Transcript Slide 1

Topic 5. Chemical Energetics
Created by S. Colgan;
Modified by K. Slater
Resources:
http://lincoln.pps.k12.or.us/lscheffler/Energetics.ppt#269,36,Sta
ndard Enthalpy Changes
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IB Topic 5: Energetics
5.1: Exothermic and Endothermic Reactions




5.1.1 Define the terms exothermic reaction, endothermic
reaction and standard enthalpy change of reaction
(ΔHo).
5.1.2 State that combustion and neutralization are
exothermic processes.
5.1.3 Apply the relationship between temperature
change, enthalpy change and the classification of a
reaction as endothermic or exothermic.
5.1.4 Deduce, from an enthalpy level diagram, the
relative stabilities of reactants and products, and the
sign of enthalpy change for the reaction.
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Heat and Temperature

Heat is energy that is transferred from one
object to another due to a difference in
temperature
 Temperature is a measure of the average
kinetic energy of a body
 Heat is always transferred from objects at a
higher temperature to those at a lower
temperature
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5.1.1 Define the terms exothermic reaction,
endothermic reaction and standard enthalpy
change of reaction (ΔHo).
Exothermic Reaction: A process that
releases heat to its surroundings. Products
have less energy than the reactants
Endothermic Reaction : A process that
absorbs heat from the surroundings.
Products have more energy than the
reactants.
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5.1.1 Define the terms exothermic reaction,
endothermic reaction and standard enthalpy
change of reaction (ΔHo).
Standard Enthalpy Change of Reaction (∆H): The heat
energy exchanged with the surroundings when a
reaction happens under standard conditions (NOT STP…
see below).
Since the enthalpy change for any given reaction will vary
with the conditions, esp. concentration of chemicals, ΔH
are measured under standard conditions:




pressure = 101.3 kPa
temperature = 25ºC = 298 K
Concentrations of 1 mol dm-3
The most thermodynamically stable allotrope (which in the case of
carbon is graphite)
Only ΔH can be measured, not H for the initial or final
state of a system.
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5.1.1 Define the terms exothermic reaction,
endothermic reaction and standard enthalpy
change of reaction (ΔHo).
Pseudonyms (other names) for H
 Heat of Reaction: Hrxn heat produced in a chemical reaction
 Heat of Combustion: Hcomb heat produced by a combustion
reaction
 Heat of Neutralization: heat produced in a neutralization reaction
(when an acid and base are mixed to get water, pH = 7)
 Heat of solution: Hsol heat produced by when something dissolves
 Heat of Fusion: Hfus heat produced when something melts
 Heat of Vaporization: Hvap heat produced when something
evaporates
 Heat of Sublimation: Hsub heat produced when something sublimes
 Heat of formation: Hf change in enthalpy that accompanies the
formation of 1 mole of compound from it’s elements (this has special
uses in chemistry…)
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5.1.2 State that combustion and
neutralization are exothermic processes.
Combustion
Exothermic reaction
General Combustion Reaction Formula:
Compound (usually hydrocarbon) + O2
 CO2 + H2O + energy
CH4 + 2O2  CO2 + 2H2O + 890kJ
∆H = -890kJ
Neutralization
Exothermic reaction
Acid + Base  Salt + Water + energy
HCl + NaOH  NaCl + H2O + 57.3 kJ
∆H = -57.3kJ
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5.1.3 Apply the relationship between temperature
change, enthalpy change and the classification of a
reaction as endothermic or exothermic.
Exothermic
Heat flows out of the
system
Surroundings heat up
Heat change (ΔH) < 0
(negative)
C8H18+ 12½O2  8CO2 +
9H2O ΔH = -5512 kJ mol-1
H2 + ½O2  H2O
ΔH = -286 kJ mol-1
Endothermic
Heat flows into the system
Surroundings cool down
Heat change (ΔH) > 0
(positive)
H2O(s)  H2O(l)
ΔH = +6.01 kJ mol-1
½N2 + O2  NO2
ΔH = +33.9 kJ mol-1
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Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic process is any process in which heat has to be
supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
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6.2
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6.5
5.1.4 Deduce, from an enthalpy level diagram, the
relative stabilities of reactants and products, and the
sign of enthalpy change for the reaction.
Exothermic Reactions
Products more stable than reactants
(lower energy).
ΔH = Hproducts – Hreactants
Since the products have less energy
than the reactants, the ΔH value is
negative.
Endothermic Reactions
Products less stable than reactants
(higher energy)
ΔH = Hproducts – Hreactants
Since the products have more energy
than the reactants, the ΔH value is
positive.
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Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
H = H (products) – H (reactants)
H = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
H < 0
Hproducts > Hreactants
H > 0
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6.4
5.1.4 Deduce, from an enthalpy level diagram, the
relative stabilities of reactants and products, and the
sign of enthalpy change for the reaction.
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REVIEW
Endothermic
Exothermic
Definition
Examples (2)
N/A
Change in Temperature
of the container
∆H value
Direction of heat flow
Stability of reactants
Stability of products
Bonding
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REVIEW
Definition
Endothermic
Exothermic
A process that absorbs
heat from the
surroundings
A process that releases
heat into the
surroundings
Examples (2)
Combustion &
Neutralization reactions
Change in
Temperature
Decreases
Increases
∆H value
Positive
Negative
Direction of heat flow
From surroundings into
system
From system into
surroundings
Stability of reactants
More stable
Less stable
Stability of products
Less stable
More stable
Bonding
Bond breaking
Bond making
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IB Topic 5: Energetics
5.2: Calculation of Enthalpy Changes




