Chapter 14 – Chemical Kinetics

The rate of a chemical reaction is the speed at which products or formed and reactants broken down. There factors that affect the rate of a chemical reaction are temperature, concentration, state of reactants, and the presence of a catalyst. A rate law can be constructed to predict the rate at which products form or reactants disappear. Some chemical reactions happen in several steps called a reaction mechanism. The slowest step of a reaction mechanism is called the rate-limiting step. A catalyst is a factor that can speed up a chemical reaction. Biological catalysts are called enzymes that are an important part of biochemical reactions within living organisms.

Chemical Kinetics

• Factors that affect rates of chemical reactionso Collision Theory – In order for two

reactants to form products, they must collide with enough kinetic energy and in the correct orientation.

Chemical Kinetics

• Factors that affect rates of chemical reactions1. -

2. -

3. -

4. -

Chemical Kinetics

• Factors that affect rates of chemical reactionso Concentration – The more abundant the

reactants are, the more successful collisions to form products.

Chemical Kinetics

• Factors that affect rates of chemical reactionso Temperature – The more kinetic energy the

reactants possess, the more likely a collision will result in a successful reaction.

Chemical Kinetics

• Factors that affect rates of chemical reactionso Catalyst – A substance the helps the

reactants collide in the correct orientation.o They are not consumed in a chemical

reaction.

Chemical Kinetics

• Factors that affect rates of chemical reactionso Inhibitors – Compete with binding sites on

a catalyst that slow, or even stop, a chemical reaction form occurring.

Chemical Kinetics

• Rates of chemical reactionso The speed at which a chemical reaction occurs is called

its ‘reaction rate’.o The rate of a chemical reaction is measured as the

change in concentration of reactant or products per unit of time.

Rate = Δ Concentration / Time = Molarity / Time

= M / s

** In chemistry [ ] means the molar concentration **

Chemical Kinetics

• Rates of chemical reactionso A flask is charged with 0.100 mol of A and allowed to

react to form B according to the hypothetical gas-phase reaction: A(g) B(g).

a.) Calculate the number of mole of B at each time in the table.

Time (s) 0 40 80 120 160

Moles of A 0.100 0.067 0.045 0.030 0.020

Chemical Kinetics

• Rates of chemical reactionso A flask is charged with 0.100 mol of A and allowed to

react to form B according to the hypothetical gas-phase reaction: A(g) B(g).

b.) Calculate the average rate of disappearance of A for each 40 s interval in units of mol / s.

Time (s) 0 40 80 120 160

Moles of A 0.100 0.067 0.045 0.030 0.020

Chemical Kinetics

• Rates of chemical reactionso A flask is charged with 0.100 mol of A and allowed to

react to form B according to the hypothetical gas-phase reaction: A(g) B(g).

c.) What additional information would be needed to calculate the rate in units of concentration per time?

Time (s) 0 40 80 120 160

Moles of A 0.100 0.067 0.045 0.030 0.020

Chemical Kinetics

• Rates of chemical reactionso Instantaneous Rate versus Average Rate of Reactions

Instantaneous Rate – Refers to the rate of a chemical reaction at a specific time.

Average Rate – Refers to the rate of a reaction over an interval of time.

Which rate do you think is usually higher?

Chemical Kinetics

• Rates of chemical reactionso The role of coefficients on reaction rates

aA + bB cC + dD

In terms of ‘A’, the rate of reaction will be

Rate = - 1 Δ[A] a Δt

Chemical Kinetics

• Rates of chemical reactionso The role of coefficients on reaction rates

aA + bB cC + dD

In terms of ‘B’, the rate of reaction will be

Rate = - 1 Δ[B] b Δt

Chemical Kinetics

• Rates of chemical reactionso The role of coefficients on reaction rates

aA + bB cC + dD

In terms of ‘C’, the rate of reaction will be

Rate = 1 Δ[C] c Δt

Chemical Kinetics

• Rates of chemical reactionso The role of coefficients on reaction rates

aA + bB cC + dD

In terms of ‘D’, the rate of reaction will be

Rate = 1 Δ[D] d Δt

Chemical Kinetics

• Rates of chemical reactionso The decomposition of N2O5 proceeds according to the

following reaction: 2N2O5(g) 4NO2(g) + O2(g)

If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2 and (b) O2?

