CHEM 2880 - Kinetics Determining Rate Order II.pdf · CHEM 2880 - Kinetics 40 ... From: Physical...

CHEM 2880 - Kinetics

40

Determining Rate Order

As mentioned earlier, the rate order of a reaction can onlybe determined experimentally. A number of methods canbe used to determine both the reaction order and the rateconstant.

1. Integration method - the concentration (of reagent orproduct) is measured at various time intervals, andthe integrated rate law equations are plotted. Theorder is determined by the plot with the best linear fit(most consistent k value).

2. Isolation method - if more than one reactant isinvolved, the reaction is run several times, each timechanging the initial concentration of one of thereagents. Any corresponding change in the rate isdue to the reagent that changed concentration and thecorresponding order can be determined. Once theorder wrt to each reagent has been determined theoverall rate can be calculated.

3. Reagents in excess - this method can be used toreduce the order of a reaction (as mentioned earlier),and make complicated rate laws easier to analyze.


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4. Differential method - for an n -order reaction, theth

rate law is:

taking logarithms of both sides yields:

Thus a plot of log v vs log[A] will be linear with aslope of n and an intercept of log k. This methodoften involves measuring the initial rate of reaction atseveral different initial concentrations. This has theadvantages of: avoiding any complications caused bybuild up of products; and measuring rate at a timewhen the concentration is known most accurately.

From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 459.


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Examples:

Integration method:

The following pharmacokinetic data were determined forthe elimination of beta adrenergic blocking agents (betablockers, used to treat hypertension) from the blood. Thedrug was administered to the patients and their bloodplasma was monitored for the drug over a period of time. Determine the order of reaction and rate constant for thisprocess.

t (min) 30 60 120 150 240 360 480

c (ng/ml) 699 622 413 292 152 60 24

From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 252.


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Using the integration method requires making plots foreach of zero-, 1 -, and 2 -order. The respective plots arest nd

c vs t, ln c vs t and 1/c vs t. The resulting table and plotlook like:

t c ln c 1/c*100

30 699 6.55 0.143

60 622 6.43 0.161

120 413 6.02 0.242

150 292 5.68 0.342

240 152 5.02 0.658

360 60 4.09 1.667

480 24 3.18 4.167

Note: The 1/c values have been multiplied by 100 so that the ln c and 1/c plots are on the same scale.

The ln c vs t is the best linear fit, indicating a 1 -orderst

process. From the slope of the line, the rate constant isdetermined to be 7.6 x 10 s . The elimination of drugs-3 -1

from the body is usually a first order process.


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Isolation method:

For the acid catalyzed bromination of acetone:

The reaction rate was determined for different initialconcentrations of the reagents. Determine the rate lawand the rate constant.

3 3 2[CH COCH ] [Br ] [H ] Rate of dis-+

2appearance of Br

(M) (M) (M) (M s )-1

1 0.30 0.050 0.050 5.7 x 10-5

2 0.30 0.10 0.050 5.7 x 10-5

3 0.30 0.050 0.10 1.2 x 10-4

4 0.40 0.050 0.20 3.1 x 10-4

5 0.40 0.050 0.050 7.6 x 10-5



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For this method, we want to compare the reaction ratesbetween two different runs where the concentration ofonly one reagent has changed from one run to the next.

Comparing runs 1 and 2, the concentrations of [H ] and+

3 3 2[CH COCH ] are constant, the [Br ] increases by afactor of 2, and the rate does not change. Therefore the

2rate is not dependant on the [Br ] and the reaction is

2zero-order in [Br ].

2Comparing runs 1 and 3, the concentrations of [Br ] and

3 3[CH COCH ] are constant, the [H ] increases by a factor+

of 2, and the rate increases by a factor of 2.2. Thereforethe rate is directly proportional to the [H ] and the+

reaction is 1 -order in [H ].st +

Comparing runs 1 and 5, the concentrations of [H ] and+

2 3 3[Br ] are constant, the [CH COCH ] increases by afactor of 1.3, and the rate increases by a factor of 1.3. Therefore the rate is directly proportional to the

3 3[CH COCH ] and the reaction is 1 -order in st

3 3[CH COCH ].