5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
5.2.2 Design suitable experimental procedures for
measuring the heat energy changes of reactions.
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water
5.2.4 Evaluate the results of experiments to determine
enthalpy changes.
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Factors Affecting Heat Quantities

The amount of heat contained by an object
depends primarily on three factors:
 The mass of material
 The temperature
 The kind of material and its ability to
absorb or retain heat.
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5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
Specific heat: The amount of heat energy required to
raise the temperature of 1 g of a substance 1 oC.
Units are: J g-1 oC-1
Specific heat of water (or aqueous solutions)
= 4.184 Joules = 1 cal
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Heat Quantities

The heat required to raise the temperature of
1.00 g of water 1 oC is known as a calorie
 Calorie (with a capital “C”): dietary
measurement of heat. Food has potential
energy stored in the chemical bonds of food.
1 Cal = 1 kcal = 1000 cal
 The SI unit for heat is the joule. It is based on
the mechanical energy requirements.
 1.00 calorie = 4.184 Joules
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5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
Heat Energy Change
q = m x c x ΔT
ΔH = q = heat (joules or calories)
m = mass (g)
c = specific heat (J g-1 oC-1)
The amount of heat required to raise the temperature of 1
g of a substance 1 oC.
Specific heat of water = 4.184 Joules /
ΔT = change in temperature
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5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
How much heat in joules will be
absorbed when 32.0 g of
water is heated from 25.0 oC
to 80.0 oC?
When 435 J of heat is added to 3.4 g of
olive oil at 21 oC, the temperature
increases to 85 oC. What is the
specific heat of olive oil?
q = m x c x ΔT
q=?
m = 32.0 g
c = 4.18 J g-1 oC-1
ΔT = 80.0-25.0 = 55.0 oC
q = m x c x ΔT
q = 435 J
m = 3.4 g
c=?
ΔT = 85-21 = 64 oC
q = 32.0 x 4.18 x 55.0 = 7,360
J
435 = 3.4 x c x 64 = 2.0 J g-1 oC-1
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Heat Transfer Problem 1
Calculate the heat gained in an aluminum cooking pan
whose mass is 400 grams, from 20oC to 200oC. The
specific heat of aluminum is 0.902 J g-1 oC-1.
Solution
q = mCT
= (400 g) (0.902 J g-1 oC-1)(200oC – 20oC)
= 64,944 J = 60,000J OR 6 x 104 J OR 60 kJ
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5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
How much heat in joules is
required to raise the
temperature of 250 g of
mercury 52oC? The specific
heat capacity of Hg is 0.14 J
g-1 oC-1
A 1.55 g piece of stainless steel
absorbs 141 J of heat when its
temperature increases by 178 oC.
What is the specific heat of the
stainless steel?
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5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
How much heat in joules is
required to raise the
temperature of 250 g of
mercury 52 oC?
A 1.55 g piece of stainless steel absorbs
141 J of heat when its temperature
increases by 178 oC. What is the
specific heat of the stainless steel?
q = m x c x ΔT
q=?
m = 250 g
c = 0.14 J g-1 oC-1
ΔT = 52 oC
q = m x c x ΔT
q = 141 J
m = 1.55 g
c=?
ΔT = 178 oC
q = 250 x 0.14 x 52 = 1800 J
OR 1.8 kJ
141 = 1.55 x c x 178 = 0.511 J g-1 oC-1
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5.2.2 Design suitable experimental procedures for
measuring the heat energy changes of reactions.

