1 rates of reaction. 2 reaction rates chemical reactions require varying lengths of time for...

1

Rates of Reaction

2

Reaction Rates

Chemical reactions require varying lengths of time for completion.

– This reaction rate depends on the characteristics of the reactants and products and the conditions under which the reaction is run. (see Figure 14.1)

– By understanding how the rate of a reaction is affected by changing conditions, one can learn the details of what is happening at the molecular level.

3

Reaction Rates

The questions posed in this chapter will be:

– How is the rate of a reaction measured?– What conditions will affect the rate of a

reaction?– How do you express the relationship of rate to

the variables affecting the rate?– What happens on a molecular level during a

chemical reaction?

4

Reaction Rates

Chemical kinetics is the study of reaction rates, how reaction rates change under varying conditions, and what molecular events occur during the overall reaction.

• Concentration of reactants. • Concentration of a catalyst• Temperature at which the reaction occurs.• Surface area of a solid reactant or catalyst.

– What variables affect reaction rate?

5

Reaction Rates

Let’s look at each in more detail.

– What variables affect reaction rate?

Chemical kinetics is the study of reaction rates, how reaction rates change under varying conditions, and what molecular events occur during the overall reaction.

6

Factors Affecting Reaction Rates

Concentration of reactants.– More often than not, the rate of a reaction increases

when the concentration of a reactant is increased.– Increasing the population of reactants increases the

likelihood of a successful collision.– In some reactions, however, the rate is unaffected by

the concentration of a particular reactant, as long as it is present at some concentration.

7


Concentration of a catalyst.– A catalyst is a substance that increases

the rate of a reaction without being consumed in the overall reaction.

– The catalyst generally does not appear in the overall balanced chemical equation (although its presence may be indicated by writing its formula over the arrow).

)(2)(2 )(

)(22 22 glaqHBr

aq OOHOH

8


Concentration of a catalyst.

)g(2)l(2 )aq(HBr

)aq(22 OOH2OH2

– Figure 14.2 shows the HBr catalyzed decomposition of H2O2 to H2O and O2.

– A catalyst speeds up reactions by reducing the “activation energy” needed for successful reaction.

– A catalyst may also provide an alternative mechanism, or pathway, that results in a faster rate.

9


Temperature at which a reaction occurs.

– Usually reactions speed up when the temperature increases.

– A good “rule of thumb” is that reactions approximately double in rate with a 10 oC rise in temperature.

10


Surface area of a solid reactant or catalyst.

– Because the reaction occurs at the surface of the solid, the rate increases with increasing surface area.

– Figure 14.3 shows the effect of surface area on reaction rate.

11

Definition of Reaction Rate

The reaction rate is the increase in molar concentration of a product of a reaction per unit time. – Always a positive value.– It can also be expressed as the decrease in

molar concentration of a reactant per unit time.

12

Definition of Reaction Rates

Consider the gas-phase decomposition of dintrogen pentoxide.

)g(O)g(NO4)g(ON2 2252 – If we denote molar concentrations using brackets,

then the change in the molarity of O2 would be represented as

where the symbol, (capital Greek delta), means

“the change in.”

]O[ 2

13


Then, in a given time interval, t , the molar concentration of O2 would increase by [O2].

– This equation gives the average rate over the time interval, t.

– If t is short, you obtain an instantaneous rate, that is, the rate at a particular instant. (Figure 14.4)

– The rate of the reaction is given by:

t][O

oxygen of formation of Rate 2

Figure 14.4 The instantaneous rate of reaction

In the reaction

The concentration of O2 increases over time. You obtain the instantaneous rate from the slope of the tangent at the point of the curve corresponding to that time.

)g(O)g(NO4)g(ON2 2252

15


Figure 14.5 shows the increase in concentration of O2 during the decomposition of N2O5.

• Note that the rate decreases as the reaction proceeds.

Figure 14.5 Calculation of the average rate.

When the time changes from 600 s to 1200 s, the average rate is

2.5 x 10-6 mol/(L.s). Later when the time changes from 4200 s to 4800 s, the average rate has slowed to

5 x 10-7 mol/(L.s). Thus, the rate of a reaction decreases as the reaction proceeds.

