Thermochemistry:

Energy Flow
and
Chemical
Reactions
 The Big Question

How are the energies involved in a

chemical reaction determined
experimentally and theoretically?
 Outline

• thermodynamics
• internal energy – definition, first law
• enthalpy – definition, energy diagrams, calorimetry,
theoretical calculation (heats of formation and
Hess’ law), stoichiometry
•applications
 Thermodynamics

Thermodynamics is
the study of heat and
its transformations.

Thermochemistry is a branch of thermodynamics

that deals with the heat involved with chemical
and physical changes.
 Internal Energy

What is the difference of the two pictures above?

Energy state

If we go from cold water to boiling water as our reaction:

H2O (solid) H2O (gas)

energy state 1 energy state 2

 Internal Energy
we define E = a variable to denote the energy of our
reacting substances
= specifically, the internal energy of the substance
undergoing reaction

∆ E = change in energy
= E (boiling water) – E (cold water)

since Energy (gas) > Energy (solid)

∆ E = positive , +

For the reaction to proceed, energy has to be supplied.

But where will it come from?
 Internal Energy
A chemical system and its surroundings.

the surroundings

the system

System = the reaction itself, the object of our study

Surroundings = the rest of the universe
Internal Energy
 Internal Energy
Energy diagrams for the transfer of internal energy (E) between a
system and its surroundings.

DE = Efinal - Einitial = Eproducts - Ereactants

Energy changes occurs because of the exchange of

energy between the system and the surroundings!
 Internal Energy
Notice that all the equations/energies refer to the system!
But we can also define it from the point of view of the
surroundings.

Case 1: surroundings supplied energy to the system

∆ E = E2surr - E1 surr = negative

We have to specify which we are referring to: system or

surroundings
E 1, sys E 2, sys
State 1: State 2:
E 1, surr E 2, surr

E 2 sys - E1 sys = ∆ E system = positive, +

E 2 surr - E1 surr = ∆ E surroundings = negative, -
 Internal Energy
So we have a discrepancy of notations and definitions.

As per convention, whenever we have E1, E2, etc., we are

always referring to the system. Henceforth, we are always
referring to the system in all our studies of energies.

Notice, however that

∆ E system + ∆ E surr = 0
since: system + surroundings = universe

∆E universe = 0 = E2 universe - E1 universe = 0

E2 universe = E1 universe

The energy of the universe is constant!

 Internal Energy

First Law of Thermodynamics:

The energy of the universe is

constant.

Thus, energy can neither be

created nor destroyed. It can
be only be converted from
one form to another.
 Internal Energy
Let’s define formally E = internal energy
= total energy of the system
= summation of all the kinetic and potential
energies of the system

Experimentally: ∆E =q+w
q = heat = energy transferred from a hotter object to a colder one
w = work = energy used to cause an object to move against a force

That means, we can change the energy of the system by

applying heat and doing work. Or,
When energy is transferred from one object to another,
it appears as work and/or as heat.
Internal Energy

A system transferring energy as heat only.

 Internal Energy

A system losing energy as work only.

Zn(s) + 2H+(aq) + 2Cl-(aq)

Energy, E
work done on
DE<0 surroundings

H2(g) + Zn2+(aq) + 2Cl-(aq)

 Internal Energy
The Sign Conventions* for q, w and DE

q + w = DE

+ + +

+ - depends on sizes of q and w

- + depends on sizes of q and w

- - -
* For q: + means system gains heat; - means system loses heat.

* For w: + means work done on system; - means work done by system.