Calorimeter: Reactions
used to heat up an
external source of
water.
Temperature change of
water, mass of material
and mass of water are
measured.
Use q = m x c x ΔT to
solve for q then find the
heat of reaction in
kJ/mol of reacting
substance.
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Calorimetry

Calorimetry involves the measurement of
heat changes that occur in chemical
processes or reactions. Determines
the ΔH by measuring temp Δ's created
from the rxn

The heat change that occurs when a
substance absorbs or releases energy is
really a function of three quantities:



The mass
The temperature change
The heat capacity of the material
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Heat Capacity and Specific Heat



The ability of a substance to absorb or retain heat
varies widely.
The heat capacity depends on the nature of the
material.
The specific heat of a material is the amount of
heat required to raise the temperature of 1 gram of
a substance 1 oC (or Kelvin)
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Specific Heat values for
Some Common Substances
Substance
CJ g-1 K-1
C J mol-1K-1
Water (liquid)
4.184
75.327
Water (steam)
2.080
37.47
Water (ice)
2.050
38.09
Copper
0.385
24.47
Aluminum
0.897
24.2
Ethanol
2.44
112
Lead
0.127
26.4
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5.2.2 Design suitable experimental procedures for
measuring the heat energy changes of reactions.
Heat Energy Change
q = -q
HUH?!?
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Heat Transfer Problem 2
What mass of water will be heated from 5.72oC to
20.65oC if it is mixed in a calorimeter with 80.72 g of
water that starts at 29.5oC? Assume that the loss of
heat to the surroundings is negligible. The specific heat
of water is 4.184 J g-1 oC-1
Solution: Q (Cold) = -Q (hot)
mcT= -mcT
(m)(4.184)(20.65 – 5.72) = – (80.72)(4.184)(20.65 – 29.5)
(m)(4.184)(14.93) = – (80.72)(4.184)(–8.85)
62.46712 (m) = 2988.932448
m = 47.84 g
Using proper sig. fig.s…
m = 47.8 g
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Heat Transfer Problem 3
What mass of water will be heated from 25oC to 38oC
when 5.0 grams of 98oC copper is added to the water?
Assume that heat loss to the surroundings is negligible.
Substance
C (J g-1 K-1)
C (J mol-1K-1)
Copper
0.385
24.47
Aluminum
0.897
24.2
Lead
0.127
26.4
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Heat Transfer Problem 3
What mass of water that will be heated from 25oC to
38oC when 5.0 grams of 98oC copper is added to the
water? Assume that the loss of heat to the
surroundings is negligible
Solution: Q (Cold: H2O) = -Q (hot Cu)
mcT= -mcT
Let T = final temperature
(x) (4.184)(38 - 25oC) = - (5.0g) (0.385)(38 - 98)
(x)(4.184)(13) = - (5.0 g)(0.385)(-60)
54.392x = 115.5
x = 2.123474 g
With proper sig. figs. x = 2.1 g
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Heat Transfer Problem 4
What is the final temperature when 50 grams of water at
20oC is added to 80 grams water at 60oC? Assume
that the loss of heat to the surroundings is negligible.
The specific heat of water is 4.184 J g-1 oC-1
Solution: Q (Cold) = -Q (hot)
mcT= -mcT
Let T = final temperature
(50 g) (4.184 J g-1 oC-1)(T – 20oC)
= –(80 g) (4.184 J g-1 oC-1)(T – 60oC)
(50 g)(T – 20oC) = –(80 g) (T – 60oC)
50T – 1,000 = 4,800 – 80T
130T = 5,800
T = 44.6 oC
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Frayer Model
Heat
 Enthalpy
 Endothermic
 Exothermic

