rate of reactions {chemical kinetics}. chemical kinetics chemical kinetics is the branch of...

Rate of Reactions{Chemical Kinetics}

Chemical Kinetics

• Chemical kinetics is the branch of chemistry concerned with the rates of chemical reactions

• A rate is a change in a measurable quantity over time

• Different chemical reactions proceed from reactants to products at different rates (e.g. combustion of propane and oxidation of silver)

Reaction Rate

• The reaction rate of a chemical reaction is the change in concentration of a reactant or product per unit time

• Δc is measured in mol/L

• Δt is measured in s

• r is measured in mol/(Ls)

Fast or Slow Reactions

• Extremely slow reactions– Iron rusting– Limestone weathering

• Extremely fast reactions– Explosion

Measuring Rate of Reactions

• Some rate of reactions have detectable change with respect to time

• Changes that are observable like– When a volume of gas is given off– When this is a change in mass during the

reaction– When there are temperature changes– When there are colour changes– When a precipitate forms– When there are pH changes

Speed= Speed=

=

= 300km/h

=

= 10km/h

According to Collision Theory what factors determine whether a reaction happens or not?

According to Thermochemistry, what two factors determine whether a reaction is favoured?

Collision theory

• Theory that explains how factors affect rate of reaction

• Before reaction can occur, particles must first collide

• Particles collide so that bonds are broken and new bonds can form

Collision theory

• Collision of particles must be effective collisions to produce result

• Effective collisions are collisions which – produce enough energy to overcome

energy of activation– correct orientation

Orientation

Collision theory

• Activation energy is the minimum energy the colliding particles must overcome so that reaction occur

• In order for particles to overcome the activation energy, several factors are involved

• So lets say collision theory is determining that a reaction can happen….what factors determine how fast it will happen?

• Mr McCormack is cooking a roast dinner but his over-sized potatoes are not boiling fast enough….what measures can he take to increase the rate of reaction???

How factors affect rate of reaction according to collision theory

How factors affect rate of reaction according to collision theory

• Size of reactant for solid reactant• Temperature of reactant mixture• Concentration of solution reactant• Presence of catalyst• Pressure on the reaction

Concentration

• Rate of reaction increases with increasing concentration

• Higher concentration, more reacting particles are present

• Greater probability of an effective collision

• Faster rate of reaction

Concentration – Same no. of molesAmt of product formed

Time/s

Higher concentration

Lower concentration

Concentration – Different no. of molesAmt of product formed

Time/s

Higher concentration

Lower concentration

Temperature

• Rate of reaction increases with increasing temperature

• High temperature, particles have greater heat energy

• Particles move faster with greater kinetic energy

• Leading to more collisions between particles

• Increased probability of effective collision• Reactions take place faster

• Speed of reaction doubles when the temperature

rises by 10 C

TemperatureAmt of product formed

Time

Higher temperature

Lower temperature

Catalyst

• Presence of catalyst increases rate of reaction

• (Presence of inhibitors decreases rate of reaction)

• Catalysts lower activation energy of reactants

• Aids the formation of unstable intermediate products

• Increases probability of formation of products

• Faster rate of reaction

POTENTIAL ENERGY DIAGRAM

REACTION COORDINATE

POTETIAL

ENERGY

REACTANTS

PRODUCTS

A + B

AB

ACTIVATIONENERGYWITHOUTCATALYST

LOWER ACTIVATONENERGY PATH

WITH CATALYST

CatalystEnergy

Time/s

Use of catalyst

Absence of catalyst

Ea

Ea

CatalystAmt of product formed

Time/s

Use of catalyst

Absence of catalyst

Catalyst

Definition: A substance which increases the rate of a chemical reaction by providing an alternative pathway with a lower activation energy but remains unchanged at the end of the reaction

Pressure

• Rate of reaction increases with increasing pressure

• Higher pressures, reacting particles are closer together

• Increasing concentration per unit volume

• Greater probability of an effective collision

• Faster rate of reaction

PressureAmt of product formed

Time/s

Higher pressure

Lower pressure

Particle Size

• Rate of reaction increases when particle size decreases

• Smaller particles has greater surface area than larger particles of the same mass

• Greater surface area for collision by another reacting particle

• Greater probability of an effective collision• Faster rate of reaction

HOW PARTICLE SIZE AFFECTS CHEMICAL REACTION RATE!

