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Chemical Kinetics Unit 11

Chemical Kinetics

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Chemical Kinetics. Unit 11. Chemical Kinetics. Chemical equations do not give us information on how fast a reaction goes from reactants to products. KINETICS : the study of reaction rates and their relation to the way the reaction proceeds, i.e. its mechanism - PowerPoint PPT Presentation

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Page 1: Chemical Kinetics

Chemical Kinetics

Unit 11

Page 2: Chemical Kinetics

Chemical Kinetics Chemical equations do not give us

information on how fast a reaction goes from reactants to products.

KINETICS: the study of reaction rates and their relation to the way the reaction proceeds, i.e. its mechanism

We can use thermodynamics to tell if a reaction is product – or reactant – favored

Only kinetics will tell us how fast the reaction happens!

Page 3: Chemical Kinetics

Rate of Reaction A rate is any change per interval of time.

Example: speed (distance/time) is a rate!

Reaction rate = change in concentration of a reactant or product with time

Page 4: Chemical Kinetics

Expressing a Rate

For the reaction A P

t

ARate

t

PRate

=

Appearance of product

Disappearance of reactant

Page 5: Chemical Kinetics

Reaction Conditions & Rates

Collision Theory of Reactants Reactions occur when molecules collide

to exchange or rearrange atoms Effective collisions occur when

molecules have correct energy and orientation

Page 6: Chemical Kinetics

Factors Affecting Rates

1. Concentrations (and physical state of reactants and products)

2. Temperature

3. Catalysts

Catalysts are substances that speed up a reaction but are unchanged by the reaction

Page 7: Chemical Kinetics

Effect of Concentration on Reaction Rate

To propose a reaction mechanism,

we study the reaction rate and its

concentration dependence.

Page 8: Chemical Kinetics

Rate Laws or Rate Expressions

The rate law for a chemical reaction relates the rate of reaction to the concentration of reactants.

For aA + bB cC + dD

The rate law is: Rate = k[A]m[B]n

The exponents in a rate law must be determined by experiment.They are NOT derived from the stoichiometry coefficients in an overall chemical equation.

Page 9: Chemical Kinetics

Rate Laws & Orders of Reactions

Rate Law for a reaction:

Rate = k[A]m[B]n[C]p

The exponents m, n, and pAre the reaction orderCan be 0, 1, 2, or fractions (may be other

whole numbers in fictional examples)Must be determined by experiment

Overall Order = sum of m, n, and p

Page 10: Chemical Kinetics

Interpreting Rate Laws If m = 1 (1st order)

Rate = k [A]1

If [A] doubles, then the rate doubles (goes up by a factor of 2)

If m = 2 (2nd order) Rate = k [A]2

If [A] doubles, then rate quadruples (increases rate by a factor of 4)

If m = 0 (zero order) Rate = k [A]0 If [A] doubles, rate does not change!

Rate = k[A]m[B]n[C]p

Page 11: Chemical Kinetics

Rate Constant, kRelates rate and concentration at a given temperature.

General formula for units of k: M(1- overall order) time-1

Overall Order Units of k

0 M time-1

1 Time-1

2 M-1 Time-1

3 M-2 Time-1

Page 12: Chemical Kinetics

Rate Law Problem:The initial rate of decomposition of acetaldehyde, CH3CHO, was measured at a series of different concentrations and at a constant temperature.

Using the data below, determine the order of the reaction – that is, determine the value of m in the equation

CH3CHO(g) CH4(g) + CO(g)

Rate = k[CH3CHO]m

CH3CHO (mol/L)

0.162 0.195 0.273 0.410 0.518

Rate (mol/L*min)

3.15 4.56 8.94 20.2 35.2

Page 13: Chemical Kinetics

Strategy

Use the equation:

Pick any two points from the given data!

m

1

2m

1

m2

AA

AA

1 Rate2 Rate

Page 14: Chemical Kinetics
Page 15: Chemical Kinetics

Deriving Rate Laws

Rate of rxn = k[CH3CHO]2

Here the rate goes up by FOUR when the initial concentration doubles.

Therefore, we say this reaction is SECOND order overall.

Page 16: Chemical Kinetics

Example:

Using the same set of data from the previous example, and knowing the order of the reaction, determine:

b) the value of the rate constant, k (w/ units!)

c) the rate of the reaction when

[CH3CHO] = 0.452 mol/L

Strategy: Use any set of data to find k. Solve for rate using k, rate order equation, and given

concentration.