17


Because the amounts of products and reactants are related by stoichiometry, any substance in the reaction can be used to express the rate.

t]O[N

ON of iondecomposit of Rate 5252 -

• Note the negative sign. This results in a positive rate as reactant concentrations decrease.

18

Definition of Reaction Rates The rate of decomposition of N2O5 and the formation of O2

are easily related.

)(t

]O[N2

1 t

][O 522 -

– Since two moles of N2O5 decompose for each mole of O2 formed, the rate of the decomposition of N2O5 is twice the rate of the formation of O2.

)g(O)g(NO4)g(ON2 2252

19

Experimental Determination of Reaction Rates

To obtain the rate of a reaction you must determine the concentration of a reactant or product during the course of the reaction.

– One method for slow reactions is to withdraw samples from the reaction vessel at various times and analyze them.

– More convenient are techniques that continuously monitor the progress of a reaction based on some physical property of the system.

20


Gas-phase partial pressures.

– Manometer readings provide the concentration of N2O5 during the course of the reaction based on partial pressures.

– When dinitrogen pentoxide crystals are sealed in a vessel equipped with a manometer (see Figure 14.6) and heated to 45oC, the crystals vaporize and the N2O5(g) decomposes.

)g(O)g(NO4)g(ON2 2252

21

This device indicates the difference between two pressures (differential pressure), or between a single pressure and atmosphere (gage pressure), when one side is open to atmosphere. If a U-tube is filled to the half way point with water and air pressure is exerted on one of the columns, the fluid will be displaced. Thus one leg of water column will rise and the other falls. The difference in height "h" which is the sum of the readings above and below the half way point, indicates the pressure .

22


Colorimetry

– The hypoiodate ion, IO-, absorbs near 400 nm. The intensity of the absorbtion is proportional to [IO-], and you can use the absorbtion rate to determine the reaction rate.

– Consider the reaction of the hypochlorite ion with iodide.

)aq(Cl)aq(IO)aq(I)aq(ClO

23

Dependence of Rate on Concentration

Experimentally, it has been found that the rate of a reaction depends on the concentration of certain reactants as well as catalysts.

– The rate of this reaction has been observed to be proportional to the concentration of nitrogen dioxide.

– Let’s look at the reaction of nitrogen dioxide with fluorine to give nitryl fluoride.

)g(FNO2)g(F)g(NO2 222

24


• When the concentration of nitrogen dioxide is doubled, the reaction rate doubles.

– The rate is also proportional to the concentration of fluorine; doubling the concentration of fluorine also doubles the rate.

– We need a mathematical expression to relate the rate of the reaction to the concentrations of the reactants..

25


A rate law is an equation that relates the rate of a reaction to the concentration of reactants (and catalyst) raised to various powers.

– The rate constant, k, is a proportionality constant in the relationship between rate and concentrations.

]][F[NOk Rate 22

26


As a more general example, consider the reaction of substances A and B to give D and E.

catalystC eEdDbBaA C – You could write the rate law in the form

pnm [C][B][A]k Rate – The exponents m, n, and p are frequently, but not

always, integers. They must be determined experimentally and cannot be obtained by simply looking at the balanced equation.

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Find the rate law expression and evaluate k for 2H2 + 2NO 2H2O + N2 @800K; all gases

H2 NO Rate (atm/min

1 .001 .006 .025

2 .002 .006 .050

3 .003 .006 .075

4 .009 .001 .0063

5 .009 .002 .025

6 .009 .003 .056

Rate = k [A]x[B]y

[A]x [B]y

Trial 2 = .002 x x .006 y= .050

Trial 1 .001 .006 .025

2x x 1y = 2

2x = 2

x = 1

28

Find the rate law expression and evaluate k for 2H2 + 2NO 2H2O + N2 @800K; all gases

H2 NO Rate (atm/min

1 .001 .006 .025

2 .002 .006 .050

3 .003 .006 .075

4 .009 .001 .0063

5 .009 .002 .025

6 .009 .003 .056

[A]x [B]y

Trial 6 = .009 1 x .003 y= .056

Trial 4 .009 .001 .0063

3y = 8.89

ylog3 =log 8.89

y = 2

Rxn is first order for x

second order for y , and

(2 + 1 =) third order overall.

