Chemical Kinetics. Kinetics Chemical kinetics is the branch of chemistry concerned with the rate of chemical reactions and the mechanism by which chemical

Chemical Kinetics

Presented by Mark Langella, PWISTA.com

Kinetics

Chemical kinetics is the branch of chemistry concerned with the rate of

chemical reactions and the mechanism by which chemical reactions occur.


Collision Theory

In order for two particles to react chemically, they must collide. Not only must they collide, but it must be an “effective collision.” That is, they must have the correct amount of energy and collide with the proper orientation in space.

Any factor which increases the likelihood that they will collide will increase the rate of the chemical reaction.


FACTORS WHICH AFFECT THE RATE OF A CHEMICAL

REACTION (Bonds Must Break)

Surface Area/ Contact Area (Opportunity for collisions)

Concentration ( Increase Frequency) Temperature ( Increase Frequency) Catalyst ( Effective Collisions) Nature of Reactants ( Effective collisions)


Surface Area /Contact Area

Sugar Cube vs. Packet of Sugar Steel vs Steel Wool Two Sided Candle Coffee Creamer Alka-tablets


Concentration

Sudsy Kinetics Magnesium in Various Concentrated Acids Which Reactant has the greater

concentration effect ( that’s another story)


Which of the following hydrogen peroxide and sodium iodide mixtures will give the

greatest rate of reaction?

30% 10% 3%


Temperature

Spoiling of Fruit Alka Seltzer Hot and Cold Light Stick Kinetics


Temp vs. Reaction rate

The effect of temperature on the rate of a chemical reaction can best be explained in terms of the kinetic molecular theory. The higher the temperature, the faster molecules move. The faster they more, the faster they collide and therefore, react at a faster rate.


Every ten degrees rate doubles

Energy, E

T0

T0 + 10


Arrhenius Equation

- Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy).

- Orientation of reactants must allow formation of new bonds.


Temperature

Activation EnergyT1

T2

Energy


12_300

T1

T2

00

EaEnergy

T2 > T1

Plot showing the number of collisions with a particular energy at T1& T2, where T2 > T1 -- Boltzman Distribution.


Arrhenius Equation (continued)

k = rate constant A = frequency factor Ea = activation energy T = temperature R = gas constant

k Ae E RT a /


Arrhenius Equation

If the natural logarithm of each side of the Arrhenius Equation is taken, the following equation results:

ln(k) = -Ea/R (1/T) + ln(A) y = mx + b

m = -Ea/R when ln(k) is plotted versus 1/T.



THE RELATIONSHIP BETWEEN k AND T

k

k RT

= A - e

or ln = ln A - E

-EaRT

a /


Since the second form of the equation follows the form, y = mx + b, then a graph of ln k vs 1/T should give a straight line.

The activation energy of the reaction may be calculated from the slope of the ln k vs 1/T plot.

Slope = - E

Therefore, E = - R (slope)a

a

/ R


If you have data of two points

1 21

ln / (1/ 1/ )2

kEa R T T

k






Final Notes on Kinetics

• As T increases, so does k.• Ea & E are independent of T.

• A catalyst lowers Ea and increases the rate of both kf & kr.


Light Stick Kinetics

Bend a light stick to break the glass tube inside mixing the hydrogen peroxide with the phenyl oxalate ester. Repeat this several times to eliminate any large sections of glass and expedite mixing. Place the light stick aside for two to three hours. Your instructor may have activated and aged the light sticks for you.


Prepare the data collection device by slipping the one-hole #00 stopper prepared for this experiment over the temperature probe. It might be helpful to lubricate the hole in the stopper with silicone oil or spray. Cut the top from an aged and activated light stick with a sharp knife or scissors. Divide its contents equally with another lab group into two small test tubes.


Place the temperature probe into the mixture. Adjust the position of the temperature probe so that it is centered in the solution but just visible when you sight into the light sensor hole. It must not extend too deep as to interfere with the light sensor readings. Clamp the foam block onto the ring stand with a large clamp. Gently force the light sensor into the hole prepared for it on the side of the foam block. It should be deep enough to fit securely but not touch the test tube.


Place the switch on the light sensor to the 600 lux sensitivity. Plug the light sensor into Channel one (CH1) of a powered Lab Pro and the temperature probe into Channel two (CH2).

Connect the LabPro to your computer by means of the appropriate cable. Double click the Logger Pro icon on your computer. It should auto-ID both sensors. If not, check your connections and try again. If your instructor has a file setup for this experiment, open it now.


Remove the test tube and temperature probe from the foam block, placing it into a beaker of hot (45 - 50°C) water. Using the temperature probe as a handle, stir the water with the test tube observing the temperature until it remains more or less constant. Remove the test tube from the water, dry the outside, and insert it into the foam block.

When everything is ready, click on the green “COLLECT” icon on the icon toolbar, and observe the data. The experiment will run for fifteen minutes.


