1. Reaction mechanisms
EQ: What is the difference between a mechanism with the RDS in the first step versus one with the RDS in the second step?

2. 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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3. 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 + N2 (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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. 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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11. 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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12. 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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13. 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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14. 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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15. 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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16. 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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17. 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
The result of the hydrolysis of an ester is the formation of an acid and an alcohol
Acid
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Alcohol

18. 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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19. 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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20. 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 500 C 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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21. 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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22. 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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23. 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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24. 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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25. 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 2.5 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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26. 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. 26 b. 1.8 c. 527 d. 4,218 e. 119
Solution on next Slide
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27. 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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28. 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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29. Rate Equations - Summary
Integrated Rate Laws
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30. 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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31. 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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