CHAPTER 15 Chemical Kinetics. KINETICS — the study of REACTION RATES and their relation to the way...

CHAPTER 15

Chemical Kinetics

KINETICSKINETICS — — the study of the study of REACTION RATESREACTION RATES and and their relation to the way the reaction their relation to the way the reaction proceeds, i.e., its proceeds, i.e., its MECHANISMMECHANISM..

Reaction rate = change in concentration of Reaction rate = change in concentration of a reactant or product with time.a reactant or product with time. Three “types” of rates Three “types” of rates

initial rateinitial rateaverage rateaverage rateinstantaneous rateinstantaneous rate

The Rate of Reaction

• Consider the hypothetical reaction,

A(g) + B(g) C(g) + D(g)

• equimolar amounts of reactants, A and B, will be consumed while products, C and D, will be formed as indicated in this graph:

0

0.2

0.4

0.6

0.8

1

1.2

Time

Con

cent

rati

ons

of

Rea

ctan

ts &

Pro

duct

s

[A] & [B]

[C] & [D]

The Rate of Reaction• Mathematically, the rate of a reaction can be written as:

aA(g) + bB(g) cC(g) + dD(g)

t d

D+

t c

C+or

t b

B-

t a

A-= Rate

The Rate of Reaction

• The rate of a simple one-step reaction is directly proportional to the concentration of the reacting substance.

• [A] is the concentration of A in molarity or moles/L.

• k is the specific rate constant.

– k is an important quantity in this chapter.

Ak = Rateor ARate

C + BA (g)(g)(g)

The Rate of Reaction

• Important terminology for kinetics.

• The order of a reaction can be expressed in terms of either:

1 each reactant in the reaction or

2 the overall reaction. Order for the overall reaction is the sum of the orders

for each reactant in the reaction.

• For example:

overall.order first and

ONin order first isreaction This

ONk= Rate

O + NO4ON 2

52

52

g2g2g52

In general, for a A + b B --> x X with a catalyst C

Rate = k [A]m[B]n[C]p

The exponents m, n, and p • are the reaction order• can be 0, 1, 2 or fractions• must be determined by experiment!

The Rate of Reaction

• Given the following one step reaction and its rate-law expression.

– Remember, the rate expression would have to be experimentally determined.

• Because it is a second order rate-law expression:

– If the [A] is doubled the rate of the reaction will increase by a factor of 4. 22 = 4

– If the [A] is halved the rate of the reaction will decrease by a factor of 4. (1/2)2 = 1/4

2ggg

Ak = Rate

CBA 2

Factors That Affect Reaction Rates

• There are several factors that can influence the rate of a reaction:

1. The nature of the reactants.

2. The concentration of the reactants.

3. The temperature of the reaction.

4. The presence of a catalyst.

• We will look at each factor individually.

Nature of Reactants

• This is a very broad category that encompasses the different reacting properties of substances.

• For example sodium reacts with water explosively at room temperature to liberate hydrogen and form sodium hydroxide.

burns. and ignites H The

reaction. rapid and violent a is This

HNaOH 2OH 2Na 2

2

g2aq2s

Nature of Reactants

• Calcium reacts with water only slowly at room temperature to liberate hydrogen and form calcium hydroxide.

reaction. slowrather a is This

HOHCaOH 2Ca g2aq22s

Nature of Reactants

• The reaction of magnesium with water at room temperature is so slow that that the evolution of hydrogen is not perceptible to the human eye.

reaction No OH Mg 2s

Nature of Reactants

• However, Mg reacts with steam rapidly to liberate H2 and form magnesium oxide.

• The differences in the rate of these three reactions can be attributed to the changing “nature of the reactants”.

g2sC100

2s HMgOOHMg o

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Flour

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Concentrations of Reactants: The Rate-Law Expression

• This is a simplified representation of the effect of different numbers of molecules in the same volume.

– The increase in the molecule numbers is indicative of an increase in concentration.