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Rate = k[H2][NO]2

From Trial 1: .025atm = k(.001M)(.006M)2

min

.025atm = k(.001M)(.006M)2

min

6.94 x 105 atm = k

min M3

30

Find the rate law expression and evaluate k for H2 + 2NO 2H2O + N2O @1100K; all gases

PNO [atm] PH2 [atm] Rate [atm/min]

1 .150 .400 .020

2 .075 .400 .005

3 .150 .200 .101

31

Find the rate law expression and evaluate k for H2 + 2NO 2H2O + N2O @1100K; all gases

ANSWERS:x = 2y = 1k = 2.22(min atm)-1

32


Reaction Order– The reaction order with respect to a given

reactant species equals the exponent of the concentration of that species in the rate law, as determined experimentally.

– The overall order of the reaction equals the sum of the orders of the reacting species in the rate law.

33


Reaction Order– Consider the reaction of nitric oxide with

hydrogen according to the following equation.

– Thus, the reaction is second order in NO, first order in H2, and third order overall.

– The experimentally determined rate law is

][H[NO]k Rate 22

)g(OH2)g(N)g(H2)g(NO2 222

34


Reaction Order

– Zero and negative orders are also possible.– The concentration of a reactant with a zero-

order dependence has no effect on the rate of the reaction..

– Although reaction orders frequently have whole number values (particularly 1 and 2), they can be fractional.

35


Determining the Rate Law.– One method for determining the order of a

reaction with respect to each reactant is the method of initial rates.

– It involves running the experiment multiple times, each time varying the concentration of only one reactant and measuring its initial rate.

– The resulting change in rate indicates the order with respect to that reactant.

36


Determining the Rate Law.– If doubling the concentration of a reactant has a

doubling effect on the rate, then one would deduce it was a first-order dependence. 2:2 correspondence

– If doubling the concentration had a quadrupling effect on the rate, one would deduce it was a second-order dependence. 2:4 correspondence

– A doubling of concentration that results in an eight-fold increase in the rate would be a third-order dependence. 2:8 correspondence

37

A Problem to Consider

Iodide ion is oxidized in acidic solution to triiodide ion, I3- , by hydrogen peroxide.

– A series of four experiments was run at different concentrations, and the initial rates of I3

- formation were determined.

– From the following data, obtain the reaction orders with respect to H2O2, I-, and H+.

– Calculate the numerical value of the rate constant.

)(2)()(2)(3)( 2322 lOHaqIaqHaqIaqOH

38


Initial Concentrations (mol/L)H2O2 I- H+ Initial Rate [mol/(L.s)]

Exp. 1 0.010 0.010 0.00050 1.15 x 10-6

Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6

– Comparing Experiment 1 and Experiment 2, you see that when the H2O2 concentration doubles (with other concentrations constant), the rate doubles.

– This implies a first-order dependence with respect to H2O2.

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Exp. 1 0.010 0.010 0.00050 1.15 x 10-6

Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6

– Comparing Experiment 1 and Experiment 3, you see that when the I- concentration doubles (with other concentrations constant), the rate doubles.

– This implies a first-order dependence with respect to I-.

40



Exp. 1 0.010 0.010 0.00050 1.15 x 10-6

Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6

– Comparing Experiment 1 and Experiment 4, you see that when the H+ concentration doubles (with other concentrations constant), the rate is unchanged.

– This implies a zero-order dependence with respect to H+.

41


– Because [H+]0 = 1, the rate law is:

]I][OH[kRate 22

– The reaction orders with respect to H2O2, I-, and H+, are 1, 1, and 0, respectively.