Page 17: Chemical Kinetics
Page 18: Chemical Kinetics

The data below is for the reaction of nitrogen (II) oxide with hydrogen at 800oC.

2NO(g) + 2H2(g) N2(g) + 2H2O(g)

Determine the order of the reaction with respect to both reactants, calculate the value of the rate constant, and determine the rate of formation of product when [NO]=0.0024 M and [H2]=0.0042 M.

Strategy:Choose two experiments where concentration of

one reactant is constant and other is changed; solve for m and n separately!

Page 19: Chemical Kinetics
Page 20: Chemical Kinetics

Example:

The initial rate of a reaction A + B C was measured with the results below. State the rate law, the value of the rate constant, and the rate of reaction when [A] = 0.050 M and [B] = 0.100 M.

Experiment [A] (M) [B] (M) Initial Rate (M/s)

1 0.1 0.1 4.0x10-5

2 0.1 0.2 4.0x10-5

3 0.2 0.1 16.0x10-5

Page 21: Chemical Kinetics
Page 22: Chemical Kinetics

Potential Energy DiagramsMolecules need a minimum amount of energy for a reaction to take place.

Activation energy (Ea) – the minimum amount of energy that the reacting species must possess to undergo a specific reaction

Activated complex - a short-lived molecule formed when reactants collide; it can return to reactants or form products.

Formation depends on the activation energy & the correct geometry (orientation)

Page 23: Chemical Kinetics

Potential Energy Diagram

Page 24: Chemical Kinetics

Potential Energy Diagrams

Page 25: Chemical Kinetics

Potential Energy Diagrams

Page 26: Chemical Kinetics

Catalyzed Pathway

Catalysts lower activation energy!!!

Page 27: Chemical Kinetics

Reaction MechanismsMechanism – how reactants are converted to products at the molecular level

Most reactions DO NOT occur in a single step! They occur as a series of

elementary steps

(a single step in a reaction).

Page 28: Chemical Kinetics

Rate Determining StepRate determining step –

the slowest step in a reaction

COCl2 (g) COCl (g) + Cl (g) fast

Cl (g) + COCl2 (g) COCl (g) + Cl2 (g) slow

2 COCl (g) 2 CO (g) + 2 Cl (g) fast

2 Cl (g) Cl2 (g) fast

Page 29: Chemical Kinetics

Getting the Overall Reaction

COCl2 (g) COCl (g) + Cl (g) fast

Cl (g) + COCl2 (g) COCl (g) + Cl2 (g) slow

2 COCl (g) 2 CO (g) + 2 Cl (g) fast

2 Cl (g) Cl2 (g) fast

2 COCl2 (g) 2 Cl2 (g) + 2 CO (g)

Adding elementary steps gives the

net (or overall) reaction!

Page 30: Chemical Kinetics

Intermediates Intermediates are produced in one

elementary step but reacted in another

NO (g) + O3 (g) NO2 (g) + O2 (g)

NO2 (g) + O (g) NO (g) + O2 (g)

O3 (g) + O (g) 2 O2 (g)

Page 31: Chemical Kinetics

Catalysts Catalyst – a reactant in an elementary step

but unchanged at the end of the reactionA substance that speeds up the reaction but is

not permanently changed by the reactionBoth an original reactant and a final product

NO (g) + O3 (g) NO2 (g) + O2 (g)

NO2 (g) + O (g) NO (g) + O2 (g)

O3 (g) + O (g) 2 O2 (g)

Page 32: Chemical Kinetics

Example

Cl2 (g) 2 Cl (g) Fast

Cl (g) + CHCl3 (g) CCl3 (g) + HCl (g) Slow

CCl3 (g) + Cl (g) CCl4 (g) Fast

Identify: The rate determining step The overall (net) reaction The identity of any intermediates The identity of any catalysts

Page 33: Chemical Kinetics

Example

H2O2(aq) + I1-(aq) H2O(l) + IO1-(aq) Slow

H2O2(aq) + IO1-(aq) H2O(l) + O2(g) + I1- (aq) Fast

Identify: The rate determining step The overall (net) reaction The identity of any intermediates The identity of any catalysts

Page 34: Chemical Kinetics

Example

O3 (g) + Cl (g) O2 (g) + ClO (g) Slow

ClO (g) + O (g) Cl (g) + O2 (g) Fast

Identify: The rate determining step The overall (net) reaction The identity of any intermediates The identity of any catalysts