unit 8 chemical kinetics & thermodynamics. chemical kinetics chemical kinetics is the study of...
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Unit 8Chemical Kinetics & Thermodynamics
Chemical Kinetics• Chemical kinetics is the study of the
factors that affect the speed of a reaction and the mechanism by which a reaction proceeds.• Rate is how much
a quantity changesin a given period of time.
• Collision Theory says that in order for reactions to occur between substances, particles must collide.• For a collision to be
effective, particles must:
1. Collide with sufficient energy.
2. Have a favorable orientation.
Collision Theory
Effect of Orientation on a Collision
Smaller pieces = larger surface area
• The rate of a chemical reaction depends on the number of collisions between particles.
• Rate-influencing factors are:• Nature of Reactants• Surface Area• Temperature• Concentration• Presence of Catalysts
Rate-Influencing Factors
Visual Concept
Effect of Concentration• The rate of a reaction often depends on the
concentration of one or more of the reactants.• Greater concentration of reactant particles =
more collisions = increase in reaction rate.
Rate Law Expressions• Rate law equations are based on the Molarity
concentrations of the reactant(s). (Indicate Molarity concentrations with brackets.)• The Rate (R) involves the product of the
concentrations of the individual reactants.• Experiments have proven that coefficients of
reactants have an exponential effect on R.
Ex: For the equation A + 2 B → C
R = k [A][B]2 (k = rate law constant)
Rate Law ExpressionsSample Problem
For the following single-step chemical reaction:
2SO2 + O2 → 2 SO3
1. Write the rate law expression.
R = k [SO2]2 [O2]
2. What happens to the Rate if the [SO2] is cut by ½?
Rate is 4x slower
3. What happens to the Rate if the [O2] is tripled?
Rate is 3x greater
Reaction Mechanism• Chemical reactions often occur as a
series of consecutive steps (this is called a reaction mechanism.)• Products in an early step that are reactants in a
later step are called intermediates.
Example: Overall reaction:
H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) Mechanism:
1) H2(g) + ICl(g) HCl(g) + HI(g)
2) HI(g) + ICl(g) HCl(g) + I2(g)
intermediate
Rate-Determining Step• One of the steps will always occur at a
slower rate than the others.• The rate-determining step is
the slowest step (bottleneck)in a reaction mechanism.• ONLY the rate-determining step
is used for rate law expressions.
Rate-Determining StepSample Problem
Equation: 2NO2 + F2 → 2 NO2F
Step 1: NO2 + F2 → NO2F + F “slow”
Step 2: F + NO2 → NO2F“fast”
1. Write the rate law expression for this reaction.
R = k [NO2] [F2]
2. Given the following data, calculate the [NO2] used:
R = 0.00905 M/s, k = 1.44 x 10-3 1/M●s, [F2] = 1.11 M
[NO2] = 5.66 M
[NO2] =R
k [F2]=
0.00905 M/s(1.44 x 10-3 1/M●s) (1.11 M)
Rate-determining step
Getting a Reaction Started• Energy is needed to overcome the repulsion
between molecules and get a reaction started.• Just as a ball cannot get
over a hill without added energy, a reactioncannot occur unless the molecules have enough energy to get over this initial energy barrier.
Activation Energy• Activation energy (Ea) is the minimum amount
of energy required to get a reaction started.
Exothermic Reactions• In an exothermic reaction,
energy is released. Therefore, the energy of the products must be less than the energy of the reactants.• The great majority of
chemical reactions innature are exothermic.
Endothermic Reactions• In an endothermic reaction, energy is
absorbed. Therefore, the energy of the products must be greater than the energy of the reactants.
Catalysts• Catalysts are substances which affect the rate
of a reaction without being consumed.• Catalysts increase the rate of a reaction by
changing the reaction mechanism and decreasing the activation energy.
Enzymes• Due to the complexity of
organic molecules, most biological reactions needa catalyst to proceed at a reasonable rate.• Protein molecules that
catalyze biological reactionsare called enzymes.• Enzymes pull the reactants onto
an active site that orients them for the reaction.
Rate Order• Experiments have shown that there are 3 ways
that a reactant’s concentration can affect the overall rate of the reaction:• Rate Order 0 ([X]0) - The concentration of a
reactant has no effect on reaction rate.• Rate Order 1 ([X]1) - The concentration of a
reactant has a direct effect on reaction rate.• Rate Order 2 ([X]2) - The concentration of a
reactant has an exponential effect on reaction rate.
• The order of a reaction can be determined only by experiment.
