Download Energy and Change - SCH4U Catherine Gordeyev

UNIT 3: ENERGY CHANGES AND RATES OF REACTION
Big Ideas
 Energy changes and rates of chemical reactions can be described quantitatively.
 Efficiency of chemical reactions can be improved by applying optimal conditions.
 Technologies that transform energy can have societal and environmental costs and benefits.
Energy
Success Criteria:
 I am able to write thermochemical equations, expressing the energy change as a ΔH value or as a
heat term in the equation.
 I am able to draw qualitative and quantitative connections between the reaction enthalpy and the
energies involved in the breaking and formation of chemical bonds.

the study of energy and energy transfer is known as thermodynamics; the study of energy
involved in chemical reactions is thermochemistry

energy (E) is defined as the capacity to do work or transfer heat; it includes kinetic energy and
potential energy

kinetic energy is the energy due to the motion of an object while potential energy is the energy
that an object possesses by virtue of position relative to other objects

the SI unit for energy is the joule (J) which is equivalent to 1 kg•m2/s2; the nutrional Calorie (Cal)
is a non-SI unit equivalent to 4.18 kJ

the system is the part of the universe that is being studied (ex.
reactants and products) and the surroundings are everything
outside the system

heat is the amount of energy transferred from one system to
another and always flows from hot to cold
Euniverse = Esystem + Esurroundings
thus, Esystem = -Esurroundings

the first law of thermodynamics states that energy cannot be
created or destroyed but only changes form; therefore, the
total energy of the universe is constant (Euniverse = 0)

temperature (T) is the average kinetic energy of all particles in a system

endothermic reactions result in an absorption of energy from
the surroundings; exothermic reactions result in a net release
of energy to the surroundings
Homework:
Read: pg. 278 – 291 (5.1)
Answer: pg. 281 Q 1, 5, 6, 7
pg.291 Q 2, 6, 7 10
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Enthalpy
Success Criteria:
 I can use appropriate terminology related to energy changes including, but not limited to: enthalpy,
endothermic, exothermic, potential energy, and specific heat capacity.
 I can compare the energy change from a reaction in which bonds are formed to one in which bonds
are broken, and explain these changes in terms of endothermic and exothermic reactions.

heat flow in processes occurring at constant pressure is called enthalpy (H); enthalpy change
(H) is the heat gained or lost at constant pressure

if the system absorbs heat (endothermic), the H will be positive; if the system loses heat
(exothermic), the H will be negative

the enthalpy change of a chemical reaction is known as the enthalpy of reaction, Hr; the standard
enthalpy of reaction, Hor occurs at SATP

in chemical reactions, enthalpy changes result from bonds being broken and formed; bond breaking
requires energy and bond making releases energy

enthalpy changes can be represented in three ways:
 a thermochemical equation
ex.
117.3 kJ + MgCO3 (s)  MgO (s) + CO2 (g)
ex.
H2 (g) + ½O2 (g)  H2O (l) + 285.8 kJ
 a regular equation with a separate enthalpy expression
ex.
MgCO3 (s)  MgO(s) + CO2(g) Ho = 117.3 kJ
ex.
H2 (g) + ½O2 (g)  H2O (l)
Ho =-285.8 kJ
 an enthalpy diagram

H directly proportional to the amounts of substances that react; thus if the amount of product
doubles, so does the H
ex.
2MgCO3 (s)  2MgO (s) + 2CO2 (g) Ho = 234.6 kJ

H is equal in magnitude but opposite in sign to the H for the reverse reaction
ex.
H2O (l)  H2 (g) + ½O2 (g)
Ho = 285.8 kJ

enthalpy changes can also be associated with physical
changes such as changes of state (i.e. boiling = enthalpy of
vaporization, Hvap, melting = enthalpy of fusion , Hfus)

a heating curve is a plot of temperature vs. time; when
energy is absorbed for a phase change, the temperature
remains constant (plateaus)

when a solute dissolves in a solvent, heat is transferred
(ex. hot packs and cold packs); this is called the enthalpy of solution, Hsoln
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Calorimetry
Success criteria:
 I can solve problems involving analysis of heat transfer in a chemical reaction, using the equation
Q = mcΔT.
 I can explain how mass, heat capacity, and change in temperature of a substance determine the
amount of heat gained or lost by the substance.

