Chemical Kinetics

Rates of chemical reactions and how they can be measured

experimentally and described mathematically

So far, we have worked with reactions that occur almost

instantaneously

• Precipitations– Ba2+ (aq) + SO4

2- (aq) BaSO4 (s)

• Acid-base reactions– HCl(aq) + NaOH (aq) NaCl (aq)+ H2O

(l)

Lots of reactions are much slower…

• Rusting– 4Fe (s) + 3O2 (g) 2Fe2O3 (s)

• Formation of ammonia– N2 (g) + 3H2 (g) 2NH3 (g)

• Formation of diamond– C (graphite) C (diamond)

In this chapter we will…

• Explore the factors that affect rates of reactions

• Quantify the influence of the above factors on the rates of reactions

• Determine what happens at the molecular level in reactions

REACTION RATES• The change in molar concentration

of a reactant or a product per unit time

• Units are Moles/L s

In an experiment, the decomposition of N2O5 in 100 mL of a 2.00 M CCl4 solution produced 0.500 L of oxygen at STP after 200 minutes. Calculate the concentration of unreacted N2O5 at this time. 2N2O5 (in CCl4) 4NO2 (in CCl4) + O2 (g)

• ? Moles oxygen produced• ? Moles N2O5 reacted

• ? Moles N2O5 initially

• ? Moles N2O5 left over• ? Moles/L

• 0.500 L x 1mole = 0.02232 mol O2

22.4 L

• 0.02232 mol O2 x 2N2O5 = 0.04464 mol N2O5

1 O2

• Initially… (0.100L) (2.00M) = 0.200 mol N2O5

• 0.200 mol – 0.04464 mol = 0.155 mol left over

• M = 0.155 mol/ 0.100 L = 1.55 M

Following this procedure we can calculate the concentration of unreacted N2O5 at any stage of the reaction and plot the data as shown

Mol/L

Time

To find the rate at any instant, draw a tangent to the curve and determine its slope

This gives the average rate over time, D[N2O5]/Dt

If you know the rate of one substance in a reaction, you can determine the rate of any other substance in the reaction by using a mole ratio2N2O5 (in CCl4) 4NO2 (in CCl4) + O2

(g)N2O5 decomposes at a rate of

0.23M/s,what is the rate of formation of NO2 ?0.23M N2O5 x 4 NO2 = 0.46 M/s NO2

s 2 N2O5

Rate Laws• Have the general form: rate =

k[A]n

• [A] = concentration of reactants and catalyst

• k = rate constant• n = order of reactant

(an integer or fraction)

Initial Rate Method• To determine the form of the rate

law, one MUST use experimental data

• Do a series of experiments in which the initial concentrations of the reactants are varied one at a time and record the initial rates of reaction

(OH-)

I- (aq) + OCl- (aq) OI- (aq) + Cl- (aq)

initial M Rate (mol/Ls)

I- OCl- OH-1) 0.010.01 0.01 6.1 x 10-4

2) 0.020.01 0.01 12.2 x 10-4

3) 0.010.02 0.01 12.3 x 10-4

4) 0.010.01 0.02 3.0 x 10-4

Rate = k[I-]x [OCl-]y [OH-]z

1: 6.1 x 10-4 = k[0.01]x[0.01]y[0.01]z

2: 12.2 x 10-4 = k[0.02]x[0.01]y[0.01]z

Solve for x…

0.5 = 0.5x

X = 1

1: 6.1 x 10-4 = k[0.01]x[0.01]y[0.01]z

3: 12.3 x 10-4 = k[0.01]x[0.02]y[0.01]z

Solve for y

0.5 = 0.5y

Y = 1

1: 6.1 x 10-4 = k[0.01]x[0.01]y[0.01]z

4: 3.0 x 10-4 = k[0.01]x[0.01]y[0.02]z

Solve for z

2 = 0.5z (ohhh…this is tricky!)Log 2 = log 0.5z

Z log 0.5 = log 2Z = log 2/log 0.5

Z = -1

Rate = k[I-][OCl-]/[OH-]

The reaction is first order with respect to

I- and OCl-, and inverse first order with

respect to [OH-].What is the overall reaction order?(sum of exponents)What are the units of k for this

reaction?(plug units into the rate law and

solve)

YOU TRY!

Do your “You Try!” section now.

Integrated Rate Law Method

First Order Rate Lawsrate = k [A]1 = - D [ A ] / D t

orrate = -D[A] = k D t

[A]

Integrating both sides gives…

-ln ( [A]t / [A]0 ) = kt

ln ( [A]0 / [A]t ) = kt

ln [A]t = -kt + ln [A]0time

ln[A]t

slope = - k

Half-life• time it takes for the reactant

concentration to reach one half of its initial value

• symbol t ½ • for first order:

t ½ = (ln 2) / k

Integrated Rate Law Method

Second Order Rate Lawsrate = k [A]2 = - D

[ A ] /D tintegrating gives

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

t1/2 = 1 / k[A]0

2nd order graph

Integrated Rate Law Method

Zero Order Rate Lawsrate = k [A]0 = k

integrating gives[A]t = -kt + [A]0

t1/2 = [A]0 / 2k

Rate data was plotted, what is the order of NO2?