Exp. 1 0.010 0.010 0.00050 1.15 x 10-6

Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6

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Exp. 1 0.010 0.010 0.00050 1.15 x 10-6

Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6

– You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:

Lmol

010.0L

mol010.0k

sLmol

1015.1 6

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Exp. 1 0.010 0.010 0.00050 1.15 x 10-6

Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6

– You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:

)/(102.1/010.0010.0

1015.1 216

smolLLmol

sk

44

Change of Concentration with Time

A rate law simply tells you how the rate of reaction changes as reactant concentrations change.– A more useful mathematical relationship

would show how a reactant concentration changes over a period of time.

45

Change of Concentration with Time

A rate law simply tells you how the rate of reaction changes as reactant concentrations change.– Using calculus we can transform a rate law

into a mathematical relationship between concentration and time.

– This provides a graphical method for determining rate laws.

46

Concentration-Time Equations

First-Order Integrated Rate Law– You could write the rate law in the form

]A[k t

]A[ Rate

47


First-Order Integrated Rate Law– Using calculus, you get the following equation.

kt- ]A[]A[

lno

t

– Here [A]t is the concentration of reactant A at time t, and [A]o is the initial concentration.

– The ratio [A]t/[A]o is the fraction of A remaining at time t.

48


The decomposition of N2O5 to NO2 and O2 is first order with a rate constant of 4.8 x 10-4 s-1. If the initial concentration of N2O5 is 1.65 x 10-2 mol/L, what is the concentration of N2O5 after 825 seconds?

– The first-order time-concentration equation for this reaction would be:

kt- ]ON[]ON[

lno52

t52

49


The decomposition of N2O5 to NO2 and O2 is first order with a rate constant of 4.8 x 10-4 s-1. If the initial concentration of N2O5 is 1.65 x 10-2 mol/L, what is the concentration of N2O5 after 825 seconds?– Substituting the given information we obtain:

s) 825()s10(4.80- L/mol1065.1

]ON[ln 1-4-

2t52

50


– Substituting the given information we obtain:

0.396- L/mol1065.1

]ON[ln 2

t52


51


– Taking the inverse natural log of both sides we obtain:

673.0 L/mol1065.1

]ON[ 0.396-2

t52 e


52


– Solving for [N2O5] at 825 s we obtain:

L/mol 0111.0)673.0()L/mol101.65(]O[N 2-52


53


Second-Order Integrated Rate Law– You could write the rate law in the form

2]A[k t

]A[ Rate

54


Second-Order Integrated Rate Law– Using calculus, you get the following equation.

kt [A]

1

][

1

o

tA

– Here [A]t is the concentration of reactant A at time t, and [A]o is the initial concentration.

55


Zero-Order Integrated Rate Law– The Zero-Order Integrated Rate Law equation

is:.

ktAA ][][ o

56


First order integrated rate law

Second order integrated rate law

Zero order integrated rate law

kt [A]

1

][

1

o

tA

kt - ][

][ln

o

t

A

A

ktAA ][][ o

57


k values

General k formula:

Units k = (L/mol)order-1

unit of time First order: s-1

Second order: L/mol•sec, or L•mol-1•sec-1

Zero order: L2/mol2•sec, or L2•mol-2•sec-1

58

Half-life

The half-life of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value.

– For a first-order reaction, the half-life is independent of the initial concentration of reactant.

21kt)ln(2

1

– In one half-life the amount of reactant decreases by one-half. Substituting into the first-order concentration-time equation, we get:

21693.)ln( 2/1 kt

59

Half-life

The half-life of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value.

– Solving for t1/2 we obtain:

k693.0

t21

– Figure 14.8 illustrates the half-life of a first-order reaction.

60

Half-life

Sulfuryl chloride, SO2Cl2, decomposes in a first-order reaction to SO2 and Cl2.

– At 320 oC, the rate constant is 2.2 x 10-5 s-1.

What is the half-life of SO2Cl2 vapor

at this temperature?

)g(Cl)g(SO)g(ClSO 2222

– Substitute the value of k into the relationship

between k and t1/2.

k693.0

t21

61

Half-life



What is the half-life of SO2Cl2 vapor at

this temperature?– Substitute the value of k into the relationship

between k and t1/2.