Particle sizeAmt of product formed

Time/s

Smaller particle size

Larger particle size

A reaction is fast , the time taken for the reaction is short .

A reaction is slow, the time taken for the reaction is long .

The rate of reaction depends to the speed of reaction .

If a reaction is fast, its rate of reaction is high .

If a reaction is slow, its rate of reaction is low .

The rate of reaction is inversely proportional with time .

Rate ά 1

time taken for reaction

Rate of reaction = change in quantity of product / reactant

time taken

For gas product , rate of reaction = volume of gas

time

From a graph , the average rate of reaction = the gradient

HOW TO DETERMINE RATE OF REACTION

FROM GRAPH

AVERAGE RATE OF

REACTION

INSTANTANEOUS RATE OF REATION

AVERAGE RATE OF REACTION

For the whole exp

From X

min to Y min

On the X min (2nd) ( from 2nd to 1st)

For first x

(3) min( from 0 to

3rd )

INSTANTANEOUS RATE OF REACTION

( rate of reaction at that time) Draw tangent to the

graph

Y

X

Rate = Y/X

b) Example from the graph, determine:i) The rate of reaction at 120 s

Instantaneous rate of reaction= Draw tangent to the graph

= 56 – 20 = 0.176 cm3 s-1

222-18

Changes to the graph

Changes to the CURVE part of graph

II

IIII

Volume of gas

Time

• Use positive catalyst• Increase temperature •Increase total surface area

Use negative catalyst

Decrease temperature

Decrease TSA

Changes to the graph

d) Curve I represents the result of the experiment using excess zinc powder and 50cm3 of 1.0 moldm-3 dilute hydrochloric acid

II

IIII

Volume of gas/ cm3

Time/s

• Use positive catalyst• Increase temperature of reactant

Lower concentration of hydrochloric acid

(h)Time (seconds) 0 30 60 90 120 150 180 210 240Burette reading (cm3) 49.5 33.5 23.5 16.0 10.5 5.0 2.0 2.0 2.0Volume CO2 (cm3) 0 16 26 33.5 39 44.5 47.5 47.5 47.5

Time(second) 0 30 60 90 120

Burette reading(cm3)

x

49.5

y

33.5

z

23.5

Total volume of gas(cm3)

x-x

0.00

x-y

16.00

x-z

26.00

Volume of CO2, cm3

Time , s

Connect the point without using ruler!Not all the point is connected

Volume of CO2 cm3

Time s

Cannot make this graphStraight lineIt’s must be

smooth graph

Average Rate Of reaction

The average rate of reaction in the first 90 seconds.= The total volume of gas released in the first 90 seconds

Time taken

(i)

Time (seconds) 0 30 60 90 120 150 180 210 240Burette reading (cm3) 49.5 33.5 23.5 16.0 10.5 5.0 2.0 2.0 2.0Volume CO2 (cm3) 0 16 26 33.5 39 44.5 47.5 47.5 47.5

=

33.5÷90=0.372

cm3s-1unit

i(ii)Time (seconds) 0 30 60 90 120 150 180 210 240Burette reading (cm3) 49.5 33.5 23.5 16.0 10.5 5.0 2.0 2.0 2.0Volume CO2 (cm3) 0 16 26 33.5 39 44.5 47.5 47.5 47.5

47.5÷180= 0.264 cm3s-1

The average rate of reaction in the whole experiment.= The total volume of gas released in the whole experiment

Time taken

=

Total volume of Hydrogen gas/cm3

Time (second)

Analysis of Data

t

p

q

Rate of reaction at t second = gradient AB

= p/q cm3 s-1

A

B

Tangent

Cannot take directly at x

Tangent is a line that touch just 1 point of graph in order to calculate gradient

Tangent

Only touch 1 point of curve

Cannot touch more than 2 points because each of point has a different gradient

tangentα

Total Volume of CO2(cm3)

Time (second)

Analysis of data

A

B

C

D

E

F

t1 t3t2

Rate of reaction at t1 = gradient ABRate of reaction at t2 = gradient CDRate of reaction at t3 = gradient EFEach of point has a

different gradient!

