THERMOCHEMISTRY:
ENERGY FLOW AND CHEMICAL FLOW
In general…
 All matter contains energy, so whenever matter undergoes
a change, the quantity of energy that the matter contains
also changes
o When a burning candle melts a piece of ice
 Higher energy reactants, wax and oxygen, form
lower energy products, carbon dioxide and water
 The difference is released as heat and light
 Some of the heat is absorbed when the lower
energy ice becomes higher energy water
o In a thunderstorm
 Lower energy N2 and O2 absorb energy from
lightning and form higher energy NO, and the
higher energy water vapor releases energy as it
condenses to lower energy water that falls as rain
o Everyday examples
 Fuels such as oil and wood release energy to heat
our houses
 Fertilizers help convert solar energy into food
 Metal wires increase the flow of electrical energy
 Polymer fibers in winter clothing limit the flow
of thermal energy away from our bodies
Thermodynamics – the study of energy and its transformations
(will be addressed in chapter 20 as well for students taking
CHEM 102!)
Thermochemistry – the branch of thermodynamics that deals
with heat in chemical and physical change
Forms of Energy and Their Interconversions
When energy is transferred from one object to another, it
appears as work and/or heat
Defining the System and Surroundings
System – part of the universe we are focusing on
Surroundings – everything that is not the system
In the above picture, the contents of the flask is the system and
everything thing else (even the flask) is the surroundings
Energy Transfer to and From a System
Internal Energy, E, is the sum of the potential and kinetic
energy found in each particle in a system
 When the reactants in a chemical system change to
products, the system’s internal energy changes
o This is represented by ∆E
 ∆E = Efinal - Einitial = Eproducts - Ereactants
 A change in the energy of the system must be accompanied
by an equal and opposite change in the energy of the
surroundings
 A system can change its internal energy in one of two ways
o By releasing some energy in a transfer to the
surroundings (diagram A below)
 Efinal < Einitial so that ∆E < 0
o By absorbing some energy in a transfer from the
surroundings (diagram B below)
 Efinal > Einitial so that ∆E > 0
 ∆E is a transfer of energy from system to surroundings, or
vice versa…
Heat and Work: Two Forms of Energy Transfer
Energy transferred from system to surroundings or vice versa
appears in two forms…
 Heat. Heat or thermal energy (symbolized as q) is the
energy transferred as a result of a difference in temperature
between the system and the surroundings. Movement from
warmer system/surroundings to cooler
system/surroundings.
 Work. All other forms of energy transfer involve some type
of work (w), the energy transferred when an object is
moved by a force.
The total change in a system’s internal energy, E, is the sum of
the energy transferred as heat and/or work.
 ∆E = q + w
o The values of q and w can have either positive or
negative signs
 We define the sign (+ or -) of the energy change
from the system’s perspective
 Energy transferred into the system is
positive because the system ends up with
more energy
 Energy transferred out from the system is
negative because the system ends up with
less energy
Energy Transferred as Heat Only
 Heat flowing out of a system… (example A)
o q is negative, so ∆E is negative
 Heat flowing into a system… (example B)
o q is positive, so ∆E is positive
 Thermodynamics in the kitchen…
o The air is your refrigerator (surroundings) has a lower
temperature than a newly added piece of food (system)
 Food releases energy to the surroundings, so
q < 0 (system loses energy)
o The hot air in an oven (surroundings) has a higher
temperature than a newly added piece of food (system)
 Food is absorbing energy from the surroundings,
so q > 0 (system gained energy)
o Just to clarify… your favorite beverage in a cooler full
of ice makes the ice warmer (the ice does not make
your beverage cooler)
 The beverage is the system and the cooler is the
surroundings. The beverage is releasing heat to
the surroundings. So… q < 0 (system lost
energy)
Energy Transferred as Work Only
 Work done by a system… (in the above figure)
o The piston is pushed out by the system, so w is
negative and ∆E is negative
 Work done on a system… (no picture) so… imagine the
piston being pushed in by the surroundings
o w is positive and ∆E is positive
Law of Conservation of Energy
Energy changes form (from chemical to kinetic to potential for
example) but does not simply appear or disappear – energy
cannot be created or destroyed.