 Internal Energy

DE = (sign), (number), (unit)

Sign = gives direction as to the transfer of energy

+ = E final > E initial = system gained energy from the surroundings
- =E final < E initial = system lost energy to the surroundings
 Internal Energy

Units of Energy

Joule (J) 1 J = 1 kg*m2/s2

calorie (cal) 1 cal = 4.184 J

British Thermal Unit 1 Btu = 1055 J

Calorie (Cal) 1 Calorie = 1 000 calorie

 Sample Problem
Hydrogen and oxygen reacts in a cylinder. As the reaction
occurs, the system loses 1150 J of heat to the
surroundings and the expanding gas product does 480 J of
work on the surroundings as it pushes against the
atmosphere. What is the change in the internal energy of
the system?

q = heat = - 1150 J
w = work = - 480 J

∆E = q + w
= (-1150 J) + (- 480 J)
= - 1630 J
 Sample Problem

A system receives 425 J of heat and delivers 400 J of

work to its surroundings. What is the change in internal
energy of the system (in joules)?

Answer: ∆E =q+w
= (+ 425) + (- 400 J)
= + 25 J
Worksheet #1-1
 Some interesting
quantities of energy.
 Enthalpy
Two different paths for the energy change of a system.
Enthalpy

Pressure-volume
work.
 Enthalpy
There is another energy variable we call enthalpy:

H = E + PV
PV = work
∆H = ∆E + P ∆V

Formal definition: Enthalpy is the thermodynamic quantity

that is the sum of internal energy and the product of the
pressure-volume work.

The change in Enthalpy equals the heat gained or lost at

constant pressure.

Most chemical reactions occur at constant pressure, so ∆H

is more relevant than ∆E.
 Enthalpy

DH ≈ DE in
w = - PDV
1. Reactions that do not
H = E + PV involve gases.
where H is enthalpy
2. Reactions in which the
DH = DE + PDV number of moles of gas does
not change.

qp = DE + PDV = DH 3. Reactions in which the

number of moles of gas does
change but q is >>> PDV.
 Enthalpy
Let’s examine enthalpy and internal energy more closely:
We defined ∆E = E2 – E1

We can calculate ∆E from the first law of thermodynamics.

What about E2, E1 or the exact energy value of a system?

E, internal energy = summation of all the energies of the

system, including KE and PE

For a number of particles (1, 2, 3 particles) energy can be

calculated, but if we are dealing with macroscopic amounts
(ex: a mole of water, 6.02 x 1023 particles), the exact value is
incalculable and impossible!

But is it important?
 Enthalpy
We do not need to calculate the energy values exactly. But
we assume that the energy of a system at a defined
condition is exact, fixed and unchanging.
Example:
1.0 mol of 2.5 L water at 90 C, 1.5 atm pressure has an
internal energy E and an enthalpy H whenever this condition
is specified

Mathematically,
we call ∆E and ∆H = state functions

state function = property of a system that is determined by

specifying its conditions or its state (ex. T, P, etc.)
= the value of a state function does not depend on the
particular history of the sample, only on its present condition
 Enthalpy
State function analogy: scaling a mountain

The height of the mountain is a state function. However, the

way you can climb the mountain is not.

H, E = state functions
q, w = not state functions, path dependent (value depends
on how the process took place)
 Enthalpy
Exothermic Reactions: ∆H = H final – H initial = negative, -
energy is released by the system

Endothermic reactions: ∆H = H final – H initial = positive, +

energy is absorbed by the system

The sign of enthalpy (like internal energy) indicates the

direction of heat transfer during a process that occurs at
constant pressure.
 Enthalpy
Enthalpy diagrams for exothermic and endothermic processes.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g)

CH4 + 2O2 H2O(g)

Hinitial Hfinal
Enthalpy, H

Enthalpy, H
DH < 0 heat out DH > 0 heat in

CO2 + 2H2O H2O(l)

Hfinal Hinitial

A Exothermic process B Endothermic process

 Enthalpy
It turns out experimentally that:
∆E = energy transferred at constant volume conditions
∆H = energy transferred at constant pressure conditions

In most chemical reactions, we are mostly concerned with

enthalpy. As most chemical reactions occur at constant
pressure.