1-5 s 1020.2

693.0t

21


62

Half-life



What is the half-life of SO2Cl2 vapor at this

temperature?– Substitute the value of k into the relationship

between k and t1/2.

s1015.3t 421


63

Half-life

For a second-order reaction, half-life depends on the initial concentration and becomes larger as time goes on.

– Again, assuming that [A]t = ½[A]o after one half-life, it can be shown that:

o]A[k1

t21

– Each succeeding half-life is twice the length of its predecessor.

64

Half-life

For Zero-Order reactions, the half-lite is dependent upon the initial concentration of the reactant and becomes shorter as the reaction proceeds.

k2

]A[t o

21

65

Half-life

First order reactions

Second order reactions

Zero order reactions

k2

]A[t o

21 2

1kt)ln(21

o]A[k1

t21

21693.)ln( 2/1 kt

66

Graphing Kinetic Data

In addition to the method of initial rates, rate laws can be deduced by graphical methods.

– If we rewrite the first-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.

ot ]Aln[kt]Aln[ y = mx + b

67



– If we rewrite the first-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.

– This means if you plot ln[A] versus time, you will get a straight line for a first-order reaction. (see Figure 14.9)

68



– If we rewrite the second-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.

ot ]A[1

kt]A[

1

y = mx + b

69



– If we rewrite the second-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.

– This means if you plot 1/[A] versus time, you will get a straight line for a second-order reaction.

– Figure 14.10 illustrates the graphical method of deducing the order of a reaction.

72

Collision Theory

Rate constants vary with temperature. Consequently, the actual rate of a reaction is very temperature dependent.

• Why the rate depends on temperature can by explained by collision theory.

73

Collision Theory

Collision theory assumes that for a reaction to occur, reactant molecules must collide with sufficient energy and the proper orientation.

• The minimum energy of collision required for two molecules to react is called the activation energy, Ea.

74

Transition-State Theory

Transition-state theory explains the reaction resulting from the collision of two molecules in terms of an activated complex.– 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.

75


Transition-state theory explains the reaction resulting from the collision of two molecules in terms of an activated complex.

– 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.

76



– 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.”

– At the instant of impact, when all three spheres are joined, we have an unstable transition-state complex.

77



– 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.

78

Potential-Energy Diagrams for Reactions

To illustrate graphically the formation of a transition state, we can plot the potential energy of a reaction versus time.

– Figure 14.13 illustrates the endothermic reaction of nitric oxide and chlorine gas.

– Note that the forward activation energy is the energy necessary to form the activated complex.

– The H of the reaction is the net change in energy between reactants and products.

Figure 14.13 Potential-energy curve for the endothermic reaction of nitricoxide and chlorine.

80

Potential-Energy Diagrams for Reactions

The potential-energy diagram for an exothermic reaction shows that the products are more stable than the reactants.

– Figure 14.14 illustrates the potential-energy diagram for an exothermic reaction.

– We see again that the forward activation energy is required to form the transition-state complex.

– In both of these graphs, the reverse reaction must still supply enough activation energy to form the activated complex.

Figure 14.14 Potential-energy curve for an exothermic reaction.

82

Collision Theory and the Arrhenius Equation

Collision theory maintains that the rate constant for a reaction is the product of three factors.

1. Z, the collision frequency

2. f, the fraction of collisions with sufficient energy to react

3. p, the fraction of collisions with the proper orientation to react

Zpfk

83


Z is only slightly temperature dependent.

– This is illustrated using the kinetic theory of gases, which shows the relationship between the velocity of gas molecules and their absolute temperature.

m

abs

MRT3

velocity absT velocity or

84


Z is only slightly temperature dependent.

– This alone does not account for the observed increases in rates with only small increases in temperature.

– From kinetic theory, it can be shown that a 10 oC rise in temperature will produce only a 2% rise in collision frequency.

85


On the other hand, f, the fraction of molecules with sufficient activation energy, turns out to be very temperature dependent.