 Law of conservation of energy (first law of
thermodynamics) – the total energy of the universe is
constant
o ∆Euniverse = ∆Esystem + ∆Esurroundings
Units of Energy
 The SI unit of energy is the joule (J)
 The calorie is an older term originally defined as the
quantity of energy required to raise the temperature of 1
gram of water 1 oC
o Now defined in terms of the joule…
 1 calorie = 4.184 J
 1 J = 0.2390 calories
 1 kJ = 1000 J = 0.2390 kcal = 239.0 calories
 British thermal unit (Btu) is a unit used for energy output
of appliances
o Quantity of energy required to raise the temperature of
1 lb. of water 1oF
 1 Btu = 1055 J
Sample Problem 6.1 p. 234
Determining the Change in Internal Energy
q = -325 J (system releases 325 J to surroundings)
w = -451 J (system does 451 J on the pistons)
∆E = q + w
∆E = -325 J + (- 451 J)
∆E = - 776 J
- 776 J 1 kJ
1000 J
= - 0.776 kJ
- 0.776 kJ 1 kcal
= - 0.185 kcal
4.184 kJ
Follow-Up Problem 6.1 p. 234
26.0 kcal 4.184 kJ
1 kcal
= 108.784 kJ
15.0 Btu
1 kJ
1000 J
1055 J
1 Btu
= 15.825 kJ
q = -108.784 kJ (absorbed by surroundings)
w = 15.825 kJ (work done on the system)
∆E = q + w
∆E = -108.784 kJ + 15.825 kJ
∆E = - 92.959 kJ
∆E = - 93.0 kJ
State Functions & the Path Independence of the Energy Change
The internal energy (E) of a system is called a state function
 A property dependent only on the current state of the
system, not on the path the system takes to reach that state.
 ∆E is also a state function because it does not depend on
how the change takes place, but only on the difference
between the final and initial states.
 q and w are not state functions because their values do
depend on the path the system takes
 In the above example, q and w for the two paths are
different, ∆E is the same
 Pressure (P), volume (V), and temperature are other state
functions
 Path independence means that changes in state functions ∆E, ∆P, ∆V, and ∆T – depend only on the initial and final
states.
Enthalpy: Chemical Change at Constant Pressure
Two most important types of chemical work…
 Electrical work
o Work done by moving charged particles
 Pressure-volume work
o The mechanical work done when the volume of a
system changes in the presence of an external pressure
The quantity of PV work equals P times
the change in V… w = - P∆V (it is
negative because the gas is doing work
on the surroundings
At constant pressure, a thermodynamic
variable called enthalpy (H) eliminates
the need to measure PV work. The
enthalpy of a system is defined as the
internal energy plus the product of pressure and volume…
H = E + PV
The change in enthalpy (∆H) is the change in internal energy
plus the product of the pressure, which is constant, and the
change in volume
∆H = ∆E + P∆V
Most important point… the change in enthalpy (∆H) equals the
heated absorbed or released at constant pressure.
 ∆H is more relevant than ∆E and easier to obtain
Comparing ∆E and ∆H
Knowing the enthalpy change of a system tells us a lot about its
energy change as well
 Many reactions involve little (if any) PV work
 Most (or all) the energy change occurs as a transfer of heat
 For most reactions, ∆H equals or is very close to ∆E
Exothermic and Endothermic Processes
Because H is a combination of the three state functions E, P, and
V, it is also a state function
∆H = Hfinal – Hinitial = Hproducts - Hreactants
Exothermic process
 An exothermic process releases heat and results in a
decrease in the enthalpy of the system
o Hproducts < Hreactants so… ∆H < O
o Heat is found on the product side of the equation
 CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + heat
 Diagram A below is showing an exothermic
reaction
Endothermic process
 An endothermic process absorbs heat and results in an
increase in the enthalpy of the system
o Hproducts > Hreactants so… ∆H > O
o Heat is found on the reactant side of the equation
o Heat + H2O(s)  H2O(l)
Sample Problem 6.2 p. 237
Drawing Enthalpy Diagrams and Determining the Sign of ∆H
Follow-Up Problem 6.2 p. 237
Enthalpy
C3H5(NO3)3(l) (reactant)
∆H = - 5.72x103 kJ Exothermic
3CO2(g) + 5/2 H2O(g) + ¼ O2(g) + 3/2 N2(g) (products)
Calorimetry: Measuring the Heat of a Chemical or Physical
Change
Temperature change depends on the object…
 Every object has its own heat capacity, the quantity of heat
required to change its temperature 1 K
o Specific heat capacity (c), the quantity of heat
required to change the temperature of 1 gram of an
object by 1 K
To calculate the heat absorbed or released…
 q = c x mass x ∆T
o q = heat absorbed or released
o c = specific heat capacity (J/g K)
o mass in grams
o ∆T in K
Molar heat capacity – the quantity of heat required to change
the temperature of 1 mole of a substance 1 K
The combination of high specific heat of water and the amount
of water that covers the Earth play an important role maintaining
an environment that is favorable to life as we know it…
If the Earth had no oceans, it
would take one-sixth as much
energy for the Sun to heat the
rocky surface, and the surface
would five off its heat six times
as much heat after sundown.