The rate law for this reaction is:

3 3v = k [CH COCH ][H ]+


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Using the rate law, the rate constant for each run can becalculated and the average rate constant determined.

3 3[CH COCH ] [H ] Rate k+

(M) (M) (M s ) (M s )-1 -1 -1

1 0.30 0.050 5.7 x 10 3.8 x 10-5 -3

2 0.30 0.050 5.7 x 10 3.8 x 10-5 -3

3 0.30 0.10 1.2 x 10 4.0 x 10-4 -3

4 0.40 0.20 3.1 x 10 3.9 x 10-4 -3

5 0.40 0.050 7.6 x 10 3.8 x 10-5 -3

6Avg 3.8 x 10-3


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Differential method:

For the binding of glucose to the enzyme hexokinase, the

0following initial rate (v ) data (in mol L s )were-1 -1

obtained:

[glucose] (mmol L )-1

1.00 1.54 3.12 4.02

[Enzyme] (mmol L )-1

1.34 5.0 7.6 15.5 20.0

3.00 7.0 11.0 23.0 31.0

10.0 21.0 34.0 70.0 96.0

Determine the rate law and the rate constant.From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 248.


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The data provided are initial rate constants for differentinitial concentrations for the two reagents, glucose andhexokinase. The rate law willbe of the form:

0 0 0v = k[glucose] [hexokinase]a b

0For a constant [hexokinase] , the order of reaction wrt toglucose can be determined using:

0 0logv = log k’ + alog[glucose]

0where k’ = k[hexokinase] b

0 0 0Plotting log v vs log[glucose] for each [hexokinase]yields straight lines with slopes of a and intercepts of logk’.

0log k’ = log k + blog[hexokinase]

0Plotting log k’ vs log[hexokinase] yields a straight linewith a slope of b and an intercept of log k.


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The table with the data converted to logs looks like:

log [glucose] (mmol L ) log[hexokinase] k’-1

-3.00 -2.81 -2.51 -2.40 (mmol L )-1

[hexokinase] (mmol L )-1

1.34 0.699 0.881 1.19 1.30 -2.87 3.69

3.00 0.845 1.04 1.36 1.49 -2.52 4.04

10.0 1.32 1.53 1.85 1.98 -2.00 4.56

The plot of this data looks like:

The three lines all have slopes of 1, so the reaction is 1st

order wrt to glucose. The intercepts are tabulated above.


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0The plot of log k’ vs log[hexokinase] looks like:

The slope and intercept are 1 and 6.55 respectively. Therefore b is 1and k is 10 = 3.6 x 10 . Thus the rate6.55 6

law for this reaction is:

v = 3.6 x 10 [glucose][hexokinase]6


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More Examples:

Sucrose (table sugar) undergoes hydrolysis (reaction withwater) to produce fructose and glucose:

12 22 11 2 6 12 6 6 12 6C H O + H O 6 C H O (fructose) + C H O (glucose)

“This reaction has particular significance in the candyindustry. First, fructose is sweeter than sucrose. Second,a mixture of fructose and glucose, called invert sugar,does not crystallize, so candy made with this combinationis chewier and bot brittle as crystalline sucrose is. Sucrose is dextrorotatory (+) whereas the mixture ofglucose and fructose resulting from inversion islevorotatory (-). Thus, a decrease in the concentration ofsucrose will be accompanied by a proportional decreasein the optical rotation.”

Given the following kinetic data, show that the reaction isfirst order and find k.

t (min) 0 7.20 18.0 27.0 4

optical rotation (") +24.08° +21.40° +17.73° +15.01° -10.73°



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The total change in optical rotation from t=0 to t=4, is

0 4(" - " ). This is proportional to the total change in

0[sucrose]. The change in optical rotation at time=t, (" -

t" ) is proportional to the change in [sucrose] at t. thusthe [sucrose] remaining at t is:

0 4 0 t t 4(" - " ) - (" - " ) = " - "

Since it is a 1 -order reaction:st

should be linear, and a plot of

vs t

will have a slope of -k.


