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 0.5Z = -1

Rate = k[I-][OCl-]/[OH-]The reaction is first order with

respect to I- and OCl-, and inverse first order

withrespect 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]0 time

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 T

f = 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 TheoryTwo 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 MechanismN2O5 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 slowStep 2: F + NO2 NO2F fast 

A 2-step mechanism was proposed for a reaction:

Step 1:NO(g) + NO(g) N2O2 (g) FASTStep 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?