Water has a high specific heat
which means it takes a lot of
energy input to change the
temperature and it takes a
considerable loss for the temperature to go down. People that
live in Chicago experience this with Lake Michigan as well.
(The lake stays cold until well into the summer and stays warm
well into the fall and early winter)
Sample Problem 6.3 p. 239
Finding the Quantity of Heat from a Temperature Change
∆T = Tfinal - Tinitial
∆T = 300. oC – 25oC = 275 oC = 275 K
q = c x mass x ∆T
q = 0.387 J/g K x 125 g x 275 K
q = 1.33x104 J
Follow-Up Problem 6.3 p. 239
∆T = Tfinal - Tinitial
∆T = 25.0 oC – 37.0oC = 12.0 oC = -12.0 K
5.50 L 1000 mL 1.11 g = 6050 g
1L
1 mL
q = c x mass x ∆T
q = 2.42 J/g K x 6050 g x -12.0 K
q = - 175,692 J = -176,000 J
-176,000 J 1 kJ
= -176 kJ
1000 J
The Two Major Types of Calorimetry
Calorimeter – a device used to measure the heat released or
absorbed by a physical or chemical process
 Two types…
o One designed to measure the heat at constant pressure
o One designed to measure the heat at constant volume
Constant Pressure Calorimetry
Often measured in a coffee-cup
calorimeter
Can be used to determine the specific
heat capacity of a solid (as long as it
does not react with or dissolve in the
water)
The solid (system) is weighed, heat
to some known temperature, and added to a known mass
and temperature of water (surroundings) in the calorimeter.
After stirring, the final water temperature is measured, which is
also the temperature of the solid. Assuming no heat escapes the
calorimeter, the heat released by the system (-qsystem or –qsolid) is
equal in magnitude but opposite in sign to the heat absorbed by
the surroundings (+qsurroundings or +qH O)
2
So… -qsolid = qH O
2
Substituting for q on both sides of the equation…
-(csolid x masssolid x ∆Tsolid) = cH O x massH O x ∆TH O
2
2
2
Solving for csolid
csolid = - (cH O x massH O x ∆TH O / masssolid x ∆Tsolid)
2
2
2
Sample Problem 6.4 p. 240
Determining the Specific Heat Capacity of a Solid
Finding ∆Tsolid and ∆Twater
∆Twater = Tfinal - Tinitial
∆Twater = 28.49oC – 25.10oC = 3.39oC = 3.39 K
∆Tsolid = Tfinal - Tinitial
∆Tsolid = 28.49oC – 100.00oC = - 71.51oC = - 71.51K
Solving for csolid
-(csolid x masssolid x ∆Tsolid) = cH O x massH O x ∆TH O
2
2
2
csolid = - (cH O x massH O x ∆TH O / masssolid x ∆Tsolid)
2
2
2
csolid = - (4.184 J/g K x 50.00 g x 3.39 K / 22.05 g x – 71.51K)
csolid = 0.450 J/g K
Follow-Up Problem 6.4 p. 240
Finding ∆Tsolid and ∆Twater
∆Twater = Tfinal - Tinitial
∆Twater = 27.25oC – 25.55oC = 1.70oC = 1.70 K
∆Tsolid = Tfinal - Tinitial
∆Tsolid = 27.25oC – 65.00oC = - 37.75oC = - 37.75K
Solving for csolid
-(csolid x masssolid x ∆Tsolid) = cH O x massH O x ∆TH O
2
2
2
csolid = - (cH O x massH O x ∆TH O / masssolid x ∆Tsolid)
2
2
2
csolid = - (4.184 J/g K x 25.00 g x 1.70 K / 12.18 g x – 37.75K)
csolid = 0.387 J/g K [unknown is Copper, Cu]
Sample Problem 6.5 p. 240
Determining the Enthalpy
(a)
Finding masssolution and ∆Tsolution
masssolution = (25.0 mL + 75.0 mL) x 1.00 g/mL = 75.0 g
∆Tsolution = 27.21oC – 25.00oC = 2.21oC = 2.21K