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t 4 0 4t " ln((" -" )-(" -" ))(min)0.00 +24.08 0.000007.20 +21.40 -0.0801118.0 +17.73 -0.2014127.0 +15.01 -0.301864 -10.73

The plot is linear, so the reaction is 1 -order and st

k = 1.11 x 10 s .-2 -1


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A certain 1 -order reaction is 34.5% complete in 49 minst

at 298 K. What is its rate constant?From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 500.


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When the concentration of A in the reaction A6B waschanged from 1.20 M to 0.60 M, the half-life increasedfrom 2.0 min to 4.0 min at 25°C. Calculate the order ofthe reaction and the rate constant.From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 500.

0 ½When [A] decreased by a factor of 2, the t increased by

½a factor of 2. Therefore t is inversely proportional to

0[A] and n=2.


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The progress of a reaction in the aqueous phase wasmonitored by the absorbance of a reactant at varioustimes. Determine the order of reaction and the rateconstant.From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 500.

Time (s) Abs ln Abs 1/Abs

0 1.67 0.5128 0.5988

54 1.51 0.4121 0.6623

171 1.24 0.2151 0.8065

390 0.847 -0.1661 1.181

720 0.478 -0.7381 2.092

1010 0.301 -1.201 3.322

1190 0.216 -1.532 4.630

The ln Abs vs t plot has the best linear fit. Therefore thereaction is 1 -order and the rate constant is 1.7 x 10 s .st -3 -1


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Iodoacetamide and N-acetylcysteine react with 1:1stoichiometry. The following data were collected at 298K for the reaction of 1.00 mmol L N-acetylcysteine with-1

1.00 mmol L iodoacetamide.-1


t (s) [ ] (mM) ln [ ] 1/[ ]

10 0.770 -0.261 1.30

20 0.580 -0.545 1.72

40 0.410 -0.892 2.44

60 0.315 -1.16 3.17

100 0.210 -1.56 4.76

150 0.155 -1.86 6.45

The 1/[ ] vs t plot has the best linear fit. Therefore thereaction is 2 -order and the rate constant is nd

3.7 x 10 mM s or 37 M s .-2 -1 -1 -1 -1


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Effect of Temperature

Intuitively most people would think that an increase intemperature would increase the rate of a reaction. Looking at a typical rate law v = k[A] [B] , changing thex y

temperature would not effect the concentrations of A andB, so any effect of temperature must reflect an effect onk. Thus k must increase with increasing temperature.

However, temperature can effect reaction rate constants ina number of ways as illustrated below:


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Plot a is the most common situation where rate constantsincrease with increasing temperatures.

Plot b follows a similar trend, but eventually reaches atemperature where the rate constant reaches a maximumand then starts to decrease with further increases intemperature. A typical example here might be anenzyme-catalyzed reaction. As long as the enzymeremains in its native state, the rate constant increases withincreasing temperature, but eventually a temperature isreached where the enzyme will denature, and the rateconstant starts to decrease.

Plot c is quite uncommon, but has been observed forreactions where one of the steps in the reaction is stronglyexothermic and this effect outweighs any increase in therate constant.

Plot d illustrates the trend in rate constants for chainreactions. The rate gradually increases with temperatureuntil a threshold is reached where the propagationreactions become significant. Any further increase intemperature causes a more dramatic increase in the rateconstant.


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The Arrhenius Equation

The Arrhenius equation was derived empirically anddescribes the temperature dependence of many rateconstants.

aWhere A is the pre-exponential factor, E is the activationenergy, R is the gas constant and T is the temperature.

aNote: the units of E and R must match.

Taking the logarithm of both sides yields:

A plot of ln k vs 1/T, or Arrhenius plot, has an intercept

aof lnA and a slope of -E /R.