A(g) + B (g) Products

A B

A B

A B BA B

A BA BA B

4 different possible A-B collisions

6 different possible A-B collisions

9 different possible A-B collisions

Concentrations of Reactants: The Rate-Law Expression

• Example 16-1: The following rate data were obtained at 25oC for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction? 2 A(g) + B(g) 3 C(g)

Experiment

Number

Initial [A]

(M)

Initial [B]

(M)

Initial rate of formation of C

(M/s)

1 0.10 0.10 2.0 x 10-4

2 0.20 0.30 4.0 x 10-4

3 0.10 0.20 2.0 x 10-4

Concentrations of Reactants: The Rate-Law Expression

Concentrations of Reactants: The Rate-Law Expression

• The following data were obtained for the following reaction at 25oC. What are the rate-law expression and the specific rate constant for the reaction?

2 A(g) + B(g) + 2 C(g) 3 D(g) + 2 E(g)

Experiment

Initial [A]

(M)

Initial [B]

(M)

Initial [C]

(M)

Initial rate of formation of D

(M/s)

1 0.20 0.10 0.10 2.0 x 10-4

2 0.20 0.30 0.20 6.0 x 10-4

3 0.20 0.10 0.30 2.0 x 10-4

4 0.60 0.30 0.40 1.8 x 10-3

Concentrations of Reactants: The Rate-Law Expression

Concentrations of Reactants: The Rate-Law Expression

• Consider a chemical reaction between compounds A and B that is first order with respect to A, first order with respect to B, and second order overall. From the information given below, fill in the blanks.

Experiment

Initial Rate

(M/s)

Initial [A]

(M)

Initial [B]

(M)

1 4.0 x 10-3 0.20 0.050

2 1.6 x 10-2 ? 0.050

3 3.2 x 10-2 0.40 ?

Concentrations of Reactants: The Rate-Law Expression

Concentration vs. Time: The Integrated Rate Equation

• The integrated rate equation relates time and concentration for chemical and nuclear reactions.

– From the integrated rate equation we can predict the amount of product that is produced in a given amount of time.

• Initially we will look at the integrated rate equation for first order reactions.

These reactions are 1st order in the reactant and 1st order overall.

Concentration vs. Time: The Integrated Rate Equation

• An example of a reaction that is 1st order in the reactant and 1st order overall is:

a A products

This is a common reaction type for many chemical reactions and all simple radioactive decays.

• Two examples of this type are:

2 N2O5(g) 2 N2O4(g) + O2(g)

238U 234Th + 4He

Concentration vs. Time: The Integrated Rate Equation

where:

[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.

k = specific rate constant. t = time elapsed since beginning of reaction.

a = stoichiometric coefficient of A in balanced overall equation.

• The integrated rate equation for first order reactions is:

k t aA

Aln 0

Concentration vs. Time: The Integrated Rate Equation

• Solve the first order integrated rate equation for t.

• Define the half-life, t1/2, of a reactant as the time required for half of the reactant to be consumed, or the time at which [A]=1/2[A]0.

A

Aln

k a

1t 0

Concentration vs. Time: The Integrated Rate Equation

• At time t = t1/2, the expression becomes:

Concentration vs. Time: The Integrated Rate Equation

• Cyclopropane, an anesthetic, decomposes to propene according to the following equation.

The reaction is first order in cyclopropane with k = 9.2 s-1 at 10000C. Calculate the half life of cyclopropane at 10000C.

CH2 CH2

CH2CH2CH3

CH

(g) (g)

Concentration vs. Time: The Integrated Rate Equation

• Refer to Previous Example: How much of a 3.0 g sample of cyclopropane remains after 0.50 seconds?

– The integrated rate laws can be used for any unit that represents moles or concentration.

– In this example we will use grams rather than mol/L.

Concentration vs. Time: The Integrated Rate Equation

• The half-life for the following first order reaction is 688 hours at 10000C. Calculate the specific rate constant, k, at 10000C and the amount of a 3.0 g sample of CS2 that remains after 48 hours.