– It can be shown that f is related to Ea by the following expression.

RTaE-

e f – Here e = 2.718… , and R is the ideal gas

constant, 8.31 J/(mol.K).

86


On the other hand, f, the fraction of molecules with sufficient activation energy turns out to be very temperature dependent.

– From this relationship, as temperature increases, f increases.

RTaE-

e f Also, a decrease in the activation energy, Ea, increases the value of f.

87


On the other hand, f, the fraction of molecules with sufficient activation energy turns out to be very temperature dependent.

RTaE-

e f

– This is the primary factor relating temperature increases to observed rate increases.

88


The reaction rate also depends on p, the fraction of collisions with the proper orientation.

– This factor is independent of temperature changes.

– So, with changes in temperature, Z and p remain fairly constant.

– We can use that fact to derive a mathematical relationship between the rate constant, k, and the absolute temperature.

89

The Arrhenius Equation

If we were to combine the relatively constant terms, Z and p, into one constant, let’s call it A. We obtain the Arrhenius equation:

– The Arrhenius equation expresses the dependence of the rate constant on absolute temperature and activation energy.

RTaE-

e A k

90


It is useful to recast the Arrhenius equation in logarithmic form.

– Taking the natural logarithm of both sides of the equation, we get:

RTEa- Aln k ln

91


It is useful to recast the Arrhenius equation in logarithmic form.

We can relate this equation to the (somewhat rearranged) general formula for a straight line.

)(- Aln k ln T1

REa

y = b + m xA plot of ln k versus (1/T) should yield a straight line with a slope of (-Ea/R) and an intercept of ln A. (see Figure 14.15)

92


A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k1, (1/T1)) and (k2, (1/T2)).– The two equations describing the relationship at

each coordinate would be

)(- Aln k ln1

a

T1

RE

1

)(- Aln k ln2

a

T1

RE

2 and

93


A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k1, (1/T1)) and (k2, (1/T2)).

– We can eliminate ln A by subtracting the two equations to obtain

)( ln21

a

1

2

T1

T1

RE

kk

94


A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k1, (1/T1)) and (k2, (1/T2)).

)( ln21

a

1

2

T1

T1

RE

kk

– With this form of the equation, given the activation energy and the rate constant k1 at a given temperature T1, we can find the rate constant k2 at any other temperature, T2.

95


The rate constant for the formation of hydrogen iodide from its elements

is 2.7 x 10-4 L/(mol.s) at 600 K and 3.5 x 10-3 L/(mol.s) at 650 K. Find the activation energy, Ea.

)(K650

1K600

1K)J/(mol 8.31

E

107.2

103.5 ln a

4

3-

– Substitute the given data into the Arrhenius equation.

)g(HI2)g(I)g(H 22

96



)g(HI2)g(I)g(H 22

– Simplifying, we get:

)1028.1(J/(mol) 8.31E

11.1 )101.30( ln 4a1

is 2.7 x 10-4 L/(mol.s) at 600 K and 3.5 x 10-3 L/(mol.s) at 650 K. Find the activation energy, Ea.

97



– Solving for Ea:

J1066.11028.1

mol/J31.811.1E 5

4a

)g(HI2)g(I)g(H 22 is 2.7 x 10-4 L/(mol.s) at 600 K and 3.5 x 10-3 L/(mol.s) at 650 K. Find the activation energy, Ea.

98

Reaction Mechanisms

Even though a balanced chemical equation may give the ultimate result of a reaction, what actually happens in the reaction may take place in several steps.– This “pathway” the reaction takes is referred to

as the reaction mechanism.– The individual steps in the larger overall reaction

are referred to as elementary reactions. (See animation: Decomposition of N2O5 Step 1)

99

Elementary Reactions

Consider the reaction of nitrogen dioxide with carbon monoxide.

)g(CO)g(NO)g(CO)g(NO 22

– This reaction is believed to take place in two steps.

)g(NO)g(NO)g(NO)g(NO 322


(elementary reaction)

(elementary reaction)

100


Each step is a singular molecular event resulting in the formation of products.