1 1 2 2Alternatively, if k , T and k , T are known:

Thus the activation energy for a reaction can bedetermined by measuring the rate constant at differenttemperatures.


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A rough rule for reactions in solution near roomtemperature is that k increases 2-3x for every 10°Cincrease in temperature.

Pre-exponential factor (A)• has the same units as k• represents the frequency of collisions between

reagent molecules - more collisions means morechances for reaction to occur

• is somewhat temperature dependent, but less so thanEa and can be considered temperature independentover short (50 K) temperature ranges


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aActivation Energy (E )• represents the minimum energy required for initiation

of the reaction

a• the exponential term exp(E /RT) is the fraction of

acollisions that have energy $E• the magnitude of the activation energy reflects the

sensitivity of the reaction to temperature

Note: The reaction coordinate represents the change ofthe reactant into products as the reaction takes place.

The Arrhenius equation is an empirical equation. We willlook at two theories: Collision Theory and TransitionState Theory which attempt to explain the effect oftemperature on chemical kinetics.


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Collision Theory of Chemical Kinetics

Is based on the kinetic theory of gases and in its simplestform applies only to bimolecular reactions in the gasphase.

Consider the reaction bimolecular, elementary reaction:

M+N 6 P

• for the reaction to occur M and N must collide

• the collision must involve some minimum kineticenergy

• the rate is proportional to the fraction of collisions

ahaving KE = E• this is reflected in the exponential term of the

Arrhenius equation

• the rate is proportional to the frequency of collisions• successful collisions require the correct orientation of

molecules (i.e. steric factor)• this is reflected in the pre-exponential term of the

Arrhenius equation


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The exponential term:

The energy distribution of the molecules is represented bythe Maxwell-Boltzmann distribution.

As the temperature increases the most probable speedincreases (the maximum in the curve), but the spread ofthe speed increases more markedly.

The fraction of molecules with the highest speedsincreases much faster with temperature than the averagespeeds.


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The pre-exponential term:

In collision theory: A = Pzwhere P = steric factor and z = collision frequency.

z can be calculated from the kinetic theory of gases.

The steric factor, P, takes into account that the moleculesmust have the correct orientation for the reaction to occur.

For example:

P values are generally very difficult to estimateaccurately.

Calculated rate constants of bimolecular gas-phasereactions: ~ 10 mol L s for small molecules.11 -1 -1

A = 10 to 10 mol L s for simple reactions.2 13 -1 -1

Few reactions proceed much faster than the ratescalculated from simple collision theory, but many areorders of magnitude slower.


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The overall result for Collision Theory is the modifiedArrhenius equation:

Drawbacks of Collision Theory• based on kinetic theory of gases - assumes reagents

are hard spheres• ignores structure of molecules• cannot be used to calculate activation energies

Transition-State Theory overcomes some of thesedrawbacks.


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Transition-State Theory

• developed by Eyring et al. in the 1930's• focuses on the relationship between the reagents and

the transition state• allows for accurate calculation of rate constants• applicable to gas and solution phase reactions• assumes reagents are in constant equilibrium with

transition state

The transition state MN has a lifetime of only a few‡

molecular vibrations, but can sometimes be observedusing femtosecond (10 s) or picosecond (10 s)-15 -12

spectroscopy.


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K is the equilibrium constant between the reagents and‡

the transition state and k is a universal rate constant:‡

These can then be related back to the observable rateconstant:

thus:k = k K‡ ‡

Statistical thermodynamics tells us:

Bwhere k is the Boltzmann constant and h is Planck’sconstant.

Thus K can be determined experimentally from the rate‡

constant.


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For any equilibrium )G° = RTlnK. Thus for theequilibrium between the reagents and the transition state:

)G° = RTlnK‡ ‡

Since K can be calculated from experimental rate‡

constants, this gives us a way to determine the free energyof activation.

reverse forward rxn)G° - )G° =)G° ‡ ‡

forward rxnGiven )G° from the rate constant and )G° ,‡

reverse)G° can also be calculated.‡


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Also:)G° = )H° - T)S°‡ ‡ ‡

So:

Compare this to the Arrhenius equation:

aGiven the E and A, the enthalpy and entropy changes foractivation can be determined.


