chemical kinetics chapter 14. summary of the kinetics reactions orderrate law concentration-time...

Chemical Kinetics

Chapter 14

Summary of the Kinetics Reactions

Order Rate LawConcentration-Time

Equation Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]2

ln[A] = ln[A]o - kt

1[A]

=1

[A]o

+ kt

[A] = [A]o - kt

t½ln 2k

=

t½ =[A]o

2k

t½ =1

k[A]o

rate = k [A][B] t½ =1

k[A]o

2

Generally, as T increases, so does the reaction rate

k is temperature dependent

Fig 14.11 Temperature affectsthe rate of chemiluminescenceIn light sticks

Fig 14.12 Dependence of rate constant on temperature

In a chemical rxn, bonds are broken and new bonds are formed

Molecules can only react if they collide with each other

Furthermore, molecules must collide with the correct orientation and with enough energy

Fig 14.13

Cl + NOCl Cl2 + NO

Activation Energy

Fig 14.14 An energy barrier

Reactant

Activated complexor

Transition state

Product

Activation energy (Ea) ≡ minimum amount of energy required to initiate a chemical reaction

Fig 14.15

Activated complex (transition state) ≡ very short-lived and cannot be removed from reaction mixture

methyl isonitrile acetonitrile

Sample Exercise 14.10 Rank the following series of reactionsfrom slowest to fastest

• The lower the Ea the faster the reaction

• Slowest to fastest: (2) < (3) < (1)

ExothermicExothermic Endothermic

How does a molecule acquire sufficient energyto overcome the activation barrier?

Fig 14.16 Effect of temperature on distribution of kinetic energies

f = e-Ea

RT

Fraction ofmolecules with

E ≥ Ea

R = 8.314 J/(mol ∙ K)T = kelvin temperature

Maxwell–Boltzmann Distributions

f = e-Ea

RTFraction of molecules with E ≥ Ea

R = 8.314 J/(mol ∙ K)T = kelvin temperature

e.g., suppose Ea = 100 kJ/mol at T = 300 K:

f = e-Ea

RT = 3.9 x 10-18 (implies very few energetic molecules)

at T = 300 K: f = 1.4 x 10-17 (about 3.6 times more molecules)

Temperature Dependence of the Rate ConstantTemperature Dependence of the Rate Constant

k = A • exp(− Ea/RT )

Ea ≡ activation energy (J/mol)

R ≡ gas constant (8.314 J/K•mol)

T ≡ kelvin temperature

A ≡ frequency factor

ln k = −Ea

R1T

+ ln A

(Arrhenius equation)

y = mx +b

Fig 14.12

k1 = A • exp(−Ea/RT1)

The Arrehenius equation can be used to relate

rate constants k1 and k2 at temperatures T1 and T2.

21

21a

2

1

TTTT

RE

kk

ln

k2 = A • exp(−Ea/RT2)combine to give:

Plot of ln k vs 1/T

Slope = −Ea/R

ln k =− Ea

R1T

+ ln A

Fig 14.17 Graphical determination of activation energy

y = m x + b

Arrhenius Equation

Sample Exercise 14.11 The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures:

(a) From these data, calculate the activation energy for the reaction.

(b) What is the value of the rate constant at 430.0 K?

Solution

Analyze: We are given rate constants, k, measured at several temperatures and asked to determine the activation energy, Ea, and the rate constant, k, at a particular temperature.

Plan: We can obtain Ea from the slope of a graph of ln k versus 1/T and the rate constant, k, at a particular temperature. Once we know Ea, we can use Equation 14.21 together with the given rate data to calculate the rate constant at 430.0 K.

Solve: (a) We must first convert the temperatures from degrees Celsius to kelvins. We then take the inverse of each temperature, 1/T, and the natural log of each rate constant, ln k.

A graph of ln k versus 1/T results in a straight line, as shown in Figure 14.17:

slope = − Ea/R = − 1.9 x 104

We use the value for the molar gas constant R in units of J/mol-K (Table 10.2). We thus obtain

(b) To determine the rate constant, k1, at T1 = 430.0 K, we can use Equation 14.21 with Ea = 160 kJ/mol, and one of the rate constants and temperatures from the given data, such as:

k2 = 2.52 × 10–5 s–1 and T2 = 462.9 K:

Thus,

Note that the units of k1 are the same as those of k2.

Practice Exercise

The first-order rate constant for the reaction of

methyl chloride with water to produce methanol and

hydrochloric acid is 3.32 x 10-10 s-1 at 25 °C. Calculate

the rate constant at 40 °C if the activation energy is

116 kJ/mol.

21

21a

2

1

TTTT

RE

kk

ln

k2 = 0.349 s-1