Thermochemical equations

2H2 (gas)+ O2 (gas)→ 2H2O (gas) ∆H = - 483.6 kJ

The enthalpy value is given at the end of the chemical

equation.
 Enthalpy
Some Important Types of Enthalpy Change
heat of combustion (DHcomb)

C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g)

heat of formation (DHf)

K(s) + 1/2Br2(l) KBr(s)

heat of fusion (DHfus)

NaCl(s) NaCl(l)

heat of vaporization (DHvap)

C6H6(l) C6H6(g)
 Sample Problem
Write a balanced equation and draw an approximate
enthalpy diagram for each of the following:
a) Combustion of one mol of methane in oxygen
b) The evaporation of liquid water
Answer:
a) CH4 + 2O2 → CO2 + 2H2O, exothermic
b) H2O(l) → H2O(g), endothermic
CH4 + 2O2 H2O(g)
Enthalpy, H

heat out Enthalpy, H heat in

CO2 + 2H2O H2O(l)

Worksheet #1-2
 Enthalpy
Properties of enthalpy:
1. Enthalpy is an extensive property. Thus its magnitude is
directly proportional to the amount of substance reacted
or produced.

Implication: in a thermochemical equation, if the equation is

multiplied by 2, the ∆H is also multiplied by 2. If the
equation is divided by 2, the ∆H is also divided by two.

Example:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ∆H = - 890 kJ
2CH4(g) + 4O2 (g) → 2CO2 (g) + 4H2O(l) ∆H = 2(- 890) kJ
 Enthalpy
2. ∆H reaction = - ∆H reverse reaction

CO2 (g) + 2H2O (l) → CH2 (g) + 2O2 (g) ∆H = 890 kJ

3. ∆H is dependent on the state of the chemical species used

in the equation. The state of the substance has to be
shown.

Example:

CO2 (g) + 2H2O (l) → CH4 (g) + 2O2 (g) ∆H = 890 kJ

CO2 (g) + 2H2O (g) → CH4 (g) + 2O2 (g) ∆H = 978 kJ
 Enthalpy
Review of chemical stoichometry

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ∆H = - 890 kJ

• How many moles of CH4 will react with 5.0 moles of O2?
(answer = 2.5 moles CH4)
• How many moles of CO2 will be produced from the
reaction of 2.8 moles CH4 with excess O2?
(2.8 moles)
• How many grams of CO2 will be produced from the
reaction of 15.7 grams of CH4 with excess O2?
(43.2 grams)
• How much energy will be produced from the reaction of
15.7 grams of CH4 with excess O2?
 Enthalpy
Enthalpy and chemical stoichometry
-The calculation of the enthalpy for a particular reaction and
its relation to the chemical equation is governed the
extensive property of enthalpy.

Example:

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ∆H = - 890 kJ

relationships: 1 mol CH4 is to ∆H = - 890 kJ

2 mol O2 is to ∆H = - 890 kJ
1 mol CO2 is to ∆H = - 890 kJ
2 mol H2O is to ∆H = - 890 k
 Sample Problem
Consider the equation:

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ∆H = - 890 kJ

1. How many moles of methane, CH4 (g) has to be burned

to generate 1500 kJ of energy?

2. How many grams of carbon dioxide is produced if the

enthalpy generated in the combustion is 1830 kJ?
 Sample Problem
Consider the following balanced thermochemical equation
for a reaction sometimes used for H2S production:

1/8 S8 + H2 → H2S ΔH = - 20.2 kJ

a) Is this an exothermic or endothermic reaction?

b) What is the ΔHrxn for the reverse reaction?
c) What is ΔH when 3.2 mol of S8 reacts?
d) What is ΔH when 20.0 g of S8 reacts?

Answers:
a) Exothermic
b) 20.2 kJ
c) -517 kJ
d) -12.6 kJ
 Enthalpy
Summary of the relationship
AMOUNT (mol) between amount (mol) of
of compound A
substance and the heat (kJ)
transferred during a reaction.