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Examples

The rate of bacterial hydrolysis of fish muscle is twice as

agreat at 2.2 °C as at -1.1 °C. Estimate the E for thisreaction. Is there any relation to the problem of storingfish for food?From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 503.

Using the Arrhenius equation:

This activation energy is relatively large, and thereforethe reaction is quite temperature sensitive. Thusrefrigeration is an essential and effective method forpreserving fish.


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The rate constant of a 1 -order reaction is 4.60 x10 s atst -4 -1

350 °C. If the activation energy is 104 kJ mol , calculate-1

the temperature at which its rate constant is 8.80 x 10 s .-4 -1


Again, using the Arrhenius equation:


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aDetermine A and E from the following data:From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 257.

T (K) k (L mol s ) 1/T (K ) lnk-1 -1 -1

300 7.90E+06 0.00333 15.9

350 3.00E+07 0.00286 17.2

400 7.90E+07 0.00250 18.2

450 1.70E+08 0.00222 19.0

500 3.20E+08 0.00200 19.6

Plotting ln k vs 1/T:

aSlope = -E /R and intercept = lnA

a-2767.2 = -E /8.314 J K mol 25.11 = lnA-1 -1

aE = 23 kJ mol A = 8.0 x 10-1 10


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The rate constant of a reaction increases by a factor of1.23 when the temperature is increased form 20 °C to 27 °C. What is the activation energy of the reaction?From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 263.



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Make an appropriate Arrhenius plot for the following datafor the binding of an inhibitor to the enzyme carbonicanhydrase and calculate the activation energy for thereaction.From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 263.

T (K) k (10 Lmol s ) 1/T (K ) lnk6 -1 -1 -1

289.0 1.04 0.00346 13.9

293.5 1.34 0.00341 14.1

298.1 1.53 0.00335 14.2

303.2 1.89 0.00330 14.5

308.0 2.29 0.00325 14.6

313.5 2.84 0.00319 14.9

Plotting lnk vs 1/T:

aSlope = -E /R

a-3620.8= -E /8.314 J K mol-1 -1

aE = 30.1 kJ mol-1


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Food rots about 40 times more rapidly at 25 °C than whenit is stored at 4 °C. Estimate the overall activation energyfor the processes responsible for decomposition.From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 263.



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The pre-exponential factor and activation energy for theuni-molecular reaction

2 3CH HC (g) 6CH CN (g)are 4.0 x 10 s and 272 kJ mol , respectively. Calculate13 -1 -1

the values of )H° , )S° and )G° at 300 K.‡ ‡ ‡


a)H° = E = 272 kJ mol‡ -1

)G° = )H° - T)S° ‡ ‡ ‡

= 272 kJ mol - (300K)(15.4 x 10 kJ K mol )-1 -3 -1 -1

= 267 kJ mol -1


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Estimate the activation Gibbs energy for thedecomposition of urea in the reaction

2 2 2 4 3 (NH ) CO (aq) + 2H O (l) 6 2NH (aq) + CO (aq)+ 2-

for which the pseudo-first-order rate constant is1.2 x 10 s at 60 °C and 4.6 x 10 s at 70 °C.-7 -1 -7 -1


BUsing : k = k K , k =k T/h and )G° = -RTln K‡ ‡ ‡ ‡ ‡

60 °C 70 °C.

k 1.2 x 10 s 4.6 x 10 s -7 -1 -7 -1

T 333 K 343 K

Bk =k T/h 7.34 x 10 7.56 x 10‡ 12 12

K = k/k 1.63 x 10 6.07 x 10‡ ‡ -20 -20

)G° = -RTln K 126 kJ mol 126 kJ mol‡ ‡ -1 -1


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Calculate the Gibbs energy, enthalpy and entropy ofactivation (at 300 K) for the binding of an inhibitor to theenzyme carbonic anhydrase by using the following data:From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 293.