the value of H can be determined experimentally; a calorimeter is a device used to measure the
amount of heat transferred between the system and the surroundings

all substances change temperature when heated, but the magnitude of the change varies for each
substance; therefore, the specific heat capacity of the calorimeter must be taken into account

c is the specific heat capacity measured in J/g•K; it the amount of heat required to raise one gram
of substance by one degree of temperature

heat transfer can be calculated by using the following equation: q = mcΔT

there are different types of calorimeters that can be used to analyse heat changes

a simple coffee-cup calorimeter can be used for
constant-pressure experiments (ex. reactions in solution)

note that when a reaction occurs in solution, the
temperature change of the water is measured

therefore, in an exothermic reaction, heat is lost by the
reaction but gained by the solution
H = qr = -qsoln

a complex bomb calorimeter can be used for constantvolume experiments (ex. combustion reactions)

to obtain precise measurements, the heat capacity (c) of the entire bomb calorimeter must be
known; as such heat transfer can be calculate using: q = mcΔT

when two objects are in contact, heat is always transferred from the object at a higher temperature
to the object at lower temperature
Homework:
Read: pg. 292 – 311 (5.2)
Answer: pg. 295 Q 9, 10, 11
pg. 299 Q 12, 15, 18, 20
pg. 305 Q 21 – 25
pg. 309 Q 31 – 36
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Hess’s Law
Success Criteria:
 I can solve problems related to energy changes in a chemical reaction, using Hess's law.
 I can state Hess's law, and explain, using examples, how it is applied to find the enthalpy changes
of a reaction.

chemists can also determine the enthalpy change of any reaction using Hess’s law; Hess’s law
allows chemists to determine enthalpy changes without directly measuring them

this law states that in a physical or chemical process, the enthalpy change
is independent of the pathway and the number of intermediate steps

in other words, if a reaction is carried out in a series of step, H for the
overall reaction will equal the sum of the enthalpy changes for the
individual steps
ex.
carbon and oxygen can form via two different pathways
1)
C (s) + ½ O2 (g)  CO (g)
CO (g) + ½ O2 (g)  CO2 (g)
Ho1 = - 110.5 kJ
Ho2 = - 283 kJ
2)
C (s) + O2 (g)  CO2 (g)
Ho3 = - 393.5 kJ
in both cases, the overall Ho is the same since Ho3 = Ho1+ Ho2

equations can be algebraically manipulated, reversed (change the sign of Ho) and multiplied by
coefficients (multiply Ho by the coefficient) to reach the target equation
Homework:
Read: pg. 312 – 314 (5.3)
Answer: pg. 316 Q 41 – 47
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Enthalpies of Formation and Bond Enthalpies
Success Criteria:
 I can calculate the heat of reaction for a formation reaction, using a table of standard enthalpies of
formation and applying Hess's law.
 I am able to draw qualitative and quantitative connections between the reaction enthalpy and the
energies involved in the breaking and formation of chemical bonds.

when a compound is formed directly from its elements, it is known as the enthalpy of formation,
Hf; for an element in its standard state (ex. O2 gas or solid iron), Hof = 0 kJ/mol

the standard enthalpy change of a reaction is defined as the enthalpy change when all the reactants
and products are at standard conditions (1 atm and 298.15 K); it is denoted with a nought, Ho

Ho for any reaction can be calculated from the standard enthalpies of formation
Ho = Σ(nHof products) - Σ(nHof reactants)
ex.
H2 (g) + Cl2 (g)  2HCl (g)
Ho
= 2 mol (-92.5 kJ/mol) – [1 mol (0 kJ/mol) + 1 mol (0 kJ/mol)] = -185 kJ

Ho can also be calculated using bond enthalpies, the energy that is absorbed or released when
breaking or making a bond

bond enthalpies are always positive quantities while double and triple bonds tend to have the
highest enthalpies

note that using enthalpies of formation is more accurate than bond enthalpies since bond enthalpies
are average values that are derived from gaseous elements
Ho = Σ(Bond energies of broken bonds) - Σ(Bond energies of bonds made)
ex.
H2 (g) + Cl2 (g)  2HCl (g)
Ho