qsolution = csolution x masssolution x ∆Tsolution
qsolution = 4.184 J/g K x 75.0 g x 2.21K
qsolution = 693 J
(b)
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
25.0 mL HCl 1 L
0.500 mol HCl
1000 mL 1L
= 0.0125 mol HCl
50.0 ml NaOH 1L
= 0.0250 mol NaOH
1000
mL
0.500 mol
NaOH
1L
**HCl is the limiting reactant, so 0.0125 mol HCl is produced
Heat absorbed by the solution was released by the reaction
qsoln = -qrxn = 693 J so qrxn = -693 J
-693 J
1 kJ
= -55.4 kJ / mol H2O
0.0125 mol H2O 1000 J
Follow-Up Problem 6.5 p. 241
Step 1
Calculate moles of reactants
50.0 mL 1L
0.500 mol Ba(OH)2 = 0.025 mol Ba(OH)2
1000 mL 1L
50.0 mL 1L
0.500 mol HCl = 0.025 mol HCl
1000 mL 1L
Step 2
Determine the limiting reactant…
0.025 mol Ba(OH)2 2 mol H2O
= 0.500 mol H2O
1 mol Ba(OH)2
0.025 mol HCl 2 mol H2O = 0.025 mol H2O
2 mol HCl
**HCl is the limiting reactant (produces 0.025 mol H2O)
- 1.386 kJ / 0.0250 mol H2O = - 55.4 kJ/mol
Constant-Volume Calorimetry
Constant-volume Calorimetry
is often carried out in a bomb
calorimeter (a device
commonly used to measure the
heat of combustion reactions,
such as for fuels and foods)
With the much more precise
bomb calorimeter, the heat
capacity of the entire
calorimeter is known (or can be determined)
Sample Problem 6.6 p. 242
Calculating the Heat of a Combustion Reaction
Step 1
Calculate ∆T
∆T = ∆Tfinal - ∆Tinitial = 26.799oC – 21.862oC = 4.937oC = 4.937K
Step 2
Calculate the heat absorbed by the calorimeter
 When the dessert (system) burns, the heat released is
absorbed by the calorimeter
 -qsystem = qcalorimeter
qcalorimeter = heat capacity x ∆T
qcalorimeter = 8.151 kJ/K x 4.937 K
qcalorimeter = 40.24 kJ
Step 3
Compare answer to 10 Calories
10 Calories 1kcal
4.184 kJ
1 Calorie 1 kcal
=41.84 kJ
40.24 kJ < 41.84 kJ The claim is correct…
Follow-Up Problem 6.6 p. 242
Step1
Calculate the amount of heat released (qsample)
0.8650 g C
1 mol C
393.5 kJ
12.01 g C 1 mol C
= 28.34 kJ
Step 2
Calculate heat capacity of bomb calorimeter
-qsample = qcalorimeter
qcalorimeter = heat capacity x ∆T
28.34 kJ = heat capacity x 2.613 K
Heat capacity = 10.85 kJ/K
Stoichiometry of Thermochemical Equations
Thermochemical equation – a balanced equation that includes
the enthalpy change of the reaction (∆H)
 ∆H refers only to the amounts (mol) of substances and their
states of matter in that equation
 The sign of ∆H depends on whether the reaction is
exothermic (-) or endothermic (+)
o A forward reaction has the opposite sign of the reverse
reaction
 2H2O(l)  2H2(g) + O2(g) ∆H = 572 kJ
 Because ∆H is positive, this reaction is
endothermic
 2H2(g) + O2(g)  2H2O(l) ∆H = -572 kJ
 Because ∆H is negative, this reaction is
exothermic
 The magnitude of ∆H is proportional to the amount of
substance
o H2(g) + ½ O2(g)  H2O(l) ∆H = -286 kJ
 Notice that this is ½ the above reaction
Two key points to understand about thermochemical
equations…
 When necessary, we use fractional coefficients to balance
the equation, because we are specifying the magnitude of
∆H for a particular amount, often 1 mol of substance
o ⅛ S8(s) + O2(g)  SO2(g)
∆H = -296.8 kJ