AMOUNT (mol)
of compound B
molar ratio from
balanced equation

HEAT (kJ)
DHrxn (kJ/mol) gained or lost
Worksheet #1 - 3
 Enthalpy

Calorimetry = measurement of heat flow

Calorimeter = apparatus that measures heat flow

∆H reaction can be determined experimentally

using a calorimeter.
 Enthalpy

How do you measure

enthalpy (ΔH)
experimentally?
 Enthalpy

Coffee-cup calorimeter.
 Enthalpy
In chemical equations, energy changes can be calculated by
monitoring the temperature changes in the reaction.

q is proportional to ∆T

In equation form: q α ∆T
q = C ∆T

where C = constant of proportionality

= heat capacity
= amount of heat required to raise the temperature of
a substance by 1.0 Kelvin
The larger the heat capacity the greater the heat required to
produce a given rise in temperature.

For 1.0 mol substance , C = C (mol)

 Enthalpy
Lets look at another experiment:

1.0 L water and 10.0 ml water are heated separately such that each would
have its temperature increase by 10.0 C. What would be the result?

The 10.0 ml water sample will have a greater increase in energy as some of
its molecules can turn into gas.

Implication: mass is also a factor in measuring enthalpies.

q = C ∆T
As there is no mass in the equation, it can be in the heat capacity factor.

C=mc
where m = mass of the substance
c = specific heat of the substance

Thus, q = m c ∆T
 Enthalpy

Heat Capacity C = q / ΔT
unit: J/K

Specific heat capacity (c) = q / (mass x ΔT)

unit: J/gK

Molar heat capacity (C mol) = q / (moles x ΔT)

unit: J/mol K
Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials

Substance Specific Heat Substance Specific Heat

Capacity (J/g*K) Capacity (J/g*K)

Elements Materials
aluminum, Al 0.900 wood 1.76
graphite,C 0.711 cement 0.88
iron, Fe 0.450 glass 0.84
copper, Cu 0.387 granite 0.79
gold, Au 0.129 steel 0.45

Compounds
water, H2O(l) 4.184
ethyl alcohol, C2H5OH(l) 2.46
ethylene glycol, (CH2OH)2(l) 2.42
carbon tetrachloride, CCl4(l) 0.864
 Enthalpy

Relation of enthalpy and q (from calorimetry experiments)

q = m c ∆T = q solution

q solution = - q reaction = ∆H reaction

∆H reaction = - m c ∆T
 Sample Problem

A 295 g aluminum engine part at an initial temperature of

3.0 C absorbs 85.0 kJ of heat. What is the final
temperature of the part (specific heat of aluminum = 0.900
J/g K)?

q = m c ∆T
85 000 J = 295 g Al (0.900 J/gK) (Tf – 3.0 C)

Answer: Tf = 323 C
 Sample Problem

When 165 ml of water at 22 C is mixed with 85 mL of water

at 82 C, what is the final temperature? Assume that no
heat is lost to the surroundings (density of water is 1.0
g/mL).

(m c ∆T) water 1 = - (m c ∆T) water 2

165 ml ( Tf – 22 C ) = - 85 ml (Tf – 82 C)

Answer: Tf = 42 C
Worksheet #1 - 4
A bomb calorimeter
 Internal Energy

In a bomb calorimeter, the conditions are

such the volume of the set-up is constant.