Chemical Kinetics

• Rates Laws of Chemical Reactionso The rate of a chemical reaction cannot be determined without

conducting an experiment.o The initial concentration of reactants are measured and the rate

of the reaction is determined.

X + Y Z

Trial Initial [X] Initial [Y] Rate (M/s)

1 0.0100 0.200 5.4 x 10-7

2 0.0200 0.200 10.8 x 10-7

3 0.0400 0.200 21.5 x 10-7

4 0.200 0.0202 10.8 x 10-7

5 0.200 0.0404 21.6 x 10-7

6 0.200 0.0808 43.3 x 10-7

Chemical Kinetics

• Rates Laws of Chemical Reactions

aA + bB cC + dD

Rate = k [A]m [B]n

k = rate constant[A] = initial molar concentration of A[B] = initial molar concentration of Bm = reaction order exponentn = reaction order exponent

Chemical Kinetics

• Rates Laws of Chemical ReactionsNH4

+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

• Notice that when the concentration of NH4+ is doubled, and NO2

- is held constant, the rate doubles.

• Notice that when the concentration of NO2- is doubled, and NH4

+ is held constant, the rate doubles.

Trial Initial [NH4+] Initial [NO2

-] Rate (M/s)

1 0.0100 0.200 5.4 x 10-7

2 0.0200 0.200 10.8 x 10-7

3 0.0400 0.200 21.5 x 10-7

4 0.200 0.0202 10.8 x 10-7

5 0.200 0.0404 21.6 x 10-7

6 0.200 0.0808 43.3 x 10-7

Chemical Kinetics

• Rates Laws of Chemical ReactionsNH4

+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

Rate = k [NH4+] [NO2

-]

• Since NH4+

and NO2- affect the rate proportionally, we do not

need to raise to concentration of the ions to any power.• The order of both reactants is 1.

Trial Initial [NH4+] Initial [NO2

-] Rate (M/s)

1 0.0100 0.200 5.4 x 10-7

2 0.0200 0.200 10.8 x 10-7

3 0.0400 0.200 21.5 x 10-7

4 0.200 0.0202 10.8 x 10-7

5 0.200 0.0404 21.6 x 10-7

6 0.200 0.0808 43.3 x 10-7

Chemical Kinetics

• Rates Laws of Chemical ReactionsNH4

+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

Rate = k [NH4+] [NO2

-]• The overall order of this reaction is the sum of the two orders

of the reactants. (1+1 = 2)• This is second order reaction overall.

Trial Initial [NH4+] Initial [NO2

-] Rate (M/s)

1 0.0100 0.200 5.4 x 10-7

2 0.0200 0.200 10.8 x 10-7

3 0.0400 0.200 21.5 x 10-7

4 0.200 0.0202 10.8 x 10-7

5 0.200 0.0404 21.6 x 10-7

6 0.200 0.0808 43.3 x 10-7

Chemical Kinetics

• Rates Laws of Chemical Reactions

Example 1: The experimentally determined rate law for the reaction 2NO(g) + 2H2(g) N2(g) + 2H2O(l) is rate = k[NO]2[H2].

a.) What are the reaction orders in this rate law?

b.) Does doubling the concentration of NO have the same effect on the rate as doubling the concentration of H2?

Chemical Kinetics

• Rates Laws of Chemical Reactionso Units of the rate constant (k)

Units of rate constant = units of rate = M/s = M-1 . s-1

(units of concentration)2 M2

Chemical Kinetics

• Rates Laws of Chemical Reactionso Example 2: The initial rate of reaction A + B C was

measured for several different starting concentrations of A and B, and the results are as follows;

a.) Determine the rate law for this reaction.

Trial [A] [B] Initial Rate(M/s)

1 0.100 0.100 4.0 x 10-5

2 0.100 0.200 4.0 x 10-5

3 0.200 0.100 16.0 x 10-5

Chemical Kinetics

• Rates Laws of Chemical Reactionso Example 2: The initial rate of reaction A + B C was

measured for several different starting concentrations of A and B, and the results are as follows;

b.) Determine the rate constant.