– Note that NO3 does not appear in the overall equation, but is formed as a temporary reaction intermediate.

101


Each step is a singular molecular event resulting in the formation of products.

– The overall chemical equation is obtained by adding the two steps together and canceling any species common to both sides.

)g(NO)g(NO)g(NO)g(NO 322


)g(CO)g(NO)g(NO)g(NO 322 )g(CO)g(NO)g(NO)g(NO 223

102

Molecularity

We can classify reactions according to their molecularity, that is, the number of molecules that must collide for the elementary reaction to occur.– A unimolecular reaction involves only one

reactant molecule.– A bimolecular reaction involves the collision of

two reactant molecules.– A termolecular reaction requires the collision of

three reactant molecules.

103

Molecularity

We can classify reactions according to their molecularity, that is, the number of molecules that must collide for the elementary reaction to occur.– Higher molecularities are rare because of

the small statistical probability that four or more molecules would all collide at the same instant.

104

• For elementary reactions, the rate is proportional to the concentrations of all reactant molecules involved.

Rate Equations for Elementary Reactions

Since a chemical reaction may occur in several steps, there is no easily stated relationship between its overall reaction and its rate law.

105


For example, consider the generic equation below.

products A

The rate is dependent only on the concentration of A; that is,

k[A] Rate

106


However, for the reaction

products B A

the rate is dependent on the concentrations of both A and B.

k[A][B] Rate

107


For a termolecular reaction

products C B A

the rate is dependent on the populations of all three participants.

k[A][B][C] Rate

108


Note that if two molecules of a given reactant are required, it appears twice in the rate law. For example, the reaction

products B 2A

would have the rate law:

[B]k[A] Rate 2k[A][A][B] Rate or

109


So, in essence, for an elementary reaction, the coefficient of each reactant becomes the power to which it is raised in the rate law for that reaction.

– Note that many chemical reactions occur in multiple steps and it is, therefore, impossible to predict the rate law based solely on the overall reaction.

110

Rate Laws and Mechanisms

Consider the reaction below.F(g)NO 2 (g)F (g)NO 2 222

– Experiments performed with this reaction show that the rate law is

]][Fk[NO Rate 22– The reaction is first order with respect to

each reactant, even though the coefficient for NO2 in the overall reaction is 2.

111

Rate Laws and Mechanisms

Consider the reaction below.

– Experiments performed with this reaction show that the rate law is

F(g)NO 2 (g)F (g)NO 2 222

]][Fk[NO Rate 22– This implies that the reaction above is not an

elementary reaction but rather the result of multiple steps.

112

Rate-Determining Step

In multiple-step reactions, one of the elementary reactions in the sequence is often slower than the rest.– The overall reaction cannot proceed any

faster than this slowest rate-determining step.

113


In multiple-step reactions, one of the elementary reactions in the sequence is often slower than the rest.– Our previous example occurs in two elementary

steps where the first step is much slower.

F(g)NO 2 (g)F (g)NO 2 222

F(g)NO F(g) (g)NO 2

22k

F(g) F(g)NO (g)F (g)NO 2

221k (slow)

(fast)

114


In multiple-step reactions, one of the elementary reactions in the sequence is often slower than the rest.– Since the overall rate of this reaction is determined

by the slow step, it seems logical that the observed rate law is Rate = k1[NO2][F2].

(slow)F(g) F(g)NO (g)F (g)NO 2

221k

115


In a mechanism where the first elementary step is the rate-determining step, the overall rate law is simply expressed as the elementary rate law for that slow step.

– A more complicated scenario occurs when the rate-determining step contains a reaction intermediate, as you’ll see in the next section.

116


Mechanisms with an Initial Fast Step– There are cases where the rate-

determining step of a mechanism contains a reaction intermediate that does not appear in the overall reaction.

– The experimental rate law, however, can be expressed only in terms of substances that appear in the overall reaction.

117


Consider the reduction of nitric oxide with H2.

