Thermochemistry
Radiant:
Solar Energy


Thermochemistry is
the study of heat
change in a chemical
reaction.
It is part of a broader
topic called
thermodynamics.

This is the scientific
study of the
interconversion of
heat and other kinds
of energy.
Thermal:
Heat
Energy
Potential:
Stored Energy
Chemical:
Stored in
Chemical
Bonds
Principles of Heat Flow



Heat is a particular form of energy that is transferred from a
body of high temperature to low temperature.
Temperature is the measure of the average kinetic energy in
particles.
When heat flow is discussed we must differentiate between
the system and it’s surroundings:


The system is that part of the universe on which attention is
focused.
The surroundings, exchange energy with the system, make up
the rest of the universe – really only include those materials in
close contact with the system
Energy cannot be created nor destroyed, simply
converted or transferred – 1st Law of Thermodynamics,
where ΔE = q + w
Systems

Open system


Closed system


Can exchange mass and energy, usually in
the form heat, with its surroundings.
Allows the transfer of energy (heat) but not
mass.
Isolated system

Does not allow the transfer of either mass or
energy
Heat (q)

When heat flows from the system to its surroundings then the heat
flow, symbolized by q, is negative. When heat flows into the system
from its surroundings then the heat flow, q, is positive.

An endothermic reaction: q > 0. Why?
2HgO (s)  2Hg (l) + O2(g)
The decomposition of mercury II oxide:

An exothermic reaction: q < 0. Why?
The combustion of methane:

CH4 (g) + 2O2 (g)

CO2 (g)
+
2H2O (l)
We are not interested only in the direction of heat flow but its
magnitude. Heat flow (q) is measured in the unit joules (J) and
kilojoules (kJ). In the past chemists used to use the calorie as an
energy unit.
1cal = 4.184 J


1 kcal = 4,184 J
Be careful! – that calorie is not the same as the dieticians calorie –
the dieticians calorie = 1 kcal.
Heat Flow & Heat Capacity

The relationship between the magnitude
of heat flow and the temperature change
is given by the following equation:
q = C x Δt



q = heat flow, unit: J
C = heat capacity represents the amount of heat
required to raise the temp. of a system 1 *C, unit:
J/*C
Δt = final temperature – initial temperature (tf – ti),
unit: *C
Heat Flow & Specific Heat

Another relationship between the magnitude of heat flow and
the temperature change is given by the following equation
q = m x c x Δt




m = mass, unit – grams
c = specific heat represents the amount of heat required to raise
1 gram of a substance 1*C, unit – J/g *C
specific heat, like density, is an intensive property and can be
used to identify a substance.
The specific heat of water = 4.18 J/g *C
Example 8.1
How much heat is given off when one
mole of liquid water cools from 100.0
*C to 10.0 *C?
Ans: -6.78 x 103 J
Coffee Cup Calorimetry



Coffee cup calorimetry is used in general
chemistry lab to determine the magnitude of
heat (q).
Consists of a polystyrene foam cup partially
filled with a water. It has a tightly fitted lid in
which an accurate thermometer is placed.
Essentially all of the heat evolved by the
reaction taking place within the calorimeter
is absorbed by the water.
So, the heat capacity of the coffee cup
calorimeter is that of the water.
Calorimetry

qreaction = -qcalorimeter

qreaction = -Ccal

qreaction = -mwater
Δt
x
x
cwater x Δt
Bomb Calorimetry
qreaction = -qcalorimeter
qreaction = -Ccal
x
Δt
Example 8.2

When 1.00 gram of calcium chloride is
added to 50.0 grams of water in a
calorimeter, it dissolves and the
temperature rises from 25.00 *C to
28.51 *C. Assuming that all the heat
given off be the reaction is transferred
to the water, calculate q for the
reaction system.

Ans: -734 J
Example 8.3

The reaction between hydrogen and
chlorine can be studied in a bomb
calorimeter. It is found that when a
1.00 g sample of hydrogen reacts
completely, the temperature rises from
20.00 *C to 29.82 *C. Taking the heat
capacity of the calorimeter to be 9.33
kJ/*C. Calculate the amount of heat
evolved.

