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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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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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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 4 Δt Δtwhere [x] is concentration of x (M) and t is time (s)

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Reaction of phenolphthalein in excess base

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 0.0050 M to 0.0045 M.

(b) During the second interval, when the concentration falls from 0.0045 M to 0.0040 M.

(c) During the third interval, when the concentration falls from 0.0040 M to 0.0035 M.

Conc. (M) Time (s)

0.0050 0

0.0045 10.5

0.0040 22.3

0.0035 35.7

0.0030 51.1

0.0025 69.3

0.0020 91.6

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Reactant Concentration by Time

Phenolphthalein Concentation in Basic Solution Over Time

0

0.001

0.002

0.003

0.004

0.005

0.006

0 10 20 30 40 50 60 70 80 90 100

Time (s)

Co

nc

. (M

)

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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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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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Finding Order of a Reactant - Example2ClO2 + 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

0.010 0.060 1.20x10-3

0.030 0.060 1.08x10-2

2x 2x

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Finding Order of a Reactant - Example2ClO2 + 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

0.010 0.060 1.20x10-3

0.030 0.060 1.08x10-23x 9x

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Finding Order of a Reactant - Example2ClO2 + 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

0.0556 = .333x(.5)

For [ClO2]x , x = 2

[ClO2] (M) [OH-] (M) Rate (mol/L-s-1)

0.010 0.030 6.00x10-4

0.010 0.060 1.20x10-3

0.030 0.060 1.08x10-2

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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-1 = 200 M-2s-1

[0.010M]2[0.030M]Overall reaction order is 2+1=3. Note units of k.

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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

Determine Reaction Order

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From the following experimental data, determine the order of the reaction and the rate constant for the reaction:  C4H8 (g) → 2C2H4 (g) The reaction was carried out in a constant volume container at 532 °C.

Time (s) P (mm Hg)

0 800

200 732

400 496

600 392

800 310

1000 244

Another Reaction Order Example

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The following data were collected during a study of its decomposition at a certain temperature:  [H2O2] (M) Time (s)0.100 00.088 1200.070 3000.050 6000.025 12000.006 2400

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Integrated Rate LawUse when time is given or requested

Relates concentration and time to rate 1st order: ln[A] = -kt + ln[A]0 or [A]=[A]0e-kT

2nd order: 1__ = kt + 1__

[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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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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Order Rate Law Concentration-Time Equation

Half-life Equation

Graphical Analysis Graph

1 Rate = k[A] ln = – kt t½ =  

ln[A] vs. time slope = –k

ln[A]

t

2 Rate = k[A]2 = kt + t½ = vs. time

slope = k t

0 Rate = k[A]0 [A] = –kt + [A]0 t½ =  

[A] vs. t  slope = –k

[A]

t

0A

A t 0.693

k

tA1

0

1

A 0

1

k A tA

1

k2

][ 0A

A1

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The Arrhenius Equation and Finding Ea

k=Ae-Ea/RT

Where A is the frequency factor– Related to frequency of collisions and favorable

orientation of collisions Ea is activation energy in J R = 8.314 J/mol-K T is temp in K k is the rate constant

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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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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.

A+B --> C+D

0

100

200

300

400

0 2 4 6

Reaction path

En

erg

y (

kJ

)

no catalyst

with catalyst

Ea

ΔH

A + B

C + D

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Temperature effects on a reaction

Two factors account for increased rate of reaction.1. 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.

2. 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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Finding Ea given info at two temperatures

ln k1 = Ea [ 1_ - 1_ ]

k2 R T2 T1

Similar to vapor pressure equation, Clausius-Clapeyron Equation:

ln PvapT1 = ΔHvap [ 1_ - 1_ ]

PvapT2 R T2 T1

In both cases, use R = 8.3145 J/K-mol and be sure Ea or ΔHvap are in J, not kJ.

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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)

1. (CH3)3CBr (CH3) 3C+ + Br- Slow step

2. (CH3)3C+ + H2O (CH3)COH2+ Fast step

3. (CH3)3COH2+ + OH- (CH3)3COH + H2O Fast step

– Rate = k((CH3)3CBr)

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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.

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch22/rateframe.html

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Substituting for an IntermediateStep 1 : N2H2O2 N2HO2- + H + fast equilibriumStep 2: N2HO2- N2O + OH- slowStep 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.

1. N2H2O2 N2HO2- + H+ fast equilibrium

2. 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+]3. Algebraically solve for the intermediate, N2HO2-

[N2H2O2] / [H+] = [N2HO2-]4. Algebraically substitute into the rate law for N2HO2-

Rate law with intermediate is: R = k [N2HO2-] , so R = k [N2H2O2] / [H+]