When the measurement is complete, disassemble the apparatus, returning all items to their designated areas. Dispose of the spent light stick solution as indicated by your instructor.

Clean both the temperature probe and the test tube with a brush using soap.


1 a) Click on the label, Lumination (lux) on the vertical axis of your graph. Select the natural log of the light intensity, “ln Illumination,” for the vertical axis. Click on the label, “Time (s),” on the horizontal axis of your graph. Select “Inverse Temperature.” Auto scale your graph. Your graph now displays the ln intensity vs 1/temperature data necessary to find the activation energy.


1 (b) Drag your curser across the most linear portion of your log I vs. 1/T plot. The linear section should be highlighted by a dark gray rectangle. Select the linear fit icon in the icon toolbar. The best fit solution will be displayed. Record these values.

1 (c) Calculate the activation energy for the light stick reaction from the slope found in Question #1b above. Express your answer in kilojoules per mole (kJ/mol).


Catalysis

Catalyst: A substance that speeds up a reaction without being consumed

Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.

Homogeneous catalyst: Present in the same phase as the reacting molecules.

Heterogeneous catalyst: Present in a different phase than the reacting molecules.


12_303

E

Reactants

Products

Catalyzedpathway

Uncatalyzedpathway

Reaction progress

Ene

rgy

Energy plots for a catalyzed and an uncatalyzed pathway for an endothermic reaction.


12_304

Ea (uncatalyzed )

Effectivecollisions(uncatalyzed)

Effectivecollisions(catalyzed)

Ea (catalyzed )

(a) (b)

Nu

mb

er

of

co

llis

ion

s

with

a g

ive

n e

ne

rgy

Nu

mb

er

of

co

llis

ion

s

with

a g

ive

n e

ne

rgy

Energy Energy

Effect of a catalyst on the number of reaction-producingcollisions. A greater fraction of collisions are effective for the catalyzed reaction.


Heterogeneous Catalysis

1. Adsorption and activation of the reactants. 2. Migration of the adsorbed reactants on the

surface. 3. Reaction of the adsorbed substances. 4. Escape, or desorption, of the products.

Steps:

Oscillating Flask Demo


Homogeneous Catalysis

Catalyst is in the same phase as the reacting molecules.

NO(g) + 1/2O2(g) ----> NO2(g)

NO2(g) ----> NO(g) + O(g)

O2(g) + O(g) ----> O3(g)

3/2 O2(g) ----> O3(g)

What is the catalyst in this reaction? What are the intermediates?


Nature of Reactant

Whoosh Bottle Closed vials with 2 cm of

White phenolphthalein in NaCl added to calcium Hydroxide

Ferric Chloride added to Sodium Acetate


Spontaneity

• tendency for a reaction to occur.• does not mean that the reaction will be

fast!• Diamonds will spontaneously change into

graphite, but the process is so slow that it is not detectable.


Reaction Mechanism

• the steps by which a chemical process occurs.

• allows us to find ways to facilitate reactions.

• can be changed by the use of a catalyst.


Reaction Rate

Change in concentration (conc) of a reactant or product per unit time.

Rate = conc of A at time conc of A at time 2 1

2 1

t tt t

A

t



Reaction Rate

It is customary to work with positive reaction rates, so a negative sign is used in some cases to make the rate positive.

Rates determined over a period of time are called average rates.

Instantaneous rate equals the negative slope of the tangent line.


12_291

0.000370s

O2

0.0025

0.005

0.0075

0.0100

0.0006

70s

0.0026

110 s

NO2

NO

50 100 150 200 250 300 350 400

Con

cent

ratio

ns (

mol

/L)

Time (s)

[NO2 ]

t

The concentrations of nitrogen dioxide, nitric oxide, oxygen plotted versus time.


12_1575

Time Time

(a) (b) (c)

Representation of the reaction of 2 NO2(g) ---> 2 NO(g) + O2(g).

a) t = 0 b) & c) with increased time, NO2 is changed intoNO and O2.


Rate Laws(differential)

Rate = k[NO2]n

k = rate constantn = rate order


Types of Rate Laws

Differential Rate Law: expresses how rate depends on concentration.

Integrated Rate Law: expresses how concentration depends on time.


Rate Laws Summary• Differential rate law -- rate of a reaction depends on

concentration.• Integrated rate law -- concentration depends on time.• Rate laws normally only involve concentrations of

reactants.• Experimental determination of either rate law is

sufficient.• Experimental convenience dictates which rate law is

determined experimentally.• Rate law for a reaction often indicates reaction

mechanism.


12_292

.20

.40

.60

.80

1.00

400 800 1200 1600 2000

Rate = 5.4 x 10-4 mol/L.s

Rate = 2.7 x 10-4 mol/L.s[N

2O5]

(mol

/L)

Time (s)

Plot of the concentration of N2O5 as a function of time for thereaction 2N2O5(soln) ---> 4NO2(soln) + O2(g). Rate at 0.90 M is twice the rate at 0.45 M.