Two methods to calculate tangent:

Total volume of Hydrogen gas/cm3

Time (second)

A

B

Tangent

number of small

boxes × value of 1 small unit box

Y

X

Total volume of Hydrogen gas/cm3

Time (second)

A

B

Tangent

Gradient of graph:

m = ΔY

ΔX

x1 x2

y2

y1

m = Y2-y1

X2-x1

First Method

Total volume of Hydrogen gas/cm3

Time (second)

Analysis of Data

t

p

q

Rate of reaction at t second = gradient AB

= p/q cm3 s-1

A

B

Tangent

Total Volume of CO2(cm3)

Time (second)

Analysis of data

A

B

C

D

E

F

t1 t3t2

Rate of reaction at t1 = gradient ABRate of reaction at t2 = gradient CDRate of reaction at t3 = gradient EF

The Rate Law: Reactant Concentration and Rate

The rate of a chemical reaction depends on several

factors ….

Relating Reactant Concentrations and Rate

Consider the general reaction below.

aA + bB cC + dD

This reaction occurs at constant temperature

The reactant formulas are represented by A and B

The stoichiometric coefficients are represented by

a and b

In this section we will look at reaction rates that are not

affected by concentrations of products

In general, the rate of a reaction increases when the

concentrations of reactants increases.

The dependence of of the rate of a reaction on the con-

centration is given by:

OR Rate=k[A]m

This relationship can be expressed in a general

equation called the rate law equation

For any reaction, the rate law equation expresses the

relationship between the concentrations of the

reactants and the rate of the equation.

The letter k represents the proportionality constant

called the rate constant

There is a different rate constant for each reaction at

any given temperature

The exponents m and n must be determined by exp-

eriment.

The do not necessarily correspond to the coefficients

of their reactants

They are usually 1 or 2, but values of 0, 3 even fractions

can occur

HOW CONCENTRATION EFFECTS REACTION RATES (REACTION ORDER)

• REACTIONS WITH RATE EQUATIONS HAVING n = 0 ARE ZERO ORDER REACTIONS. THOSE WITH n = 1 ARE FIRST ORDER AND THOSE WITH n = 2 ARE SECOND ORDER.

• IN ZERO ORDER REACTIONS, CHANGING THE CONCENTRATION OF THE REACTANT HAS NO EFFECT ON THE RATE.

• IN FIRST ORDER REACTIONS, RATE CHANGES ONE FOR ONE WITH CONCENTRATION CHANGE. FOR EXAMPLE, DOUBLING CONCENTRATION DOUBLES THE RATE.

HOW CONCENTRATION EFFECTS REACTION RATES (REACTON ORDER)

• IN SECOND ORDER REACTIONS, RATE CHANGES RELATIVE TO THE SQUARE OF THE CONCENTRATION CHANGE. FOR EXAMPLE, DOUBLING THE CONCENTRATION OF THE REACTANT RESULTS IN THE RATE INCREASING 4 TIMES.

• THIS KNOWLEDGE OF HOW RATE CHANGES WITH CONCENTRATION DEPENDING ON THE ORDER LETS US FIND REACTION ORDERS BY AN EXPERIMENTAL PROCESS CALLED “METHOD OF INITIAL RATES”

• REACTION ORDERS MUST BE DETERMINED EXPERIMENTALLY. THEY CAN NEVER BE DETERMINED FROM THE CHEMICAL EQUATION.

RXN RATE?

RXN RATE ?FORWARD OR

REVERSE RXN?