 For a particular reaction, a certain amount of substance is
thermochemically equivalent to a certain quantity of
energy. For example…
o 296.8 kJ is thermochemically equivalent to ⅛ mol of
S8
o 286 kJ is thermochemically equivalent to ½ mol of O2
o 286 kJ is thermochemically equivalent to 1 mol H2O
Sample Problem 6.7 p. 243
Using the Enthalpy Change of a Reaction to Find
Amounts of Substance
1.000x103 kJ 2 mol Al 26.98 Al = 32.20 g Al
1676 kJ 1 mol Al
Follow-Up Problem 6.7 p. 243
C2H4(g) + H2(g)  C2H6(g) ∆H = -137 kJ
15 kg C2H6 1000 g 1 mol C2H6 137 kJ
= 6.83x104 kJ
1 kg
30.07 g C2H6 1 mol C2H6
Hess’s Law: Finding ∆H of Any Reaction
In some cases, a reaction is difficult, even impossible, to carry
out…
 May be part of a complex biochemical process
 May take place under extreme conditions
 May need a change of conditions to take place
Even if we can’t run a reaction in the lab, we can still find its
enthalpy change…
 We can find the ∆H for any reaction for which we can write
an equation
Hess’s law – the enthalpy change of an overall process is the
sum of the enthalpy changes of the individual steps
∆Hoverall = ∆H1 + ∆H2 + …. + ∆Hn
We apply Hess’s Law by…
 Imagining that an overall reaction occurs through a series
of individual reaction steps, whether or not it actually does.
Adding the steps must give the overall reaction.
 Choosing individual reaction steps that each have a known
∆H
 Adding the known ∆H values for the steps to get the
unknown ∆H of the overall reaction. We can also find the
unknown ∆H of any step by subtraction (if we know the
∆H values for the overall reaction and all the other steps.
Example of the use of Hess’s law – the oxidation of sulfur to
sulfur trioxide
Equation 1: S(s) + O2(g)  SO2(g)
∆H1 = -296.8 kJ
Equation 2: 2SO2(g) + O2(g)  2SO3(g)
∆H2 = -198.4 kJ
Equation 3: S(s) + 3/2 O2(g)  SO3(g) ∆H3 = ????
Need to manipulate equations 1 and/or 2 so that they add up to
equation 3
 We need to have the same coefficients for SO2 so that they
will cancel (this will mean that we have to divide equation
2 by 2)
 Equation 2: SO2(g) + ½ O2(g)  SO3(g) ∆H2 = -99.2 kJ
Add equations 1 and 2 together…
Equation 1: S(s) + O2(g)  SO2(g)
∆H1 = -296.8 kJ
Equation 2: SO2(g) + ½ O2(g)  SO3(g)
∆H2 = -99.2 kJ
Equation 3: S(s) + O2(g) + SO2(g) + ½ O2(g)  SO2(g) + SO3(g)
∆H1 = -396.0 kJ
 Delete SO2 because it is common to both sides
 Add O2 together on reaction side
Final answer…
S(s) + 3/2 O2(g)  SO3(g)
∆H1 = -396.0 kJ
Summarize calculating an unknown ∆H
1. Identify the target equation, the step whose ∆H is unknown,
and note the amount (mol) of each reactant and product
2. Manipulate each equation with known ∆H values so that
the target amount (mol) of each substance is on the correct
side of the equation. Remember to:
a. Change the sign of ∆H when you reverse an equation
b. Multiply amount (mol) and ∆H by the same factor
Sample Problem 6.8
Using Hess’s Law to Calculate an Unknown ∆H
Equation A:
CO(g) + ½ O(g)  CO2(g)
∆H = -283.0 kJ
Equation B:
N2(g) + O2(g)  2NO(g)
∆H = 180.6 kJ
Equation C:
CO(g) + NO(g)  CO2(g) + ½ N2(g)
∆H = ?