= 1 mol (H-H) + 1mol (Cl-Cl) – 2 mol (H-Cl)
= 1 mol (436 kJ/mol) + 1 mol (243 kJ/mol) – 2 mol (432 kJ/mol)
= - 185 kJ
as bond enthalpies increase, bond lengths decrease
ex.
Homework:
C – C enthalpy: 348 kJ/mol
C = C enthalpy: 614 kJ/mol
C ≡ C enthalpy: 839 kJ/mol
length: 1.54 Ǻ
length: 1.34 Ǻ
length: 1.20 Ǻ
Read: pg. 317 – 324 (5.3)
Answer: pg. 323 Q 52 – 60
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Science, Technology, Society and the Environment: Thermochemistry

the combustion of foods in our bodies, and fuels in engines and factories, are exothermic reactions
that release lots of energy

our bodies constantly need energy to maintain body temperature, to contract muscles, and to
construct and repair tissues

carbohydrates such as glucose are broken down to supply the body with
energy quickly
ex.
C6H12O6 (s) + 6O2 (g)  6CO2 (g) + 6H2O (l)
Ho = -2803 kJ

interestingly, it is not the combustion of carbohydrates that
produces the more exothermic reaction, but the combustion
of fats (ex. tristearin)
ex.
C57H110O6 (s) + 81.5O2 (g)  57CO2 (g) + 55H2O (l)
Ho = -37,760 kJ

the body usually stores excess energy as fats; this is practical
because fats are insoluble in water and produce more energy
per gram than either proteins or carbohydrates

energy is also needed to run many of our technologies; the
combustion of natural gas and gasoline produce more energy
per g than wood or coal

interestingly, it is the combustion of H2 gas to form water which produces the most energy per g!
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Thermodynamics

a spontaneous process is a change that has a natural tendency to happen under certain conditions
and proceeds on its own without any outside assistance (ex. a ball rolling down a hill)

spontaneous reactions occur regardless of the speed; some happen quickly others slowly

a non-spontaneous process cannot happen without an input of energy (ex. rolling a ball up a hill)

processes that are spontaneous in one direction (ex. a nail rusting, a gas expanding into a vacuum)
are non-spontaneous in the opposite direction
Entropy
Success Criteria:
 I can use representations and models to predict the sign and relative magnitude of the entropy
change associated with chemical or physical processes.
 I can determine qualitatively the change in entropy of various reactions by examining the states and
number of moles of the reactants and products.

entropy, S, is the tendency toward randomness or disorder in a system

the entropy of a pure substance depends on its state; an
ordered arrangement of particles (ex. solids) has lower entropy
than a disordered arrangement (ex. gases)

the entropy of a system increases with increasing temperature

unlike energy, entropy is not conserved; the second law of thermodynamics states that the total
entropy of the universe is constantly increasing

all spontaneous changes involve an increase in the total amount of entropy; the change in entropy,
S, in system depends on the initial and final sates of the system

qualitative predictions about S in a chemical reaction can be made; the entropy of the system will
increase if:
 gases are formed from solids or liquids
ex.
H2O (l)  H2O (g)
 liquids or solutions are formed from solids
ex.
AgCl (s)  Ag+ (aq) + Cl- (aq)
 the number of moles of gas increases, the entropy of the system will increase
ex.
2SO3 (g)  2SO2 (g) + O2 (g)

the entropy of a pure crystalline substance at absolute zero is 0; standard molar entropies are
greater for gases than liquids and generally increase with molar mass

entropy can be calculated using standard molar entropies; entropy is measured in J/Kmol
So = (nSoproducts) - (nSoreactants)
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Gibbs Free Energy
Success Criteria
 I can predict whether or not a physical or chemical process is thermodynamically favored by
determination of (either quantitatively or qualitatively) the signs of both ΔH° and ΔS°, and
calculation or estimation of ΔG° when needed.
 I can determine whether a chemical or physical process is thermodynamically favorable by
calculating the change in standard Gibbs free energy.

there are three factors that determine the spontaneity of a chemical reaction:
entropy, enthalpy, and temperature

the relationship between enthalpy, temperature and entropy is known as
Gibbs Free energy, G (named in honour of Harvard professor Josiah
Willard Gibbs)
G = H - TS
(T is measured in K)

free energy is a measure of the useful work that can be obtained from a reaction

the sign of G shows which reactions are
spontaneous, which are not, and how temperature
affects the direction of the reaction
 if G is negative (< 0), the forward
reaction is spontaneous
 if G = 0, the reaction is at equilibrium or
a phase change is occurring
 if G is positive (> 0), the forward reaction
is non-spontaneous, but the reverse
reaction is spontaneous

most exothermic reactions are spontaneous because
H is negative; however, if S is negative and the
temperature is large enough, G becomes positive