RXN RATE?FORWARD OR REVERSE?

RATE = 0

RXN RATE?FORWARD OR REVERSE?

REACTION RATES

TIME

MOLES

A

TIME

MOLES

A

TIME

MOLES

A

TIME

MOLES

A

A B + C

FORWARD RXNRATE = CONSTANT

FORWARD RXNRATE = VARIABLE REVERSE RXN

RATE = VARIABLE

SLOPE OF A TANGENT LINE TO AN AMOUNT VS. TIME GRAPH = RATE

GRAPH 1 GRAPH 2

GRAPH 3 GRAPH 4

HOW CONCENTRATION EFFECTS REACTION RATES (INITIAL RATES)

• USING THE METHOD OF INITIAL RATES REQUIRES THAT A REACTION BE RUN AT SERIES OF DIFFERENT STARTING CONCENTRATIONS AND THE RATE BE DETERMINED FOR EACH.

• GIVEN THE FOLLOWING DATA FOR THE REACTION A B + C(TABLE 1)• EXPT [A] RATE (M/SEC) 1 1 x 10 -3 4 x 10 -1

2 2 x 10 -3 8 x 10 -1

3 4 x 10 -3 16 x 10 -1

• AS CONCENTRATION OF A DOUBLES, RATE DOUBLES. THE REACTION IS FIRST ORDER IN REACTANT A

• RATE = k[A]1 OR RATE = k[A]

HOW CONCENTRATION EFFECTS REACTION RATES (INITIAL RATES)

• FOR THE REACTION: A + B C + D• (TABLE 2) 1 1 x 10 -3 1 x 10 –3 4 x 10 -1

2 2 x 10 -3 1 x 10 -3 8 x 10 -1

3 1 x 10 -3 2 x 10 -3 16 x 10 –1

• USING EXPT 1 AND 2, [A] DOUBLES AND [B] IS CONSTANT. THE DOUBLING OF THE RATE IS THEREFORE CAUSED BY REACTANT A AND THE ORDER WITH RESPECT TO A IS FIRST.

• USING EXPT 1 AND 3, [A] IS CONSTANT AND [B] IS DOUBLED. THE FOUR TIMES RATE INCREASE IS THEREFORE CAUSED BY REACTANT B AND THE ORDER WITH RESPECT TO B IS SECOND.

• RATE = k[A]1[B]2 OR RATE = k[A][B]2

HOW CONCENTRATION EFFECTS REACTION RATES (RATE CONSTANTS)

• FROM INITIAL RATES DATA TABLE 1,

• RATE = k [A] , k CAN BE CALCULATED BY SUBSTITUTING ANY DATA SERIES INTO THE RATE EQUATION, FOR EXAMPLE, FROM EXPT 1 ON TABLE 1

• [A] = 1 x 10 –3 M , RATE = 4 x 10 –1 M/SEC• 4 x 10 –1 M/SEC = k (1 x 10 –3 M ) • k = 1 x 102 SEC-1 OR 1 x 102 / SEC

HOW CONCENTRATION EFFECTS REACTION RATES (RATE CONSTANTS)

• FROM INITIAL RATES DATA TABLE 2,

• RATE = k [A] x [B]2, k CAN BE CALCULATED BY SUBSTITUTING ANY DATA SERIES INTO THE RATE EQUATION, FOR EXAMPLE, FROM EXPT 1 ON TABLE 2

• [A] = 1 x 10 –3 M , [B] = 1 x 10 –3 M

• RATE = 4 x 10 –1 M/SEC

• 4 x 10 –1 M/SEC = k (1 x 10 –3 M ) x (1 x 10 –3 M )2

• k = 4 x 105 M-1 SEC-1 OR 4 x 105 / M x SEC

Let us see how the initial rates method works

2N2O3(g) 2NO2(g) + O2(g)

The general rate law equation for this reaction is:

Rate = k [N2O3]m

To determine the value of m, a chemist performs three

experiments. A different initial concentration of [N2O35]0

Is used for each experiment. The 0 represents t=0

Experiment Initial [N2O3]0 (mol/L) Initial rate (mol/(L.s))