CS2(g) CS(g) + S(g)

Concentration vs. Time: The Integrated Rate Equation• For reactions that are second order with respect to a

particular reactant and second order overall, the rate equation is:

• Where:

[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.

k = specific rate constant. t = time elapsed since beginning of

reaction.

a = stoichiometric coefficient of A in balanced overall equation.

k t aA

1

A

1

0

Temperature: The Arrhenius Equation

• Svante Arrhenius developed this relationship among (1) the temperature (T), (2) the activation energy (Ea), and (3) the specific rate constant (k).

k = Ae

or

ln k = ln A -ERT

-E RT

a

a

• If the Arrhenius equation is written for two temperatures, T2 and T1 with T2 >T1.

ln k ln A -ERT

and

ln k ln A -E

RT

1a

1

2a

2

Temperature: The Arrhenius Equation

1. Subtract one equation from the other.

ln k k A - ln A -E

RTERT

ln k kERT

-E

RT

2 1a

2

a

1

2 1a

1

a

2

ln ln

ln

2. Rearrange and solve for ln k2/k1.

ln kk

ER T T

or

ln kk

ER

T - TT T

2

1

a

1 2

2

1

a 2 1

2 1

1 1

Temperature: The Arrhenius Equation

• Consider the rate of a reaction for which Ea=50 kJ/mol, at 20oC (293 K) and at 30oC (303 K).

– How much do the two rates differ?

Temperature: The Arrhenius Equation

• For reactions that have an Ea50 kJ/mol, the rate approximately doubles for a 100C rise in temperature, near room temperature.

• Consider:2 ICl(g) + H2(g) I2(g) + 2 HCl(g)

• The rate-law expression is known to be R=k[ICl][H2].

At 230 C, k = 0.163 s

At 240 C, k = 0.348 s

k approximately doubles

0 -1 -1

0 -1 -1

M

M

Catalysts

• Catalysts change reaction rates by providing an alternative reaction pathway with a different activation energy.

• Homogeneous catalysts exist in same phase as the reactants.

• Heterogeneous catalysts exist in different phases than the reactants.

– Catalysts are often solids.

General rate expression

Differential rate law (with respect to concentration)Rate = k [A]m[B]n[C]p

If zero order Rate = k[A]0 = k first order Rate = k[A]1 = k[A] second order Rate = k[A]2

Integrated rate law (with respect to time)If zero order [A]0 - [A] = ak t

first order

second order

KineticsInitial rateInstantaneous rateAverage rateRate of reactionOrder of reactionOverall order of reactionCatalystFour factors that affect the rate of reaction

nature of reactantconcentrationtemperaturepresence of a catalyst

t d

D+

t c

C+or

t b

B-

t a

A-= Rate

[A] vs t ln [A] vs t 1/[A] vs t

k t aA

Aln 0

k t aA

1

A

1

0

[A] ln [A] 1/[A]

t t t

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55. A reaction has the following experimental rate equation: Rate = k[A]2[B]. If the concentration of A is doubled and the concentration of B is halved what happens to the rate?

87.The following statements relate to the reaction with the following rate law: Rate = k[H2][I2] H2(g) + I2(g) → 2 HI(g)

Determine which of the following statements are true. If a statement is false, indicate why it is incorrect.

a) The reaction must occur in a single stepb) This is a second-order reaction overallc) Raising the temperature will cause the value of k to decreased) Raising the temperature lowers the activation energy for this reactione) If the concentration of both reactants are doubled, the rate will doublef) Adding a catalyst in the reaction will cause the initial rate to decrease

89.Describe each of the following statements as true or false. If false, rewrite the sentence to make it correct.

a) The rate determining elementary step in a reaction is the slowest step in a mechanism

b) It is possible to change the rate constant by changing the temperaturec) As a reaction proceeds at constant temperature, the rate remains constantd) A reaction that is third order overall must involve more than one stepe) Reactions are faster at a higher temperature because activation energies

are lowerf) Rate increase with increasing concentration of reactants because there are

more collisions between reactant moleculesg) At higher temperatures a larger fraction of molecules have enough energy

to get over the activation energy barrierh) Catalyzed and uncatalyzed reactions have identical mechanisms