Determining the Order of a Reaction
[A] (M)
Rate (M/s)
0.10 0.015
0.20 0.030
0.40 0.060
x2 x2
[A] (M)
Rate (M/s)
0.10 0.015
0.20 0.015
0.40 0.015
x2 x2
[A] (M)
Rate (M/s)
0.10 0.015
0.20 0.060
0.40 0.240
x1 x4
When [A] doubles, the rate doubles
When [A] doubles, rate doesn’t change
When [A] doubles, the rate quadruples
Rate = k[A]1
First Order
Rate = k[A]0 = k
Zero Order
Rate = k[A]2
Second Order
Rate Order of a ReactionSample Problem
From the data above, determine the following:1. Rate order of each reactant.
second order in [NO2] and zero order in [CO]
2. Overall rate order of the reaction.2 + 0 = 2 second order reaction
3. The rate law for the reaction.Rate = k [NO2]2
[NO2] (M) [CO] (M) Rate (M/s)
0.10 0.10 0.0021
0.20 0.10 0.0082
0.20 0.20 0.0083
0.40 0.10 0.033
x2 x4
constant
constant
x2 x1
Finding the Rate ConstantSample Problem
Using the data above, and the rate law we just found, (R = k [NO2]2), determine the following:
1. Value of k.
2. Reaction rate when the concentration of [NO2] is 0.50 M and [CO] is 1.00 M.
[NO2] (M) [CO] (M) Rate (M/s)
0.10 0.10 0.0021
0.20 0.10 0.0082
0.20 0.20 0.0083
0.40 0.10 0.033
k =R
[NO2]2=
0.0021 M/s(0.10 M)2
= 0.21 M-1s-1
R = k [NO2]2 = (0.21 M-1s-1) (0.50 M)2 = 0.053 M/s
Use data from any one line of chart to find k
Thermodynamics• Thermodynamics is the study of energy
transfer in reactions.
Energy• Energy is the ability to do work.• Some forms of energy are:• Mechanical (kinetic and potential)• Electrical• Heat or thermal• Light or radiant• Nuclear• Chemical
• Energy is commonly measured in joules.• 1 joule is the amount of energy needed to move a
1 kg mass at a speed of 1 m/s. (1 J = 1 )2
2
s
mkg
First Law of Thermodynamics• The First Law of Thermodynamics is the
Law of Conservation of Energy which states that energy cannot be created or destroyed.• However, energy can
be transferred betweenobjects, or between forms. (Ex: heat → light → sound)
Energy Exchange• Conservation of Energy requires that the total
energy change in a system and its surroundings must be zero.• Energy is exchanged between system
and surroundings through either heat exchange or work being done.• Heat is the exchange of thermal energy between
a system and its surroundings.
Enthalpy• the Enthalpy change (DH) of a
reaction is the heat exchangedin a reaction under conditionsof constant pressure.
• DH for a reaction is equal to the difference between the DHof formation of the products andthe reactants.
DHrxn = S nDHf(products) - S nDHf(reactants)
• S means sum• n is the coefficient• Elements in their standard state have a DHf of zero.
Calculating Enthalpy of ReactionSample Problem
Calculate the Enthalpy Change in the Reaction: 2C2H2(g) + 5O2(g) ® 4CO2(g) + 2H2O(l)
Reactant/Product
DHf
(kJ/mol)
C2H2(g) +227.4
CO2(g) -393.5
H2O(l) -285.8
DHrxn = S nDHf(products) - S nDHf(reactants)
DHrxn = [(4•DHCO2 + 2•DHH2O) – (2•DHC2H2 + 5•DHO2)]
DHrxn = [(4•(-393.5) + 2•(-285.8)) – (2•(+227.4) + 5•(0))]
DHrxn = -2145.6 kJ – 454.8 kJ
DHrxn = -2600. kJ
Stoichiometry Involving ∆H• The amount of heat generated or absorbed
during a reaction depends on the amount of the substances reacting.
Example:C3H3(g) + 5O2(g) → 3CO2(g) + 4H2O(g) ∆H = -2044 kJ
Means 2044 kJ of heat is given off when 1 mole of C3H3 reacts, or when 3 moles of CO2 is produced.
• These relationships can be used as ratios in stoichiometric conversions.