Thus, the internal energy, ∆E is the one

obtained.
 Enthalpy of Formation

∆H formation = ∆H f = enthalpy associated with the

formation of a compound from its constituent elements

Examples of formation equations:

C (graphite) + O2 (gas) → CO2 (gas)

2C (graphite) + H2 (gas) → C2H2 (gas)
C (graphite) + 2H2 (gas) + 1/2 O2 (gas) + N2 (gas) → CH4N2O(urea)

The reactants should be elements in their natural form

 Enthalpy of Formation

Write balanced formation equations at standard conditions

for each of the following substances:
a) CaCl2
b) NaHCO3
c) CCl4
d) HNO3

Answer:
a) Ca(s) + Cl2(g) → CaCl2(s)
b) Na(s) + ½ H2(g) + Cgraph + 3/2 O2(g) → NaHCO3(s)
c) Cgraph + 2Cl2(g) → CCl4(l)
d) ½ H2(g) + ½ N2(g) + 3/2 O2(g) → HNO3(l)
 Enthalpy of Formation
∆H = depends of various conditions such as T, P, state of matter, etc., of the
reactants and products

To compare enthalpies, it is convenient to define a set of conditions called a

standard state at which the enthalpies are tabulated.

Standard state = pure form of the substance at 1.0 atmospheric pressure

and at 25 C (298 K)

∆Ho = standard enthalpy = enthalpy at standard conditions of all species

∆Hof = standard enthalpy of formation

= change in enthalpy of reaction that forms 1.0 mol of compound
from its elements with all the substances in their standard states

If the element exists in more than one form under standard conditions, the
most stable form of that element is used for ∆Hof
 Enthalpy of Formation

2C (graphite) + 3H2 (gas) + ½ O2 (gas) → C2H5OH ∆Hof = -277.7 kJ

C (graphite) + O2 (gas) → CO2 (gas) ∆Hof = - 393.5 kJ

2C (graphite) + 3H2 (gas) → C2H6 (gas) ∆Hof = - 84.68 kJ

The standard enthalpy of formation, ∆Hof values are tabulated in books

and can be consulted every time it is needed.
Table 6.5 Selected Standard Heats of Formation at 250C(298K)

Formula DH0f(kJ/mol) Formula DH0f(kJ/mol) Formula DH0f(kJ/mol)

calcium silver
Cl2(g) 0
Ca(s) 0 Ag(s) 0
HCl(g) -92.3
CaO(s) -635.1 AgCl(s) -127.0
CaCO3(s) -1206.9 hydrogen
H(g) 218 sodium
carbon
H2(g) 0 Na(s) 0
C(graphite) 0
C(diamond) 1.9 nitrogen Na(g) 107.8
CO(g) -110.5 N2(g) 0 NaCl(s) -411.1
CO2(g) -393.5 NH3(g) -45.9 sulfur
CH4(g) -74.9 NO(g) 90.3 S8(rhombic) 0
CH3OH(l) -238.6 S8(monoclinic) 2
oxygen
HCN(g) 135 SO2(g) -296.8
O2(g) 0
CSs(l) 87.9
O3(g) 143 SO3(g) -396.0
chlorine H2O(g) -241.8
Cl(g) 121.0 H2O(l) -285.8
 Enthalpy of Formation

Using the tabulated values, we can calculate the

enthalpy of reaction

DH0rxn = S mDH0f(products) - S nDH0f(reactants)

where m = number of moles of each product

n = number of moles of each reactant
 Enthalpy of Formation
The general process for determining DH0rxn from DH0f values.

Elements

decomposition
Enthalpy, H

-DH0f DH0f

formation
Reactants
Hinitial

DH0rxn

Products
Hfinal

DH0rxn = S mDH0f(products) - S nDH0f(reactants)

 Sample Problem

Calculate the enthalpy of reaction of the equation below

using standard enthalpies of formation.

C3H8 (g) + 5 O2 → 3 CO2 (g) + 4 H2O (l)

Solution:
∆Ho reaction = 3∆Hof [CO2(g)] + 4Hof [H2O (l)] - ∆Hof [C3H8 (g)] -
5∆Hof [O2 (g)]
= 3 (-393.5 kJ) + 4 (-285.8 kJ) – (-103.85 kJ) – 5( 0 kJ)
= - 2220 kJ
 Sample Problem

Calculate the enthalpy of reaction of the equation below

using standard enthalpies of formation.