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5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermochemical Equations
Mg(s) + ½ O2(g)  MgO(s) + heat

Mg(s) + 2HCl(aq)
 MgO(s) + 2HCl(aq)
 H2(g) + ½ O2(g)
Equation 1
 MgCl2(aq) + H2(g) Equation A
 MgCl2(aq) +H2O(l) Equation B
 H2O(l)
Equation C
∆H = -A
∆H = -B
∆H = +C
35
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermochemical Equations
Mg(s) + ½ O2(g)  MgO(s) + heat

Mg(s) + 2HCl(aq)
 MgO(s) + 2HCl(aq)
 H2(g) + ½ O2(g)

Equation 1
 MgCl2(aq) + H2(g) Equation A
 MgCl2(aq) +H2O(l) Equation B
 H2O(l)
Equation C
∆H = -A
∆H = -B
∆H = +C
The final ∆H = -A + B + C
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5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermochemical Equations
Calcium oxide combines with water to produce calcium hydroxide and heat
(exothermic reaction).

CaO(s) + H2O(l)  Ca(OH)2(s) + 65.2 kJ OR

CaO(s) + H2O(l)  Ca(OH)2(s)
H = -65.2 kJ

These H values assume 1 mole of each compound (based on the coefficients)
How many kJ of heat are produced when 7.23 g of CaO react?
1)
Write out and balance equation: already balanced
2)
Determine the number of moles: 7.23 g/56.01 gmol-1 = 0.129 mol
3)
Multiply: 0.129 mol CaO x 65.2 kJ mol-1
* Notice that mol / mol cancel out and you’re left with kJ
4)
Solve = 8.41 kJ
37
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermochemical Equations
Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to
sodium carbonate, water, and carbon dioxide.
How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)?
1)
Write out and balance the equation
2)
Determine the number of moles of NaHCO3(s)
3)
Set up the ratio
4)
Solve for x
5)
State, with justification, whether the reaction is endothermic or
exothermic.
38
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermochemical Equations
Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to
sodium carbonate, water, and carbon dioxide.

