1. Kinetics Concept of rate of reaction
Use of differential rate laws to determine order of reaction and rate constant from experimental data 
Effect of temperature change on rates 
Energy of activation; the role of catalysts 
The relationship between the rate-determining step and a mechanism 
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2. Rate of Reaction Rate = Δ[concentration] or d [product] Δ time dt
Rate of appearance of a product = rate of disappearance of a reactant
Rate of change for any species is inversely proportional to its coefficient in a balanced equation.
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3. Rate of Reaction Assumes nonreversible forward reaction
Rate of change for any species is inversely proportional to its coefficient in a balanced equation.
2N2O5  4NO2 + O2
Rate of reaction = -Δ[N2O5] = Δ[NO2] = Δ[O2]
2 Δt Δt Δt
where [x] is concentration of x (M) and t is time (s)
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4. Reaction of phenolphthalein in excess base
Conc. (M)
Time (s)
0.0050
0.0045
10.5
0.0040
22.3
0.0035
35.7
0.0030
51.1
0.0025
69.3
0.0020
91.6
Use the data in the table to calculate the rate at which phenolphthalein reacts with the OH- ion during each of the following periods:
(a) During the first time interval, when the phenolphthalein concentration falls from M to M.
(b) During the second interval, when the concentration falls from M to M.
(c) During the third interval, when the concentration falls from M to M.
Answer
(a) 4.8 x 10-5 M/s
(b) 4.2 x 10-5 M/s
(c) 3.7 x 10-5 M/s
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5. Reactant Concentration by Time
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6. Finding k given time and concentration
Create a graph with time on x-axis.
Plot each vs. time to determine the graph that gives the best line:
[A]
ln[A]
1/[A]
(Use LinReg and find the r value closest to 1)
k is detemined by the slope of best line (“a” in the linear regression equation on TI-83)
1st order (ln[A] vs. t): k is –slope
2nd order (1/[A] vs t: k is slope)
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7. Rate Law Expression
As concentrations of reactants change at constant temperature, the rate of reaction changes. According to this expression.
Rate = k[A]x[B]y…
Where k is an experimentally determined rate constant, [ ] is concentration of product and x and y are orders related to the concentration of A and B, respectively. These are determined by looking at measured rate values to determine the order of the reaction.
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8. Finding Order of a Reactant - Example 2ClO2 + 2OH-  ClO3- + ClO2- + H2O
Start with a table of experimental values:
To find effect of [OH-] compare change in rate to change in concentration.
When [OH-] doubles, rate doubles. Order is the power: 2x = 2. x is 1. This is 1st order for [OH-].
[ClO2] (M)
[OH-] (M)
Rate (mol/L-s)
0.010
0.030
6.00x10-4
0.060
1.20x10-3
1.08x10-2
same 2x
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9. Finding Order of a Reactant - Example 2ClO2 + 2OH-  ClO3- + ClO2- + H2O
Start with a table of experimental values:
To find effect of [ClO2] compare change in rate to change in concentration.
When [ClO2] triples, rate increases 9 times. Order is the power: 3y = 9. y is 2. This is 2nd order for [ClO2].
[ClO2] (M)
[OH-] (M)
Rate (mol/L-s-1)
0.010
0.030
6.00x10-4
0.060
1.20x10-3
1.08x10-2
same 9x 3x
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10. Finding Order of a Reactant - Example 2ClO2 + 2OH-  ClO3- + ClO2- + H2O
Can use algebraic method instead. This is useful when there are not constant concentrations of one or more reactants. This example assumes you found that reaction is first order for [OH-] .
6.00 x 10-4=k(0.010)x(.030)1
1.08 x 10-2 = k (0.030)x(.060)1
= .333x(.5)
For [ClO2]x , x = 2
[ClO2] (M)
[OH-] (M)
Rate (mol/L-s-1)
0.010
0.030
6.00x10-4
0.060
1.20x10-3
1.08x10-2
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11. Rate Law: 2ClO2 + 2OH-  ClO3- + ClO2- + H2O
Rate = k[ClO2]2[OH-]
To find k, substitute in any one set of experimental data from the table. For example, using the first row:
k = rate/[ClO2]2[OH-]
k = 6.00x10-4Ms = 200 M-2s-1
[0.010M]2[0.030M]
Overall reaction order is 2+1=3. Note units of k.