Reverse equation B so that N2 is on the correct side of equation
and divide by 2 so that you have the correct number of mol of
N2 , O2 and NO
Equation B:
NO(g) 1/2 N2(g) + ½ O2(g)
∆H = -90.3 kJ
Add equations A & B together…
Equation A:
CO(g) + ½ O(g)  CO2(g)
Equation B:
NO(g) 1/2 N2(g) + ½ O2(g) ∆H = -90.3 kJ
∆H = -283.0 kJ
Equation C: CO(g) + ½ O(g) + NO(g) 
CO2(g) + 1/2 N2(g) + ½ O2(g) ∆H = -373.3 kJ
Equation C:
CO(g) + NO(g)  CO2(g) + 1/2 N2(g)
∆H = -373.3 kJ
Follow-Up Problem 6.8 p. 245
Equation 1: N2O5(s)  2NO(g) + 3/2 O2(g) ∆H = 223.7 kJ
Equation 2: NO(g) + ½ O2(g)  NO2 (g)
∆H = -57.1 kJ
Equation 3: 2NO2(g) + ½ O2(g)  N2O5(s) ∆H = ???
Reverse equation 1…
Equation 1: 2NO(g) + 3/2 O2(g)  N2O5(s) ∆H = -223.7 kJ
Multiply equation 2 x 2 and reverse…
Equation 2: 2NO2(g)  2NO (g) + O2(g)
∆H = 114.2 kJ
Equation 3: 2NO(g) + 3/2 O2(g) + 2NO2(g) 
2NO (g) + O2(g) + N2O5(s)
∆H = -109.5 kJ
Equation 3: 2NO2(g) + ½ O2(g)  N2O5(s) ∆H = -109.5 kJ
Standard Enthalpies of Reaction - ∆H orxn
We can use Hess’s Law to determine the ∆H values of an
enormous number of reactions…
 ∆H varies somewhat with conditions… so chemist have
established a set of specific conditions called standard
states
o For a gas, it is 1 atm and ideal behavior
o For a substance that aqueous solution, it is 1 M
concentration
o For a pure substance (element or compound), it is
usually the most stable form of the substance at 1 atm
and the temperature of interest (usually 25oC or 298K)
 The standard state symbol (shown as a degree sign)
indicates that the variable has been measured with all the
substances in their standard states
o Standard enthalpy of formation, ∆H of is the
enthalpy change if measured at the standard state
 Also called the standard heat of formation
o C (graphite) + 2H2(g)  CH4(g) ∆H of = -74.9 kJ
o Fractional coefficients are often used with reactants to
obtain 1 mol of product
 Na(s) + ½ Cl2(g)  NaCl(s) ∆H of = -411.1 kJ
o Table 6.3 p. 246 shows selected standard enthalpy of
formations at 298K
 For an element in its standard state, ∆H of = 0
 The standard state for molecular elements, such
as the halogens, is the molecular form, not
separate atoms. An example would be…
Cl2 = ∆H of = 0
 Some elements exist in different forms, but only
one is the standard state…
 Carbon ∆H of = 0, not diamond ∆H of = 1.9
kj/mol
 Oxygen ∆H of = 0, not ozone ∆H of = 143
kJ/mol
 Sulfur (S8) ∆H of = 0, not S ∆H of = 0.3
kJ/mol
 Most compounds have a negative ∆H of
 Most compounds have exothermic formation
reactions (under standard conditions, heat is
released when most compounds form from
their elements)
Sample Problem 6.9 p. 246
Writing Formation Equations
(a)