T (K) k (10 Lmol s ) 1/T lnk6 -1 -1

289.0 1.04 0.00346 13.9

293.5 1.34 0.00341 14.1

298.1 1.53 0.00335 14.2

303.2 1.89 0.00330 14.5

308.0 2.29 0.00325 14.6

313.5 2.84 0.00319 14.9

aSlope = -E /R = -36020.8 K

a)H° = E ‡

= -(-36020.8 K)(8.314 J K mol ) -1 -1

= 30.1 kJ mol-1


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intercept = lnA = 26.403A = 2.93 x 1011

)G° = )H° - T)S° ‡ ‡ ‡

= 30.1 kJ mol - (300K)(-25.9 x 10 kJ K mol )-1 -3 -1 -1

= 37.9 kJ mol -1


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The thermal isomerization of cyclopropane to propene inthe gas phase has a rate constant of 5.95 x 10 s at -4 -1

500 °C. Calculate the value of )G° for the reaction.‡


BUsing : k = k K , k =k T/h and )G° = -RTln K‡ ‡ ‡ ‡ ‡

k 5.95 x 10 s-4 -1

T 500 °C = 773 K

Bk =k T/h 1.70 x 10‡ 13

K = k/k 3.49 x 10‡ ‡ -17

)G° = -RTln K 243 kJ mol‡ ‡ -1


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The rate of the elctron-exchange reaction betweennaphthalene and its anion radical are diffusion controlled:

10 8 10 8 10 8 10 8C H + C H º C H + C H- -

The reaction is bimolecular and second order. Using therate constants below, calculate the values of Ea, )H° ,‡

)S° and )G° at 307 K for the reaction.‡ ‡


T (K) k (10 Lmol s ) 1/T lnk9 -1 -1

307 2.71 0.00326 21.7

299 2.40 0.00334 21.6

289 1.96 0.00346 21.4

273 1.43 0.00366 21.1

aSlope = -E /R = -1592.1 K

a)H° = E ‡

= -(-1592.1 K)(8.314 J K mol ) -1 -1

= 13.2 kJ mol-1


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intercept = lnA = 26.912A = 4.87 x 1011

)G° = )H° - T)S° ‡ ‡ ‡

= 13.2 kJ mol - (307K)(-21.9 x 10 kJ K mol )-1 -3 -1 -1

= 20.0 kJ mol -1


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Catalysis

• catalysts are species that increase the rate of areaction but are not consumed by the reaction• forms an intermediate with the reagent(s) in the

first step of the mechanism • is released in the product-forming step• does not appear in the overall reaction

• catalysts lower the )G° of the reaction by providing‡

a different mechanism for the reaction whichincreases the rate of reaction

a• lowering E• improving collision efficiency by improving the

orientation of the molecules• helping to break bonds• can also have a concentration effect

• the increase in rate applies to both the forward andreverse reactions

• used throughout history in food-preparation andwine-making


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rxn• catalysts effect the )G° not the overall )G° . ‡

Equilibrium may be achieved more quickly, but thethermodynamic equilibrium constant is not effected


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Three types of catalysts:

• Heterogenous• reactants and catalyst are in different phases

usually gas/solid or liquid/solid• examples:

• Haber synthesis of ammonia - solid iron • Ostwald synthesis of nitric acid - solid Pt/Rh• hydrogenation of liquid fatty acids to

saturated fatty acids - solid Pd, Pt or Ni

• Homogenous• reactants and catalyst are in the same phase

• Enzyme• subset of homogenous• very complex type of catalyst


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Example

The activation energy for the decomposition of hydrogenperoxide in solution is 76 kJ mol . When iodide ions are-1

added, the activation energy falls to 57 kJ mol . -1

Assuming that the pre-exponential factor does not changeupon addition of a catalyst, calculate the change in therate constant.

The enzyme catalase reduces the activation energy for thesame reaction to 8 kJ mol which corresponds to a rate-1

increase of 10 .15

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