2NaHCO3(s) + 129 kJ  Na2CO3(s) + H2O(g) + CO2(g) OR

2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g) H = 129 kJ
How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)?
1)
Balance equation
2)
Moles NaHCO3(s) = 2.24 mol
3)
Ratio:
x kJ
=
129 kJ
2.24 NaHCO3(s)
2 mol
4)
Solve for x = 144 kJ
5)
The reaction is endothermic because energy is being absorbed / the
H is positive.
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5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Experimental Data
In a neutralization reaction, 25.0 mL of water containing 0.025 mol HCl is
added to 25.0 mL of water containing 0.025 mol NaOH in a foam cup
calorimeter. At the start, the solutions and the calorimeter are all at 25.0
oC. During the reaction, the highest temperature observed is 32.0 oC.
Calculate the heat (in kJ) released during this reaction. Assume the
densities of the solutions are 1.00 g mL-1.
40
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Experimental Data
In a neutralization reaction, 25.0 mL of water containing 0.025 mol HCl is
added to 25.0 mL of water containing 0.025 mol NaOH in a foam cup
calorimeter. At the start, the solutions and the calorimeter are all at 25.0
oC. During the reaction, the highest temperature observed is 32.0 oC.
Calculate the heat (in kJ) released during this reaction. Assume the
densities of the solutions are 1.00 g mL-1.
Use q = m x c x ΔT
m = mass of solution = 50.0 mL x 1.00 g mL-1 = 50.0 g
C = 4.18 j g-1 oC-1
ΔT = 32.0 – 25.0 = 7.0 oC
q = 50.0 g x 4.18 J g-1 oC-1 x 7.0 oC = 1463 J = 1.5 kJ
41
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermodynamic Quantities (Standard Heats of Formation)

Heat of reaction can be found by: sum the heats of formation of all
the products – sum of heats of formation of all the reactants
Hrxn = Hf products – Hf reactants



Hf = standard enthalpy of formation. Energy required to form a
compound from its elements.
“standard” is a term used a lot in chemistry. It usually means that
the values are experimentally determined and compared to an
agreed upon reference value
Since the Hf is given per mole, we must multiply by coefficients
42
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermodynamic Quantities (Standard Heats of
Formation)
Using the table of thermodynamic quantities, calculate the heat of
reaction for 2SO2(g) + O2(g)  2SO3(g)
Heat of reaction = Hf products – Hf of reactants
Heat of products: Hf (SO3) = -395.2 kJ/mol x 2 = -790.4 kJ
Heat of reactants = Hf (SO2) + Hf (O2)
(-296.9 kJ/mol x 2) + (0) = -593.8 KJ
Heat of reaction = -790.4 kJ – (-593.8 kJ) = -196.6 kJ
43
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
N2O4 + 3 CO  N2O + 3CO2
Reactants
Hf
Products
Hf
N 2 O4
9.7 kJ mol-1
N 2O
81 kJ mol-1
CO
-110 kJ mol-1
CO2
-393 kJ mol-1
Hf products = 1(81 kJ mol-1) + 3(-393 kJ mol-1) = -1098 kJ/mol
Hf reactants = 1(-9.7 kJ mol-1)+ 3(-110 kJ mol-1) = -320.3 kJ mol-1
Hf products – Hf reactants = (-1098 kJ mol-1) – (-320.3 kJ mol-1)
= -778 kJ mol-1
Hrxn = -778 kJ mol-1
Therefore it is exothermic
44
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Ca(OH)2(s) + CO2 (g)  H2O(g) + CaCO3 (s)
Reactants
Hf
Products
Hf
Ca(OH)2
-986.1 kJ mol-1
H 2O
-241.8 kJ mol-1
CO2
-393.5 kJ mol-1
CaCO3
-1206.9 kJ mol-1
Hf products = 1(-241.8 kJ mol-1) + 1(-1206.9 kJ mol-1) = -1448.7 kJ/mol
Hf reactants = 1(-986.1 kJ mol-1)+ 1(-393.5 kJ mol-1) = -1379.6 kJ mol-1
Hf products – Hf reactants = (-1448.7 kJ mol-1) – (-1379.6 kJ mol-1)
= -69.1 kJ mol-1
Hrxn = -69.1 kJ mol-1
Therefore it is exothermic
45
IB Topic 5: Energetics
5.3 Hess’s Law & 5.4 Bond Enthalpies