1 0.010 4.8 x 10 -6

2 0.020 9.6 x 10 -6

3 0.030 1.5 x 10-5

Value of m can be determined with at least two different methods

by inspection rate law equation

1. By inspection

When the [N2O5] is doubled expts 1 and 2 doubles

the rate also doubles

when [N2O5] is tripled, (expts 1 and 3) the rate triples

this indicates a first-order relationship as follows

Rate = k [N2O5]1

2. Compare rate law equation using ratiosthis method very useful when relation between conc and rate are not

immediately obvious from data

Write the rate expressions for expts 1 and 2 as follows:

Rate1 = k [0.010]m

= 4.8 x 10-6 mol/(L.s)

Rate2 = k [0.020]m

= 9.6 x 10-6 mol/(L.s)

Create a ratio to compare the two rates

Rate1 = k(0.010 mol/L)m = 4.8 x 10-6 mol/L.s

Rate 2 k(0.020 mol/L)m 9.6 x 10-6 mol/L.s

Since k is a constant at constant temp you can cancel out

k(0.010 mol/L)m = 4.8 x 10-6 mol/L.s k(0.020 mol/L)m 9.6 x 10-6 mol/L.s

(0.5)m = 0.5 m = 1 (by inspection)

Determining the Rate Constant

Once you know the rate law equation for a reaction,

you can calculate the rate constant using results from

any of the experiments

Rate = k [N2O5]1

You can use data from any of the three experiments to

calculate k

4.8 x 10-6 mol/(L.s) = k(0.010 mol/L)

k = 4.8 x 10-6 mol/ (L.s) 0.010 mol/L

= 4.8 x 10-4 s-1

TEMPERATURE & REACTION RATE• AT ANY TEMPERATURE THE MOLECULES IN A SYSTEM HAVE A

DISTRIBUTION OF KINETIC ENERGIES (SIMILAR TO THE DISTRIBUTION OF THE SPEEDS OF CARS ON A HIGHWAY).

• AS THE TEMPERATURE INCREASES, THE AVERAGE KINETIC ENERGY OF THE MOLECULES INCREASE (THEY MOVE FASTER AT HIGHER TEMPERATURES) AND THEREFORE COLLIDE WITH EACHOTHER MORE FREQUENTLY AND HIT HARDER WHEN THEY DO COLLIDE.

• AS A RESULT, REACTION RATE INCREASES WITH TEMPERATURE.

KINETIC ENERGY DISTRIBUTION CURVE

TEMPERATURE = T1

KINETIC ENERGY

NUMBER

OF

MOLES

LOW ENERGYMOLECULES

HIGH ENERGYMOLECULES

AVERGE ENERGYMOLECULES

KINETIC ENERGY DISTRIBUTION CURVE

TEMPERATURE = T1

TEMPERATURE = T2

KINETIC ENERGY

NUMBER

OF

MOLES

TEMP 2 > TEMP 1AT HIGHER

TEMPERATURES, MOLECULES HAVE HIGHER ENERGIES

ON AVERAGE

TEMPERATURE & REACTION RATE

• IN ORDER TO REACT, MOLECULES MUST COLLIDE WITH SUFFICIENT ENERGY. THIS MININIUM ENERGY FOR REACTION IS CALLED ACTIVATION ENERGY.

• AS THE TEMPERATURE OF A SYSTEM IS INCREASED, THE NUMBER OF MOLECULES WITH THE NECESSARY ENERGY FOR REACTION (THE ACTIVATION ENERGY) INCREASES.

KINETIC ENERGY DISTRIBUTION CURVE

TEMPERATURE = T1

TEMPERATURE = T2

KINETIC ENERGY

NUMBER

OF

MOLES

ACTIVATION ENERGY

MOLECULES WITHSUFFICIENT ENERGY

TO REACT AT T1

AS TEMPERATURE , REACTION RATE

MOLECULES WITHSUFFICIENT ENERGY

TO REACT AT T2

TEMPERATURE & REACTION RATE• THE ENERGY CHARACTERISTICS OF A CHEMICAL

REACTION CAN BE SHOWN ON A POTENTIAL ENERGY DIAGRAM.