What is the order with respect to A?

YOU TRY!

• Do your “You Try!” section now.

Collision Theory

The rate of a reaction depends on the

1. concentration of reactants2. temperature3. presence/absence of a

catalyst

The value of the rate constant is

dependent on the temperature

How can we explain the effect of

temperature on the rate of areaction?

NO(g) + Cl2(g) --> NOCl(g) + Cl- (g)

At 25oC: k= 4.9 x 10-6 L/mol sAt 35oC: k= 1.5 x 10-5 L/mol s

k has increased by a factor of THREE!

WHY?

The Collision Theory states that in order to react, molecules have to collide….

• with the proper orientation

• with an energy at least equal to Ea

activation energy (Ea ) =

required minimum energy for areaction to occur

k for a reaction depends on 3 things:

• Z = collision frequency (# collisions/second)higher temperature means more collisions

• f = fraction of collisions that occur with E > Ea

*this factor changes rapidly with Tf = e –(Ea/RT)

• p = fraction of collisions that occur with the proper orientation (independent of T)

Overall: k = pfz = A e -(Ea/RT)

• where A = pz• Arrhenius Equation: ln k = -Ea 1 +

ln AR T

• ln k2 = Ea 1 1 k1 R T1 T2

YOU TRY!

Calculate the activation energy for the reaction: 2HI (g) --> H2 (g) + I2 (g)

GIVEN: k at 650. K = 2.15 x 10-8 L/mol s and k at 700. K = 2.39 x 10-7 L/mol s

R = 8.3145 J / mol K1.82 x 105 J

Transition State Theory

Two reactants come together to form an

Activated Complex, or TRANSITION STATE which then separates to form theproducts.

Potential Energy Diagram2NO + Cl2 2NOCl

PE

Reaction

In the activated complex, the N—Cl

bond has partially formed, while the

Cl—Cl bond has partially broken.

• Breaking bonds requires an input of energy while forming bonds

RELEASES energy.

• If E reactants > E products then the reaction is EXOTHERMIC

• If E reactants < E products then the reaction is ENDOTHERMIC

Catalysis Uncatalyzed Catalyzed

• A catalyst increases the rate of reaction by…LOWERING THE ACTIVATION ENERGY

• Homogeneous Catalysis- catalyst is in the same phase as the reactants

• Heterogeneous Catalysis- catalyst is in a different phase than the reactants

The Haber Process: N2 + 3 H2 2NH3

Reaction Mechanisms

The overall balanced equation usually

represents the SUM of a series of simple

reactions called ELEMENTARY STEPSbecause they represent the progress

ofthe reaction at the molecular level.

The sequence of elementary steps iscalled the REACTION MECHANISM

NO2 (g) + CO (g) NO (g) + CO2 (g)Rate = k [NO2]2

The above reaction actually takes place in two steps:

1. NO2 + NO2 NO3 + NO SLOW

2. NO3 + CO NO2 + CO2 FAST

• Intermediate-

• Unimolecular step:

• Bimolecular step:

• Termolecular step:

The reaction mechanism must satisfy two requirements:

1) Sum of elementary steps must be the overall reaction

2) The rate law indicated by the mechanism must match the experimentally determined rate law

2 H2O2(aq) O2(g) + 2 H2O(l)by experiment Rate = k[H2O2] [I-]

H2O2(aq) + I-(aq) IO-(aq) + H2O(l) SLOW

H2O2(aq) + IO-(aq) I-(aq) + H2O(l) + O2(g)

The overall rate of the reaction is controlled by the slow step also known as the…..

RATE DETERMINING STEP

H2O2(aq) + I-(aq) IO-(aq) + H2O(l) SLOW

H2O2(aq) + IO-(aq) I-(aq) + H2O(l) + O2(g)

When the slow step is used to determine

the rate law, we get:

What is the intermediate in this reaction?

What is the catalyst?

2N2O5 (g) ----------> 4NO2 (g) + O2 (g)

Rate = k[N2O5]

Why can’t this be a one-step reaction?

If it was a one-step, the overall reaction

would be that one step, and the rate law

would be rate = k[N2O5]2

Proposed Mechanism

N2O5 NO2 + NO3 FAST

NO2 + NO3 NO + O2 + NO2

SLOW

NO + NO3 2NO2 FAST

Rate = k[ ]

A general example:

E + S MfastM E + P slow

 Work out the rate law in terms of

reactants and catalyst:

Is this an acceptable mechanism? Why or why not?

Overall Reaction = 2NO2 (g) + F2 (g) 2NO2F (g) Rate = k[NO2] [F2]

 Proposed MechanismStep 1: NO2 + F2 NO2F + F slow

Step 2: F + NO2 NO2F fast

 

A 2-step mechanism was proposed for a reaction:

Step 1:NO(g) + NO(g) N2O2 (g) FAST

Step 2:N2O2 (g) + O2 (g) 2NO2 (g) SLOW

 a) what is the overall reaction? b) what is the rate law? c) what were the reactants? the product?

the intermediate? the catalyst?