endothermic reactions can be spontaneous if S is
positive and the temperature is large enough to overcome the positive H
H
S
G
+
+
-
+
+
-
+
Comment
The reaction is favourable at all T.
The reaction is unfavourable at all T.
The reaction is favourable at high T.
The reaction is favourable at low T.
ex. 2H2O2 (l)  2H2O (l) + O2 (g)
H = -197 kJ
 both enthalpy (-H) and entropy (+S) favour the reaction; the reaction will be spontaneous at all
temperatures
ex. 3O2 (g)  2O3 (g)
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H = 285 kJ
8
 both enthalpy (+H) and entropy (-S) are unfavourable; this reaction will not spontaneously occur
at any temperature
ex. 2H2O (g)  2H2 (g) + O2 (g)
H = 484 kJ
 entropy favours the reaction (+S) but enthalpy (+H) does not; once the value of TS exceeds
H, the reaction will be spontaneous
ex. NH3 (g) + HCl (g)  NH4Cl (s)
H = - 176 kJ
 enthalpy favours the reaction (-H) but the entropy does not (-S); thus, the reaction will be
spontaneous at low temperatures

G can also be calculated using standard free energies of formation which is analogous to
enthalpies of formation; Gibbs free energy is measured in kJ/mol

Go (at standard conditions) for any reaction can be calculated from the standard enthalpies of
formation
Go = (nGofproducts) - (nGofreactants)

the values for enthalpy and entropy change with temperature, however, the variations are small
enough that they can be ignored
Science, Technology, Society and the Environment: Free-Energy

many desirable reactions that occur in biological systems are non-spontaneous; therefore,
spontaneous reactions are used to drive non-spontaneous ones

spontaneous reactions (ex. the combustion of glucose) yield enough free energy to drive nonspontaneous (ex. the conversion of ADP to ATP)
C6H12O6 + 6O2  6CO2 + 6 H2O
ADP + Pi  ATP
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ΔG° = – 2880 kJ
ΔG° = 30 kJ
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Rates of Reaction
Success Criteria:
 I can interpret the results of an experiment regarding the factors (i.e., temperature, concentration,
surface area) that may influence the rate of a reaction.

kinetics is the field of chemistry that deals with the rate at which chemical reactions occur

the change in the amount of reactants or products over time is called the reaction rate
Rate of Reaction 
[ A] final  [ A]initial
t final  t initial

[ A]
t

reaction rates are not usually constant, are always positive
by convention, and are expressed in mol/(L∙s) for reactions
in gases or solutions

rates can be plotted on a kinetic curve (also called a
concentration versus time curve)

the average rate is the change in the concentration of a
reactant or product over a time interval; this can be found by drawing a secant on a graph

the instantaneous rate is the rate of reaction at a particular time; this can be found by drawing a
tangent line at that time

the rate of appearance or disappearance of any substance in a reaction can be determined
stoichiometrically

reaction rates can be determined by measuring or
observing the change of mass, pH, conductivity, pressure,
colour and volume during a reaction

reactions rate can be increased by increasing:
 the temperature of the reaction (as a rule, rates
double for every increase of 10oC)
 the concentration of the reactants
 the surface area of the reactants

the rate of reaction can also be increased by adding a catalyst; the catalyst is not consumed in the
reaction and is regenerated after the reaction

catalysts are widely used in industries; enzymes are
biological catalysts

a homogeneous catalyst exists in the same phase as the
reactants, whereas a heterogeneous catalyst exists in a
different phase from the reactants
Homework:
Read: pg. 354 – 363 (6.1), Answer: pg. 360 Q 6 – 8
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The Rate Law
Success Criteria:
 I can analyze concentration vs. time data to determine the rate law for a zeroth-, first-, or secondorder reaction.
 I am able to use data tables to calculate the order of each reactant, the overall reaction order, the
rate law equation and the rate constant.

the rate law equation expresses the relationship between the concentrations of the reactants and the
rate of the reaction: when aA + bB  products
Rate = k[A]m[B]n

where k is the rate constant; there is a different rate constant for each reaction at any given
temperature

the magnitude of the rate constant indicates the speed of a reaction; a large rate indicates a fast
reaction, a small rate a slow reaction

m and n must be determined experimentally; the sum of their values is the overall reaction order