• THIS GRAPH SHOWS THE ENERGY STATE OF THE SYSTEM AS REACTANTS PROCEED THROUGH THE ACTIVATED COMPLEX TO FORM THE PRODUCTS. THE ACTIVATED COMPLEX IS THE INTERMEDIATE STATE (MOLECULAR FORM) WHICH REACTANTS GO THROUGH AS THEY CONVERT INTO THE PRODUCTS.

• THE ACTIVATION ENERGY IS THE ENERGY REQUIRED TO FORM THE INTERMEDIATE ACTIVATED COMPLEX MOLECULE.

POTENTIAL ENERGY DIAGRAM

REACTION COORDINATE

POTETIAL

ENERGY

REACTANTS

PRODUCTS

ACTIVATEDCOMPLEX

A + B

AB

H

ACTIVATIONENERGY

EXOTHERMIC

H = (-)ENERGY ISRELEASED

The Effect of Surface Area on Reaction Rate

Which will cook faster?

One 60g whole potato…

…or 60g of small potato

pieces?

The small potato pieces…

Why do they cook faster?

Because if you compare a 60g potato with 60g of small potato pieces, the small potato pieces

have a LARGER SURFACE AREA

So what?

• The larger the surface area, the more collisions that can happen between the boiling water and the potato at the same time

• This means the potato cooks faster

SURFACE AREA & REACTION RATE

REACTANT THAT IS INTERIOR CANNOT BE ATTACKED UNTIL THE EXTERIOR REACTANT IS CONSUMED. THE REACTION RATE IS SLOW!

SURFACE AREA & REACTION RATE

WHEN SURFACE AREA IS INCREASED MORE REACTANTS ARE EXPOSEDTO EACHOTHER SIMULTANEOUSLY AND THE REACTION IS RAPID.

REACTION MECHANISMS• ALL REACTIONS, NO MATTER HOW SIMPLE FOLLOW A PARTICULAR

REACTION PATHWAY CALLED A MECHANISM. IT IS A SERIES OF STEPS WHICH LEAD TO THE FORMATION OF PRODUCTS.

• DURING THE PROCESS OFTEN MOLECULES CALLED INTERMEDIATES ARE FORMED AND SUBSEQUENTLY CONSUMED.

• THE SLOWEST STEP IN THE SERIES OF STEPS THAT MAKE UP THE MECHANISM IS CALLED THE RATE DETERMINING STEP.

• THE SPEED OF THE RATE DETERMINING STEP DEPENDS ON THE COMPLEXITY OF THE STEP (NUMBER OF MOLECULES INVOLVED CALLED THE MOLECULARITY) AND THE BOND STRENGTHS OF THE REACTING COMPONENTS

NO

NONO2

NO2

O2

REACTION MECHANISM FOR;2 NO + O2 2 NO2

STEP I 2 NO N2O2 (fast)STEP II N2O2 + O2 2 NO2 (slow)

N2O2 INTERMEDIATE

REACTION MECHANISMS• UNLIKE OVERALL REACTIONS, THE REACTON ORDERS FOR

THE STEPS IN A MECHANISM ARE BASED ON THE NUMBER OF MOLECULES REACTING IN THAT STEP.

IN THE REACTION MECHANISM FOR:2 NO + O2 2 NO2

STEP I 2 NO N2O2 (fast)STEP II N2O2 + O2 2 NO2 (slow)

SINCE THERE IS ONLY ONE N2O2 AND ONE O2 IN THE RATE DETERMING STEP (THE SLOW STEP), THE RATE EQUATION FOR THE REACTION IS:

RATE = k [N2O2] x [O2]

THE REACTION IS FIRST ORDER IN BOTH REACTANTS