Trial [A] [B] Initial Rate(M/s)

1 0.100 0.100 4.0 x 10-5

2 0.100 0.200 4.0 x 10-5

3 0.200 0.100 16.0 x 10-5

Chemical Kinetics

• Rates Laws of Chemical Reactionso Example 2: The initial rate of reaction A + B C was

measured for several different starting concentrations of A and B, and the results are as follows;

b.) Determine the rate of reaction when [A] = 0.050 M and [B] = 0.100 M

Trial [A] [B] Initial Rate(M/s)

1 0.100 0.100 4.0 x 10-5

2 0.100 0.200 4.0 x 10-5

3 0.200 0.100 16.0 x 10-5

Chemical Kinetics

• Rates Laws of Chemical Reactionso Integrated Rate Laws

o An integrated rate law allows use to calculate the rate of a chemical reaction during a specific time interval.

integrate rate law (first-order reactions):

ln [A]t = -kt or ln [A]t - ln [A]0 = -kt [A]0

ln = natural logarithm[A]t = concentration of A a specific time[A]0 = initial concentration of Ak = rate law constantt = time interval

Chemical Kinetics

• Rates Laws of Chemical Reactionso Integrated Rate Law ln [A]t - ln [A]0 = -kt

rearrange into

ln [A]t = -k t + ln [A]0

y = m . x + b

ln = natural logarithm[A]t = concentration of A a specific time[A]0 = initial concentration of Ak = rate law constantt = time interval

Chemical Kinetics

• Rates Laws of Chemical Reactionso Integrated Rate Laws – The graph on the left shows [A]

versus time. The graph in the middle shows the ln [A] versus time. A linear relationship of ln [A] proves that it is a first-order reaction.

Chemical Kinetics

• Rates Laws of Chemical Reactionso Integrated Rate Laws

o Example: The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45y-1 at 12°C. A quantity of this insecticide is seeps into a lake on June 1st, leading to a concentration of 5.0 x 10-7 g/cm3. Assuming that the lake remains at 12°C:

a.) What is the concentration of the insecticide in the lake on June 1st of the following year?

Chemical Kinetics

• Rates Laws of Chemical Reactionso Integrated Rate Laws

o Example 1: The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45y-1 at 12°C. A quantity of this insecticide is seeps into a lake on June 1st, leading to a concentration of 5.0 x 10-7 g/cm3. Assuming that the lake remains at 12°C:

a.) What is the concentration of the insecticide in the lake on June 1st of the following year?

Chemical Kinetics

• Rates Laws of Chemical Reactionso Integrated Rate Laws

o Example 1: The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45y-1 at 12°C. A quantity of this insecticide is seeps into a lake on June 1st, leading to a concentration of 5.0 x 10-7 g/cm3. Assuming that the lake remains at 12°C:

b.) How long will it take for the concentration of the insecticide in the lake to drop to 3.0 x 10-7 g/cm3?

Chemical Kinetics

• Rates Laws of Chemical Reactionso Integrated Rate Laws

o Example 2: The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of 6.8 x 10-4 s-1:

(CH3)2O(g) CH4(g) + H2(g) + CO(g)

If the initial pressure of (CH3)2O is 135 torr, what is the pressure after 1420 s?

Chemical Kinetics

• Rates Laws of Chemical Reactionso Integrated Rate Laws (second-order reactions)

1 = kt + 1 [A]t [A]0

[A]0 = initial concentration [A]t = concentration at time t k = rate law constant t = time

Chemical Kinetics

• Rates Laws of Chemical Reactionso Integrated Rate Laws (second-order reactions)

Example: The following is a second-order reaction2NO2(g) 2NO(g) + O2(g) k = 0.543 M-1s-1

If the initial concentration of NO2 in a closed vessel is 0.0500M, What is the remaining concentration after 0.500h?

Chemical Kinetics

• Rates Laws of Chemical Reactionso Half-Life (t1/2) – The time required for the concentration

to fall to half of its original amount.

[A]t1/2 = ½[A]0

Calculating the the half life of a first-order reaction:

t1/2 = ln1/2 = 0.693 k k

Chemical Kinetics

• Rates Laws of Chemical Reactionso Half-Life (t1/2) –

Example: Calculate the half-life for the decomposition of the insecticide used in the previous example (slide 31).

How long does it take for the concentration of the insecticide to reach ¼ of its original value?

Chemical Kinetics

• Temperature and Rates of Chemical Reactions• An increase in the temperature of a chemical reaction

will also increase the rate constant, k.

Chemical Kinetics

• Temperature and Rates of Chemical Reactions• An increase in the temperature is also an increase in

the kinetic energy of the reactants.• More kinetic energy = more velocity = more collisions

= more successful collisions.