Ammonia reacts with oxygen as follows:4NH3(g) + 5O2(g)→ 4NO(g) + 6 H2O(g) ∆H = -906 kJ
Calculate the heat (in kJ) associated with the complete reaction of 155 g of NH3.
mol NH3
mol NH3
Stoichiometry Involving ∆HSample Problem
155 g NH3g NH3
-2070 kJ NH3=17.0
1
4
kJ-906
Hess’s Law• Hess’s Law – the overall enthalpy change in a
reaction is equal to the sum of enthalpy changes for the individual steps in the process.• Possible steps in Hess’s Law:
1. Reverse equation/change sign on ΔH.
2. Multiply or Divide coefficients/multiply or divide ΔH.
Hess’s LawSample Problem
Calculate the enthalpy of formation for CH4:
C(s) + 2H2(g) → CH4(g) ∆Hf = ?
The component reactions are:C(s) + O2(g) → CO2(g) ∆Hc = -393.5 kJ
H2(g) + ½O2(g) → H2O(l) ∆Hc = -285.8 kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆Hc = -890.8 kJ
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ∆Hc = +890.8 kJ
- 2 moles of H2 are used to make CH4,
so multiply the 2nd equation by 2 (including ∆H.)- CH4 is on the products side, not the reactants side, so
reverse the 3rd reaction and change the sign on ∆H.- Cancel unwanted terms and add the ∆H’s.
2 2 -571.6 kJ
-74.3 kJ
-74.3 kJ
Sign of ∆H• In an exothermic reaction, the energy of
the products is less than the energy of the reactants, so ∆H is negative.• In an endothermic reaction, the energy of
the products is greater than the energy of the reactants, so ∆H is positive.
Exothermic∆H < 0
Endothermic∆H > 0
Spontaneous Processes• A spontaneous process is one that occurs
without ongoing outside intervention.• A nonspontaneous process is not impossible,
but it requires energy input.• if a process is spontaneous in one direction,
it must be nonspontaneous in the other.
Enthalpy and Reaction Tendency
• Most spontaneous reactions are exothermic (∆H is negative).• Products have less energy than reactants.• Less energy = greater stability.
• But not all spontaneous reactions are exothermic.• Ex: The melting of ice
is endothermic andspontaneous (at >0oC).
• Enthalpy must not be thesole criteria for spontaneity.
Entropy• Entropy (S) is a measure of the
degree of disorder in a system.• Changes that increase entropy:• Physical state (phase) changes
(solid < liquid < gas)• A larger number of moles
of products than reactants• Increase in temperature• Solids dissociating into
ions upon dissolving• Expansion (greater
volume) of gases
Second Law of Thermodynamics
• The Second Law of Thermodynamics states that for any spontaneous process, the entropy of the universe increases (DSuniv is positive).• The entropy of the
system could decreaseas long as the entropyof the surroundings increases by a greateramount.(DSuniv = DSsys + DSsurr)
Temperature Dependence of DS • The increase in DSsurr often comes from the
heat released in an exothermic reaction.• the amount the entropy
of the surroundings changes depends on the temperature it is at originally.• the higher the original temperature, the less
effect addition or removal of heat has.
T
HS system
gssurroundin
Gibbs Free Energy• The total amount of energy available in the system
to do work on the surroundings is called the Gibbs Free Energy (DG).• DG can be calculated using the following equation
(T is the Kelvin temperature):
• Systems tend toward lower Gibbs free energy (lower chemical potential.)• A negative DG = a spontaneous reaction.• A positive DG = a nonspontaneous reaction.
DG = DH – TDS
Gibbs Free EnergySample Problem
Given: CCl4(g) C(s, graphite) + 2 Cl2(g)
DH = +95.7 kJ and DS = +142.2 J/K at 25°C.
1. Calculate DG and determine if it is spontaneous:
DG = DH – TDS
DG = 95.7 x 103 J – (298K) (142.2 J/K)
DG = +5.33 x 104 J
2. Determine at what temperature (if any) the reaction becomes spontaneous:
DG = DH – TDS < 0
95.7 x 103 J – T (142.2 J/K) < 0
95.7 x 103 J < T (142.2 J/K)
Nonspontaneous
T > 673 K
Spontaneous at:
Effect on Spontaneity• When DH and DS have opposite signs,
spontaneity does not depend on the temperature, but when they have the same sign it does.
DG = DH – TDS
Third Law of Thermodynamics• The Third Law of Thermodynamics states
that for a perfect crystal at absolute zero, absolute entropy (S) = 0.• So every substance that is not a
perfect crystal at absolute zero has some energy from entropy.• Perfect crystals never occur in
practice; imperfections just get"frozen in" at low temperatures.• Scientists have achieved
temperatures extremely close to absolute zero.