5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
5.4.1 Define the term average bond enthalpy
5.4.2 Explain, in terms of average bond enthalpies, why
some reactions are exothermic and others are
endothermic
46
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
Hess’s Law: Reactions can be added together in order to determine heats of
reactions that can’t be measured in the lab.
C(diamond)  C(graphite)
This reaction is too slow to be measured in the lab. Two reactions can be used
that can be measured in the lab:
a) C(graph) + O2(g)  CO2(g)
H = -393.5 kJ
b) C(diam) + O2(g)  CO2(g)
H = -395.4 kJ
Since C(graphite) is a product, write equation a) in reverse to give:
c) CO2(g)  C(graph) + O2(g)
H = 393.5 kJ
Now add equations b) and c) together:
C(diam) + O2(g) + CO2(g)  C(graph) + O2(g) + CO2(g)
H = -395.4 kJ + 393.5 kJ = -1.9 kJ
Final equation: C(diamond)  C(graphite)
H = - 1.9 kJ
47
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction
for:
C2H4(g) + H2O(l)  C2H5OH(l)
a) C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
b) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
H = -1367 kJ
H = -1411 kJ
48
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction
for
C2H4(g) + H2O(l)  C2H5OH(l)
a) C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
b) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
H = -1367 kJ
H = -1411 kJ
Since C2H5OH(l) is a product, write equation a) in reverse order:
c) 2CO2(g) + 2H2O(l)  C2H5OH(l) + 3O2(g)
H = 1367 kJ
Add equations b) & c) together, cancelling out substances on opposite sides of
the arrow. Add the heat values to obtain the heat of reaction.
C2H4(g) + H2O(l)  C2H5OH(l)
H = -44 kJ
49
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction
for:
C(s) + 2H2(g)  CH4(g)
a) C(s) + O2(g)  CO2(g)
b) H2(g) + ½ O2(g)  H2O(l)
c) CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = -393 kJ mol-1
H = -286 kJ mol-1
H = -890 kJ mol-1
50
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction
for:
C(s) + 2H2(g)  CH4(g)
C(s) + O2(g)  CO2(g)
H = -393 kJ
2(H2(g) + ½ O2(g)  H2O(l))
2H2(g) + O2(g)  2H2O(l))
H = 2(-286 kJ mol-1)
H = -572 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
H = -890 kJ mol-1
H = 890
C(s) + 2H2(g)  CH4(g)
H = -75 kJ
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5.4.1 Define the term average bond enthalpy
In a chemical reaction
 Chemical bonds are broken
 Atoms are rearranged
 New chemical bonds are formed
 These processes always involve
energy changes
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52
Energy Changes

Enthalpy changes of reactions are the
result of bonds breaking and new bonds
being formed.
53
53
Energy Changes

Breaking chemical bonds requires energy
 Endothermic

Forming new chemical bonds releases
energy
 Exothermic
54
54
5.4.1 Define the term average bond enthalpy
Remember…

Breaking bonds requires energy

Forming new bonds releases energy
Bond enthalpy is the energy required to break one mole
of a certain type of bond in the gaseous state
averaged across a variety of compounds.
FYI: Bond enthalpies for unlike atoms will be affected by
surrounding bonds and will be slightly different in
different compounds so average bond enthalpies
are used.
55
5.3.2 Using bond enthalpy to determine
enthalpy change of a reaction
The average bond enthalpies for several types of
chemical bonds are shown in the table below:
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56
5.3.2 Using bond enthalpy to determine
enthalpy change of a reaction
 H
=
∑ (energy required
to break bonds)
OR
 H
–
∑(energy released
when bonds are formed)
–
∑(bond enthalpy
of products)
=
∑ (bond enthalpy
of reactants)
57
5.3.2 Using bond enthalpy to determine
enthalpy change of a reaction
H =
∑ (bond enthalpy
of reactants)
–
∑(bond enthalpy
of products)
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5.3.2 Using bond enthalpy to determine
enthalpy change of a reaction