Chemical Kinetics

• Temperature and Rates of Chemical Reactions• Not only do molecules need to collide with enough

kinetic energy, but they also need to be in the correct orientation.

Chemical Kinetics

• Temperature and Rates of Chemical Reactions• Activation energy (Ea) – The minimum amount of

energy required to initiate a chemical reaction.

Chemical Kinetics

• Temperature and Rates of Chemical Reactions• Activation energy (Ea) – The minimum amount of

energy required to initiate a chemical reaction.

Chemical Kinetics

• Temperature and Rates of Chemical Reactions• Energy diagrams

Chemical Kinetics

• Temperature and Rates of Chemical Reactions• Activation Energy

• Arrhenius Equation – Incorporates the fraction of molecules that posses enough kinetic energy (Ea), the number of molecular collisions, and the fraction of reactants that are in the correct orientation.

k = Ae-Ea/RT

k = reaction constantAe = frequency factor constant

Ea = activation energy

R = gas law constantT = Temperature (K)

Chemical Kinetics

• Temperature and Rates of Chemical Reactions• Determining Activation Energy

ln k1 = Ea ( 1 - 1 ) k2 R ( T2 T1 )

K1 = rate constant at T1K2 = previously known rate constant at T2Ea = activation energyT1 = Temperature of reaction with k1T2 = Temperature of reaction with k2

Chemical Kinetics

• Reaction Mechanisms • A detailed step-by-step explanation as to how a

reaction occurs.• Some reactions occur only as 1 step.• Other reactions occur in several steps.

Chemical Kinetics

• Reaction Mechanisms • Elementary reactions (1 step)

NO(g) + O3(g) NO2(g) + O2(g)

• The rate of this reaction is only determined in this step.

Chemical Kinetics

• Reaction Mechanisms • Multistep Reaction Mechanisms (2 or more steps)

NO2(g) + CO(g) NO(g) + CO2(g)

This reaction is the result of two steps;

NO2(g) + NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)

The sum of these reactions gives us the overall reaction.

Chemical Kinetics

• Reaction Mechanisms • Multistep Reaction Mechanisms (2 or more steps)

NO2(g) + NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)

2NO2(g) + NO3(g) + CO(g) NO3(g) + NO(g) + NO2(g) + CO2(g)

NO2(g) + CO(g) NO(g) + CO2(g)

NO3(g) is called an intermediate, it is neither a reactant nor a product.

Chemical Kinetics

• Reaction Mechanisms • Example: For the reaction

Mo(CO)6 + P(CH3)3 Mo(CO)5P(CH3)3 + CO

The proposed mechanism isMo(CO)6 Mo(CO)5 + CO

Mo(CO)5 + P(CH3)3 Mo(CO)5P(CH3)3

a.) Is the proposed mechanism consistent with the overall reaction?

Chemical Kinetics

• Reaction Mechanisms • Example: For the reaction

Mo(CO)6 + P(CH3)3 Mo(CO)5P(CH3)3 + CO

The proposed mechanism isMo(CO)6 Mo(CO)5 + CO

Mo(CO)5 + P(CH3)3 Mo(CO)5P(CH3)3

b.) Identify any intermediate(s).

Chemical Kinetics

• Reaction Mechanisms • Example: For the reaction

Mo(CO)6 + P(CH3)3 Mo(CO)5P(CH3)3 + CO

The proposed mechanism isMo(CO)6 Mo(CO)5 + CO

Mo(CO)5 + P(CH3)3 Mo(CO)5P(CH3)3

b.) Identify the molecularity of each step of the mechanism.

Chemical Kinetics

• Reaction Mechanisms and Rate Determining Stepso Single step reactions – The rate of the reaction is

determined by the speed of one step.

o Multiple Step Reactions – The rate of the reaction will be determine by the slowest step (rate determining step).

NO2(g) + NO2(g) NO3(g) + NO(g) (slow)

NO3(g) + CO(g) NO2(g) + CO2(g) (fast)

---------------------------------------------------------NO2(g) + CO(g) NO(g) + CO2(g)

Chemical Kinetics

• Catalyts

o A substance that speeds up a chemical reaction without being consumed as a reagent.

o Catalysts speed up a chemical reaction by lowering the activation energy.

Chemical Kinetics

• Catalytso Homogenous Catalyts – Catalysts in the same state as

the reagents.

Chemical Kinetics

• Catalytso Heterogeneous Catalyts – Catalysts that are not in the

same state as the reagents.