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12. Determining k Given Overall Reaction Order
Rate(M/s) = k[A]x
x = overall order of reaction
[A] = the reactant concentration (M)
Overall reaction order
Example
Units of k 1
Rate=k[A]
(M/s)/M = s-1 2
Rate=k[A]2
(M/s)/M2 = M-1s-1 3
Rate=k[A]3
(M/s)/M3 = M-2s-1
1.5
Rate=k[A]1.5
(M/s)/M1.5 = M-0.5s-1
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13. Integrated Rate Law Use when time is given or requested
Relates concentration and time to rate
1st order: ln[A] = -kt + ln[A] or [A]=[A]0e-kT
2nd order: 1__ = kt __
[B] [B]0
Wow! y = mx b
Both equations can be used with linear regression to solve for slope, or k.
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14. Half life for 1st vs 2nd Order Reactions
1st order: t1/2 = 0.693 k
2nd order: t1/2 = 1__
k[A]0
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15. The Arrhenius Equation and Finding Ea
k=Ae-Ea/RT
Where A is the frequency factor
Related to requency of collisions and favorable orientation of collisions
Ea is activation energy in J
R = J/mol-K
T is temp in K
k is the rate constant
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16. Using the Arrhenius equation
As Ea increases, rate decreases.
Fewer molecules have the needed energy to react.
As temp increases, rate increases
More collisions occur and increased kinetic energy means more collisions have enough energy to react.
Mathematically, T is in denominator of the power –Ea/T.
ln k = -Ea + lnA T
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17. Activation Energy: Energy vs. Reaction Progress
Ea is lowered with the addition of a catalyst.
Peak is where collisions of reactants have achieved enough energy to react
The arrangement of atoms at the peak is activated complex or the transition state.
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18. Temperature effects on a reaction
Two factors account for increased rate of reaction.
Energy factor: When enough energy is in collision for formation of activation complex, bonds break to begin reaction. With higher temp, more collisions have this energy.
Frequency of collision: Particles move faster and collide more frequently with higher temp, increasing chance of reaction.
Increasing temperature increases the rate of a reaction more if the reaction is endothermic to start with.
K increases according to k=Ae-Ea/RT
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19. Finding Ea given info at two temperatures
(or to find a temperature given conditions at one temp and Ea)
ln k1 = Ea [ 1_ - 1_ ]
k2 R T2 T1
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20. Mechanisms: Multistep Reactions
The following reaction occurs in a single step.
CH3Br (aq) + OH-(aq)  CH3OH(aq) + Br-(aq)
Rate = k(CH3Br)(OH-)
This one occurs in several steps:
(CH3) 3CBr(aq) + OH-(aq) (CH3) 3COH (aq) + Br- (aq)
(CH3)3CBr  (CH3) 3C+ + Br Slow step
(CH3)3C+ + H2O  (CH3)COH2+ Fast step
(CH3)3COH2+ + OH-  (CH3)3COH + H2O Fast step
Rate = k((CH3)3CBr)
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21. Mechanisms: Multistep Reactions
The overall rate of reaction is more or less equal to the rate of the slowest step. (rate-limiting step)
If only one reagent is involved in the rate-limiting step, the overall rate of reaction is proportional to the concentration of only this reagent.
Ex. For the reaction with Rate = k((CH3) 3CBr)
Although the reaction consumes both (CH3) 3CBr and OH-, the rate of the reaction is only proportional to the concentration of (CH3)3CBr.
The rate laws for chemical reactions can be explained by the following general rules.
The rate of any step in a reaction is directly proportional to the concentrations of the reagents consumed in that step.
The overall rate law for a reaction is determined by the sequence of steps, or the mechanism, by which the reactants are converted into the products of the reaction.
The overall rate law for a reaction is dominated by the rate law for the slowest step in the reaction.
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22. N2H2O2 N2HO2- + H+ fast equilibrium
Substituting for an Intermediate Step 1 : N2H2O2  N2HO2- + H + fast equilibrium Step 2: N2HO2-  N2O + OH- slow Step 3: H+ + OH-  H2O fast
Requirement: A fast equilibrium prior to the rate determining (slow) step that contains the intermediate for which you wish to substitute.
N2H2O2 N2HO2- + H fast equilibrium
For the fast equilibrium, write the rate law (leaving out the k and R) for the reactants and set it equal to the rate law for the products. This can be done because in an equilibrium reaction the forward rate must be equal to the reverse rate.
[N2H2O2] = [N2HO2-] [H+]
Algebraically solve for the intermediate, N2HO2-
[N2H2O2] / [H+] = [N2HO2-]
Algebraically substitute into the rate law for N2HO2-
Rate law with intermediate is: R = k [N2HO2-] , so R = k [N2H2O2] / [H+]
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