Ans: -91.6 kJ
Enthalpy (H)

Enthalpy is the measurement of heat (really
energy) stored in chemical bonds. The heat flow
for a reaction system is equal to the difference
in enthalpy (H) between products and reactants


Enthalpy is a state function, depends only on
the state of the system the way the system
reached the state.


keep in mind that this is at a constant pressure
ΔE and ΔH are both state functions unlike q and w.
qreaction =
ΔH = Hproducts
-
Hreactants
Exothermic vs Endothermic
-Exothermic reactions:
CH4 (g) + 2O2 (g)
-Endothermic reactions:
H2O (s)  H2O (l)

CO2 (g)
ΔH > 0
+
2H2O (l)
ΔH < 0
Thermochemical Equations


Equations that show the enthalpy
relationship between products and reactants
is called a thermochemical equation.
For example:
NH4NO3 (s)  NH4+(aq) + NO3-(aq) ΔH = +28.1 kJ
Is the above reaction exothermic or endothermic?
Why?
H2 (g) + Cl2 (g)  2HCl (g)
ΔH = -185 kJ
Is the above reaction exothermic or endothermic?
Why?
Thermochemical Equations
Important Points




The sign of ΔH indicates endothermic or exothermic
reaction.
The coefficients represent numbers of moles (i.e.
ΔH = -185 kJ when there is 1 mole of hydrogen, 1
mole of chlorine, and 2 moles of hydrogen chloride).
The phases of all species must be specified.
Enthalpies will differ for the same substance in a
different state of matter.
The value given for ΔH applies when products and
reactants are at the same temperature, generally 25
*C – unless otherwise stated.
Stoichiometry using Enthalpy

The major source of aluminum in the
world is bauxite (mostly aluminum oxide).
Its thermal decomposition can be
represented by
If aluminum is produced this way, how
many grams of aluminum can form when
1.000 × 103 kJ of heat is transferred?
Steps to using enthalpy:
Example 8.4

Consider the thermochemical
equation for the production of
hydrogen chloride gas from it
foundational elements. In this reaction
ΔH = - 185 kJ. Calculate ΔH when:

1.00 mol of HCl is formed


1.00 g grams of chlorine reacts


Ans: -92.5 kJ
Ans: -2.61 kJ
2.50 L of HCl at 50.0 *C and 725 mmHg
is formed.

Ans: -8.32 kJ
Various Enthalpy Changes

Some enthalpy changes are frequently studied
and, therefore, they have special names:

When 1 mol of a compound is produced from its
elements, the heat of reaction is called the heat of
formation (ΔHf):

When 1 mol of a substance vaporizes, the enthalpy
change is called the heat of vaporization (ΔHvap):

When 1 mol of a substance melts, the enthalpy
change is called the heat of fusion (ΔHfus):
Rules of Thermochemsitry

1. The magnitude of ΔH is directly proportional to the amount of
reactant or product. For example, the more water you have the
more heat it will take to boil it.

2. ΔH for a reaction is equal in magnitude but opposite in sign to
ΔH for the reverse reaction.
H2O (s)  H2O (l)
ΔH = +6.00 kJ
Based on the above reaction how much heat is evolved (or
released) when liquid water freezes?

3. The value of ΔH for a reaction is the same whether it occurs in
one step or in a series of steps. This is known as Hess’s law and is
conveniently used for reactions that are difficult to carry out in a
calorimeter:
ΔH = ΔHequation 1 + ΔHequation 2 + …..
Example 8.5
Given the equation:
2H2 (g) + O2 (g)  2H2O (l) ΔH = -571.6 kJ
Calculate the ΔH for the equation:
H2O (l)  H2 (g) + ½ O2 (g)


Ans: +285.8 kJ
Hess’s Law
First, we identify Equation 3 as our “target” equation, the one whose ΔH we want
to find, and we carefully note the number of moles of each reactant and product in
it. We also note that ΔH1 and ΔH2 are the values for Equations 1 and 2 as written.
Now we manipulate Equations 1 and/or 2 as follows to make them add up to
Equation 3:
•Equations 1 and 3 contain the same amount of S, so we leave Equation 1
unchanged.
•Equation 2 has twice as much SO3 as Equation 3, so we multiply it by 1/2, being
sure to halve ΔH2 as well.
•With the targeted amounts of reactants and products now present, we add
Equation 1 to the halved Equation 2 and cancel terms that appear on both
sides:
To Summarize Hess’s Law
Calculating an unknown ΔH basically involves three steps:
1.
Identify the target equation, the step whose ΔH is unknown, and
note the number of moles of each reactant and product.
2. Manipulate the equations with known ΔH values so that the target
numbers of moles of reactants and products are on the correct
sides. Remember to:
•
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. All
substances except those in the target equation must cancel. Add
their ΔH values to obtain the unknown ΔH.
Example 8.6
Given:
1. C (s) + O2 (g)  CO2 (g) ΔH = -393.5 kJ
2. 2CO (g) + O2 (g)  2CO2 (g) ΔH = -566.0 kJ
Calculate the enthalpy for:
C (s) + ½ O2 (g)  CO (g) ΔH = ?


