collisions that are => activation energy p, fraction of collisions in proper orientation A = frequency factor (pZ) Ea = activation energy (J) R = gas constant (8.314 J/molK) T = temperature (K)
Arrhenius Equation
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