[A ]
Rate 2
the reaction order can be found using the equation:  2  
Rate1
 [ A1 ] 
n

the order of the rate law can also be determined from graphs of concentration (or the values of
some function of concentration) versus time; the rate constant is the slope of the line

in a zero-order reaction, concentration does not affect rate; [A] versus t is linear (negative slope)

a first-order reaction has an overall reaction order of 1; ln [A] versus t is linear (negative slope)
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ex. 2N2O5  4NO2 + 5O2
ex. (CH3)3CBr + H2O  (CH3)3OH + H+ + Br
Rate = k[N2O5]1
Rate = k[(CH3)3CBr]1[H2O]0
a second-order reaction has an overall reaction order of 2; 1/[A] versus t is linear (positive slope)
ex. 2HI  H2 + I2
ex. NO + O3  NO2 + O2
Rate = k[HI]2
Rate = k[NO]1[O3]1

many reactions have an overall reaction order higher than 2; some have orders that are fractions;
some reactions are independent of concentration

the half-life (t1/2) of a reaction is the time needed for the reactant mass or concentration to decrease
by half

the half-life of any first-order reaction and the initial or final concentrations can be calculated using
the equations:
t1 / 2 
Homework:
0.693
k
1
[ A]t1 / 2  ( ) n [ A]initial
2
Read: pg. 380 - 383 (6.3)
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Collision Theory
Success Criteria:
 I can explain, using collision theory and potential energy diagrams, how factors such as
temperature, the surface area of the reactants, the nature of the reactants, the addition of catalysts,
and the concentration of the solution control the rate of a chemical reaction.
 I can describe simple potential energy diagrams of chemical reactions (e.g., the relationships
between the relative energies of reactants and products and the activation energy of the reaction)

for a reaction to occur, reacting particles must collide with each other; an increase in concentration
and temperature increases the rate of reaction because there will be more collisions per unit time

however, not every collision results in a reaction; there are two criteria that must be satisfied in
order for a collision to be effective
 Orientation: reacting particles must have the correct collision geometry; that is, they must
collide with the proper orientation relative to one another
 Activation Energy: the energy of the collision must be sufficient to begin to break the
bonds; activation energy, Ea, is the minimum energy required for a successful reaction

catalysts work by lowering the activation energy of a reaction so that a larger fraction of the
reactants have the required collision energy

in most reactions, only a small fraction of collisions have
sufficient activation energy; collision energy is dependent on
the kinetic energy (i.e. temperature) of the particles

the number of collisions and kinetic energy can be shown on a
graph called a Maxwell-Boltzmann distribution

the potential energy during a chemical reaction and the reaction progress can be shown on a
diagram called a potential energy diagram

the maximum potential energy of a reaction is called the transition state; here, the activated
complex forms

the activated complex is an intermediate chemical species that is neither product nor reactant; it
has partial bonds and is highly unstable

the difference between the energy of the reactants and the transition state is the activation energy
Homework:
Read: pg. 365 - 377 (6.2), Answer: pg. 371 Q 12 – 17, pg. 377 27, 29
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Reaction Mechanisms
 I can explain, with reference to a simple chemical reaction, how the rate of a reaction is determined
by the series of elementary steps that make up the overall reaction mechanism.
 I can analyze reaction mechanisms to determine the overall reaction, the molecularity, any reaction
intermediates and catalysts, and the rate-determining step.

most chemical reactions take place via a series of steps; a reaction mechanism is the series of
steps that make up an overall reaction
ex.
Step 1:
Step 2:
Overall:
NO + O2  NO3
NO3 + NO  2NO2
2NO + O2 2NO2

each step involves a collision and is known as an elementary reaction; molecules that are
produced in one step and used in another are called reaction intermediates

reaction mechanisms are difficult to determine; most of the evidence is obtained indirectly or by
proving the existence of reaction intermediates

a catalyst actually works by providing an alternative reaction mechanism which has a lower
activation energy; catalysts always appear as a reactant in the first step and product in the final step
ex.

Step 1:
Step 2:
Overall:
A + catalyst  A-catalyst
A-catalyst + B  AB + catalyst
A + B  AB
the number of reactants colliding in an elementary reaction is called molecularity; most steps are
unimolecular or bimolecular; termolecular reactions are possible but rare
ex.
Step 1:
Step 2:
Overall:
O3  O2 + O unimolecular
O3 + O  2O2 bimolecular
2O3 3O2

for an elementary reaction, the exponents in the rate law are the same as the stoichiometric
coefficients
A  products
Rate: k[A]
A + B  products
Rate: k[A][B]
2A  products
Rate: k[A]2
2A + B  products
Rate: k[A]2[B]

every reaction has a rate-determining step; this is always the slower step and it determines the
overall rate law; intermediates cannot be included in the rate law
ex.
Homework:
Step 1:
Step 2:
Overall:
NO2 + Cl2  NO2Cl + Cl
NO2 + Cl  NO2Cl
2NO2 + Cl2  2NO2Cl
(slow)
(fast)
Rate: k[NO2][Cl2]
Read: pg. 383 – 387 (6.3), Answer: pg. 387 Q 5, 6, 10 11
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