Ag(s) + ½ Cl2(g)  AgCl(s)
∆H of = -127.0 kJ
(b)
Ca(s) + C(graphite) + 3/2 O2 CaCO3(s)
∆H of = -1206.9 kJ
(c)
½ H2(g) + C(graphite) + ½ O2(g) 
HCN(g) ∆H of = 135.0 kJ
Follow-Up Problem 6.9 p. 247
(a)
C(graphite) + H2(g) + ½ O2(g)  CH3OH(l) ∆H of = -238.6 kJ
(b)
Ca(s) + ½ O2(g)  CaO(s)
∆H of = -635.1 kJ
(c)
C(graphite) + 1/4S8(rhombic)  CS2
∆H of = 87.9 kJ
Determining ∆H orxn from ∆H of Values for Reactants & Products
We can use ∆H of values to determine ∆H orxn for any reaction
Step 1
Each reactant decomposes to its elements. This is the reverse of
the formation reaction for the reactant, so the standard enthalpy
change is -∆H of
Step 2
Each product forms from its elements. This step is the
formation reaction for the product, so the standard enthalpy
change is ∆H of
Text example… p. 247
TiCl4(l) + 2H2O(g)  TiO2(s) + 4HCl(g)
TiCl4(l)  Ti(s) + 2Cl2(g)
-∆H of [TiCl4(l)]
2H2O(g)  2H2(g) + O2(g)
-2∆H of [H2O(g)]
Ti(s) + O2(g)  TiO2(s)
∆H of [TiO2(s)]
2H2(g) + 2Cl2(g)  4HCl(g)
4∆H of [HCl(g)]
TiCl4(l) + 2H2O(g) + Ti(s) + O2(g) + 2H2(g) + 2Cl2(g) 
Ti(s) + 2Cl2(g) + 2H2(g) + O2(g) + TiO2(s) + 4HCl(g)
TiCl4(l) + 2H2O(g)  TiO2(s) + 4HCl(g)
It is important to realize that when titanium (IV) chloride and
water react, the reactions don’t actually decompose to their
elements, which then recombine to form the products. ∆H orxn is
the difference between two state functions, H oproducts minus
H oreactants, so it doesn’t matter how the reaction actually occurs.
We simply need to add the individual enthalpy changes to find
∆H orxn
∆H orxn = ∆H of [TiO2(s)] + 4∆H of [HCl(g)] + {-∆H of [TiCl4(l)]
+ {-2∆H of [H2O(g)]}
OR
∆H orxn = {∆H of [TiO2(s)] + 4∆H of [HCl(g)]} {∆H of [TiCl4(l)] + 2∆H of [H2O(g)]}
The standard enthalpy of reaction (∆H orxn) is the sum of the
standard enthalpies of formation of the products minus the sum
of the standard enthalpies of formation of the reactants
∆H orxn = ∑m∆H of(products) - ∑n∆H of(reactants)
Sample Problem 6.10 p. 248
Calculating ∆H orxn from ∆H of Values
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
∆H orxn = ∑m∆H of(products) - ∑n∆H of(reactants)
∆H orxn = {4∆H of [NO(g)] + 6∆H of [H2O(g)]} –
{4∆H of [NH3(g)] + 5∆H of [O2(g)]}
∆H orxn = {(4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol)} –
{(4 mol)(-45.9 kJ/mol) + (6mol)(0 kJ/mol)
∆H orxn = [361 kJ + (-1451 kJ)] – [(-184 kJ) + 0 kJ]
∆H orxn = - 906 kJ
Follow-Up Problem 6.10 p. 248
CH3OH(l) + 3/2O2(g)  CO2(g) + 2H2O(g) ∆H orxn = -683.5 kJ
∆H orxn = ∑m∆H of(products) - ∑n∆H of(reactants)
-683.5 kJ = {(1 mol)(-393.5 kJ) + (2 mol)(-241.8) –
(3/2 mol)(0.0 kJ) + ∆H of [CH3OH(l)]
∆H of [CH3OH(l)] = 638.5 kJ + (-393.5 kJ) + (-483.6)
∆H of [CH3OH(l)] = -238.6 kJ