C6H6 (l) + 15/2 O2 → 6 CO2 (g) + 3 H2O (l)

Solution:
∆Ho reaction = 6∆Hof [CO2(g)] + 3Hof [H2O (l)] - ∆Hof [C6H6 (g)] -
15/2 ∆Hof [O2 (g)]
= 6 (-393.5 kJ) + 3 (-285.8 kJ) – (49.0 kJ) – 15/2(0 kJ)
= - 3267.4 kJ
 Sample Problem

Acetylene burns in air according to the following equation:

C2H2 + 5/2 O2 → 2CO2 + H2O ΔH = -1255.8 kJ

Given that ΔHfo of CO2 = - 393.5 kJ/mol and ΔHfo of H2O =

-241.8 kJ/mol, what is the ΔHfo of acetylene?

Solution:

∆Ho reaction = 2∆Hof [CO2(g)] + Hof [H2O (l)] - ∆Hof [C2H2 (g)] -
5/2 ∆Hof [O2 (g)]

∆Hof [C2H2 (g)] = 227 kJ/mol

Worksheet #1 - 5
 Enthalpies of Reaction

Which is a more accurate calculation?

• the one using the enthalpies of formation

(state of substances indicated, position of
atoms indicated, values are not averaged, etc..)

• bond energies are used to only estimate the

enthalpy of reaction and if enthalpy of formation
of some substances are unknown
 Hess Law
Hess’s Law of Heat Summation

= the enthalpy change of an overall process is the sum

of the enthalpy changes of its individual steps
Steps in calculating an unknown ΔH
1. Identify the target equation, the step whose ΔH is unknown and note
the number of moles of reactants and products.
2. Manipulate the equations with known ΔH values so that the target
numbers of moles of reactants and products are on the correct side.
Remember:
- change the sign of ΔH when you reverse an equation
- multiply numbers of moles and ΔH by the same factor
3. Add the manipulated equations to obtain the target equation. Add their
ΔH values to obtain the unknown ΔH .
 Sample Problem

Calculate ΔHrxn for:

Ca + ½ O2 + CO2 → CaCO3

Given the following set of reactions:

Ca + ½ O2 → CaO ΔH = - 635.1 kJ
CaCO3 → CaO + CO2 ΔH = 178.3 kJ

Answer: - 813.4 kJ
 Sample Problem

Calculate ΔHrxn for:

C2H4 + 6F2 → 2CF4 + 4HF

Given the following set of reactions:

H2 + F2 → 2HF ΔH = - 537 kJ
C + 2F2 → CF4 ΔH = - 680 kJ
2C + 2H2 → C2H4 ΔH = + 52.3 kJ

Answer: - 2490 kJ
 Heats of Formation and Hess’s Law

The common lead-acid car battery produces a large burst of

currrent, even at low temperatures, and is rechargeable. The
reaction that occurs while recharging a dead battery is:

2PbSO4(s) + 2H2O(l) → Pb(s) + PbO2(s) + 2H2SO4(l)

a. Use ΔHfo values to calculate ΔHrxn.

b. Use the following equations to check your answer in part (a).

Pb(s) + PbO2(s) + 2SO3(g) → 2PbSO4(s) ΔH = -768 kJ

SO3(g) + H2O(l) → H2SO4(l) ΔH = -132 kJ

Answer: = 2(-813.99) + (-276.6) + 0 - 2(- 918.39) – 2(-285.84)

= 503.88 kJ
Worksheet #1 - 6
 Energy from Foods
Heats of Combustion of Some Fats and Carbohydrates

Substance DHcomb(kJ/g)

Fats
vegetable oil -37.0
margarine -30.1

butter -30.0

Carbohydrates
table sugar (sucrose) -16.2
brown rice -14.9

maple syrup -10.4

 Points to ponder…….

Which is a better source of energy for our body?

Proteins

Carbohydrates

Fats
 Points to ponder…….

How do cold packs and hot packs work?