Method of Initial Rates

Initial Rate: the “instantaneous rate” just after the reaction begins.

The initial rate is determined in several experiments using different initial concentrations.


Overall Reaction Order

Sum of the order of each component in the rate law.

rate = k[H2SeO3][H+]2[I]3

The overall reaction order is 1 + 2 + 3 = 6.


First-Order Rate Law

Integrated first-order rate law is

ln[A] = kt + ln[A]o

y = mx + b If a reaction is first-order, a plot of ln[A] versus time

is a straight line.

Rate = A

A

t

k

For aA Products in a 1st-order reaction,


Half-Life of a First-Order Reaction

t1/2 = half-life of the reactionk = rate constant

For a first-order reaction, the half-life does not depend on concentration.

tk1/2

0 693.


12_294

[N2O5]0 0.1000

0.0100

0.0200

0.0300

0.0400

0.0500

0.0600

0.0700

0.0800

0.0900

[N2O5]02

[N2O5]04

[N2O5]08

50 150 250 350100 200 300 400

t1/2 t1/2 t1/2

Time (s)

[N2O

5](m

ol/L

)

Plot of [N2O5] versus time for the decomposition reaction of N2O5. Note that the half-life for a 1st order reaction isconstant.


Second-Order Rate Law For aA products in a second-order

reaction,

Integrated rate law is

Rate = A

A

t

k 2

1

A +

1

A o

kt

y = mx + bIf a reaction is second-order, a plot of 1/[A]

versus time is a straight line.


Half-Life of a Second-Order Reaction

t1/2 = half-life of the reactionk = rate constantAo = initial concentration of A

The half-life is dependent upon the initial concentration.

tk1/2

oA

1


Zero-Order Rate Law For aA---> products in zero-order reaction,

Rate = k[A]o = k The integrated rate law is

[A] = -kt + [A]0

y = mx + b If a reaction is zero-order, the plot of [A] versus time is a straight

line. Example -- surface of a solid catalyst cannot hold a greater

concentration of reactant.

12_06T

Table 12.6 Summary of the Kinetics for Reactions of the Type aA Products That AreZero, First, or Second Order in [A]

Order

Zero First Second

Rate law Rate = k Rate = k[A] Rate = k[A]2

Integrated rate law [A] = -kt + [A]0 ln[A] = -kt + ln[A]0 [A]1

= kt + [A]0

1

Plot needed to give a straight line [A] versus t ln[A] versus t [A]1

versus t

Relationship of rate constantSlope = -k Slope = -k Slope = kto the slope of straight line

Half-life

2kt1/2 =

[A]0k

t1/2 =0.693

k[A]0t1/2 =

1

KNOW THIS TABLE!!!!Presented by Mark Langella,

PWISTA.com


A Summary

1. Simplification: Conditions are set such that only forward reaction is important.

2. Two types: differential rate law integrated rate law 3. Which type? Depends on the type of

data collected - differential and integrated forms can be interconverted.


A Summary (continued)

4. Most common: method of initial rates. 5. Concentration v. time: used to

determine integrated rate law, often graphically.

6. For several reactants: choose conditions under which only one reactant varies significantly (pseudo first-order conditions).


Reaction Mechanism

- The series of steps by which a chemical reaction occurs.

- A chemical equation does not tell us how reactants become products - it is a summary of the overall process.


Reaction Mechanism (continued)

The reaction

has many steps in the reaction mechanism.

6CO 6H O C H O O2 2light

6 12 6 2 6


Often Used TermsIntermediate: formed in one step and used up in a subsequent step and so is never seen as a product.Molecularity: the number of species that must collide to produce the reaction indicated by that step.Elementary Step: A reaction whose rate law can be written from its molecularity.

uni, bi and termolecular



Reaction Mechanism Requirements

1. The sum of the elementary steps must give the overall balanced equation for the reaction.

2. The mechanism must agree with the experimentally determined rate law.


Rate-Determining Step

In a multistep reaction, it is the slowest step. It therefore determines the rate of reaction.


Fast Equilibrium Reaction When the 1st step is not the slow step, but a

fast equilibrium, the rate can be determined as follows:

2A + B ---> C 2nd order in B, 1st order in A. Step 1 A + B <----> D (Fast equilibrium) Step 2 D + B ----> E Slow Step 3 E + A ----> C + B Fast A + B <---> D D + B ---> E E + A ---> C + B 2 A + B ---> C

1st requirement that elementary stepsequal overall reaction is met.


Fast Equilibrium Reaction(continued)

kf[A][B] = kr[D] (fast)

[D] = kf/kr([A][B])

Rate = k2[D][B] (slow) Substitute fast equation in terms of [D] into

slow reaction. Rate = k2kf/kr([A][B][B]) Rate = k[A][B]2

2nd requirement is also met, this is, then, a possible mechanism.

Documents

Chemical Kinetics. Kinetics Chemical kinetics is the branch of chemistry concerned with the rate of chemical reactions and the mechanism by which chemical