Bond enthalpies can be used to calculate the
enthalpy change for a chemical reaction.
 Energy is required to break chemical bonds.
Therefore when a chemical bond is broken
its enthalpy change carries a positive sign.
 Energy is released when chemical bonds
form. When a chemical bond is formed its
enthalpy change is expressed as a negative
value
 By combining the enthalpy required and the
enthalpy released for the breaking and
forming chemical bonds, one can calculate
the enthalpy change for a chemical reaction
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Exothermic and Endothermic
Processes

Exothermic processes release energy
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4H2O (g)
+ 2043 kJ

Endothermic processes absorb energy
C(s) + H2O (g) +113 kJ  CO(g) + H2 (g)
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60
Energy Changes in endothermic
and exothermic processes
In an
endothermic
reaction there is
more energy
required to
break bonds
than is released
when bonds are
formed.
The opposite is
true in an
exothermic
reaction.
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61
5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
If the amount of energy required to break the
bonds in the reactants is greater than the
amount of energy released when bonds are
formed in the products, the reaction is
endothermic.
bond enthalpy reactants > bond enthalpy products
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
If the amount of energy required to break the
bonds in the reactants is less than the amount
of energy released when bonds are formed in
the products, the reaction is exothermic.
bond enthalpy reactants < bond enthalpy products
63
5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
Use the following average bond enthalpies (kJ mol-1) to determine the
heat of reaction for 3F2 + NH3  3HF + NF3
F-F = 158; N-H = 388; H-F = 562; N-F = 272
Energy in (kJ mol-1)
3F-F is 3(158) + 3N-H is 3(388) = 1638 kJ
Energy out (kJ mol-1)
3H-F is 3(562) + 3N-F is 3(272) = 2502 kJ
Since Energy in < Energy out, the reaction is exothermic
Heat of reaction is -864 kJ mol-1
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
Using the Bond Enthalpy Table, determine the heat of reaction for:
CO(g) + 2H2(g)  CH3OH
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
Using the Bond Enthalpy Table, determine the heat of reaction for:
CO(g) + 2H2(g)  CH3OH
Energy in (kJ)
1CO (triple bond) 1074 + 2H-H is 2(436) = 1946 kJ mol-1
Energy out (kJ)
3C-H is 3(413) + 1C-O is 358 + 1 O-H is 464 = 2061 kJ mol-1
Hrxn = 1946 – 2061 = -115 kJ mol-1
Since Energy in < Energy out, reaction is exothermic
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
Using the Bond Enthalpy Table, determine the heat of reaction for:
CO(g) + 2H2(g)  CH3OH
Energy in (kJ)
1CO (triple bond) 1074 + 2H-H is 2(436) = 1946 kJ mol-1
Energy out (kJ)
3C-H is 3(413) + 1C-O is 358 + 1 O-H is 464 = 2061 kJ mol-1
Hrxn = 1946 – 2061 = -115 kJ mol-1
Since Energy in < Energy out, reaction is exothermic
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Bond Enthalpy Calculations
Calculate the enthalpy change for the
reaction N2 + 3 H2  2 NH3
Bonds broken
1 N=N:
3 H-H:
= 945
3(435) = 1305
Total = 2250 kJ
Bonds formed
2x3 = 6 N-H:
6 (390) = - 2340 kJ
Net enthalpy change
= + 2250 - 2340 = - 90 kJ
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
In the 1960s NASA considered the relative merits of using hydrogen
and oxygen compared with hydrogen and fluorine as rocket fuels.
Assuming all the reactants and products are in the gaseous state,
use bond enthalpies to calculate the enthalpy change of reaction
(in kJ mol-1 of product) for both fuels. As mass is more important
than amount in the choice of rocket fuels, which reaction would
give more energy per kilogram of fuel?
Bond enthalpies (kJ mol-1):
H-H: 435; O O: 496; H-O: 464 kJ; F-F: 158; H-F: 562
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Terms to Know










Endothermic
Exothermic
Temperature
Heat
Standard enthalpy change of reaction
Standard enthalpy of formation
Enthalpy of combustion
Average bond enthalpy
Hess’ Law
Standard conditions
70