– It has been experimentally determined that the rate law is Rate = k [NO]2[H2]

)g(OH2)g(N)g(H2)g(NO2 222 – A proposed mechanism is:

OHONHON 22 k

2222

OHNHON 22 k

223

22ON NO2k1

k-1

(fast, equilibrium)

(slow)

(fast)

118


The rate-determining step (step 2 in this case) generally outlines the rate law for the overall reaction.

– As mentioned earlier, the overall rate law can be expressed only in terms of substances represented in the overall reaction and cannot contain reaction intermediates.

]H][ON[k Rate 2222(Rate law for the rate-determining step)

119



– It is necessary to reexpress this proposed rate law after eliminating [N2O2].


120



– We can do this by looking at the first step, which is fast and establishes equilibrium.


121



– At equilibrium, the forward rate and the reverse rate are equal.


]ON[k]NO[k 2212

1

122



– Therefore, 2

1122 ]NO)[k/k(]ON[ – If we substitute this into our proposed rate

law we obtain:


123



]H[]NO[k

kkRate 2

2

1

12

– If we replace the constants (k2k1/k-1) with k, we obtain the observed rate law: Rate = k[NO]2[H2].


124


Determine the overall reaction and the rate expression that corresponds to the following reaction mechanism.

2A + B ↔ D

D + B → E + F

F → GWhitten 6th Ed. Ch 16 # 66a

125


Determine the overall reaction and the rate expression that corresponds to the following reaction mechanism.

2A + B ↔ DD + B → E + F

F → GAnswer:

2A + 2B → E + GRate = k[A]2[B]2

126

Catalysts

A catalyst is a substance that provides a good “environment” for a reaction to occur, thereby increasing the reaction rate without being consumed by the reaction.

– To avoid being consumed, the catalyst must participate in at least one step of the reaction and then be regenerated in a later step.

127

Catalysts

A catalyst is a substance that provides a good “environment” for a reaction to occur, thereby increasing the reaction rate without being consumed by the reaction.

– Its presence increases the rate of reaction by either increasing the frequency factor, A (from the Arrhenius equation) or lowering the activation energy, Ea.

128

Catalysts

Homogeneous catalysis is the use of a catalyst in the same phase as the reacting species.

– The oxidation of sulfur dioxide using nitric oxide as a catalyst is an example where all species are in the gas phase.

)g(SO2)g(O)g(SO2 3 )g(NO

22

129

Catalysts

Heterogeneous catalysis is the use of a catalyst that exists in a different phase from the reacting species, usually a solid catalyst in contact with a liquid or gaseous solution of reactants.

– Such surface catalysis is thought to occur by chemical adsorbtion of the reactants onto the surface of the catalyst.

– Adsorbtion is the attraction of molecules to a surface.

130

Enzyme Catalysis

Enzymes have enormous catalytic activity.

– The substance whose reaction the enzyme catalyzes is called the substrate. (see Figure 14.20)

– Figure 14.21 illustrates the reduction in acivation energy resulting from the formation of an enzyme-substrate complex.

131

Operational Skills

Relating the different ways of expressing reaction rates

Calculating the average reaction rate Determining the order of reaction from the rate law Determining the rate law from initial rates Using the concentration-time equation for first-

order reactions Relating the half-life of a reaction to the rate

constant

132

Operational Skills

Using the Arrhenius equation• Writing the overall chemical equation from a

mechanism• Determining the molecularity of an elementary

reaction• Writing the rate equation for an elementary

reaction• Determining the rate law from a mechanism

133

Figure 14.1: Combining product of formaldehyde with hydrogen sulfite ion.Photo courtesy of James Scherer.

Return to Slide 2

134

Figure 14.9: A plot of ln [N2O5] versus time.

Return to Slide 58

135

Figure 14.15: Plot of ln k versus 1/T.

Return to Slide 80

136

Animation: Decomposition of N2O5

Return to Slide 87

(Click here to open QuickTime animation)

137

Figure 14.20: Enzyme action (lock-and-key model).

Return to Slide 117

138

Figure 14.21 Potential-energy curves for the reaction of substrate, S, to products, P.

Return to Slide 117

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