Ans: -110.5 kJ
Finding ΔH from ΔHf

Enthalpies of formation may be used to
calculate the ΔH for the reaction. To do this
apply this rule:


The standard enthalpy change for a given
thermochemical equation is equal to the sum of
the standard enthalpies of formation of the
product compounds minus the sum of the
standard enthalpies of formation of the reactant
compounds.
ΔH = ∑ ΔHf products - ∑ ΔHf reactants
Using Heats of Formation
When applying this equation, elements in their standard
states can be omitted because their heats of formation
equal zero. So for the following reaction:
2Al (s) + Fe2O3 (s)  2Fe (s) + Al2O3 (s)
We can write:
ΔH = ∑ ΔHf products - ∑ ΔHf reactants =
ΔHf Al2O3 (s) - ΔHf Fe2O3 (s)

The coefficients of the products and reactants must also
be taken into account. So for the following reaction:
2Al (s) + 3Cu+2 (aq)  2Al+3 (aq) + 3Cu (s)
We can write:
ΔH = 2 ΔHf Al+3 (aq) - 3 ΔHf Cu+2 (aq)

Example 8.7

Using table 8.3 calculate (please refer
to page 208 for balanced equations of
the following reactions):

ΔH for the combustion of methane


The ΔHf of benzene using the equation
for the combustion of benzene when ΔH
= -3267.4 kJ


-2219.9 kJ
49.0 kJ/mol
The ΔH for the combustion of methane.

-3169.o kJ
Example 8.8

The following equation represents the
reaction between hydrochloric acid and a
solution of sodium carbonate:
2H+ (aq) + CO32- (aq)  CO2 (g) + H2O (l)
Calculate ΔH for this thermochemical equation.
 Ans: -2.2 kJ
Bond Enthalpy

Bond enthalpy is defined as ΔH when one mole of bonds is broken in
the gaseous state.
H2 (g)


2H (g)
ΔH = +436 kJ
The above equation means that the ΔHf for 1 mole of hydrogen =
218 kJ
Using bond enthalpies it is possible to explain why gas phase
reactions are endothermic. Generally, a reaction is expected to be
endothermic if:




The bonds in the reactants are stronger than those in the products.
There are more bonds in the reactants than in the products.
-Generally enthalpies of formation are used rather than bond
enthalpies because they are more accurate (within + or – 0.1 kJ).
1st Law of Thermodynamics



Energy can be converted from one form to another
but cannot be created nor destroyed.
The change in energy is always energy transfer
from system to surroundings, or vice versa.
ΔEsystem = -ΔEsurroundings
The total change in a system’s internal energy is
the sum of the energy transferred as heat and/or
work:
ΔE = q + w
Example 8.9

Calculate ΔE of a gas for a process in
which the gas:

Absorbs 20 J of heat and does 12 J of
work by expanding.


Ans: +8 J
Evolves 30 J of heat and has 52 J of
work done on it as it contracts.

Ans: +22 J
Enthalpy vs Energy

ΔH = ΔE + ΔngRT



R = 8.31 J/ mol K
Δng = n gaseous products – n of gaseous reactants
This equation is derived from the
following, keeping in mind that work for
a gas is is PV:
H = E + PV
* In this relationship H is really the q in the
first law and PV is the w in the first law.
Example 8.10
Calculate ΔH and ΔE at 25 *C for the reaction
that takes place when an oxyacetylene torch is
used:
C2H2 (g) + 5/2 O2 (g)  2CO2 (g) + H2O (g)


Ans: ΔH = -1255.5 kJ and ΔE = -1254.3 kJ