Thermochemistry
Objective: to examine heat changes that takes place in chemical and physical properties.
Energy is weightless, odorless, and tasteless. How do we detect it? By its effects.
What is energy?
The capacity to do work or to supply heat.
Chemical potential energy — energy stored in chemicals because of their compositions.
What is heat?
Heat is a form of energy that flows from a warm object to a cooler object. When heat is
added, the temperature of the object increases (except during phase changes).
What are some examples of this?
Radiator
Hair drier (book example)
Motors of various devices
Energy can take a number of forms —
Useable energy — work
Stored energy — available for later use
Unusable energy — heat
Thermochemistry - the study of heat changes that occur during chemical reactions and physical changes
of state.
All chemical reactions and changes of state involve absorption or release of heat.
There are several units commonly used for measuring heat. What are they?
o
o
o
calorie — the amount of heat that raises the temperature of 1 g of water 1ºC.
Calorie = 1000 calories (1kcal) used in measuring energy content of food.
Joule — the SI unit of heat and energy. 1 kcal = 4186 J and 1 J = 0.239 cal
The amount of heat it takes to change an object’s temperature by exactly 1ºC is the heat capacity of that object.
Heat capacity depends on; the mass of the object and the chemical composition of the object.
Let’s look at an example:
You’re at the beach standing in 30 kg of water. You decide to go over to your best friend, who is building a 30
kg pyramid of dry sand. Each is absorbing the same amount of radiant energy from the sun that day. Which is
warmer?
The sand has a lower specific heat capacity than the water, thus it is warmer than the water.
Specific Heat (C ) — is the amount of heat it takes to raise the temperature of 1 g of a substance 1ºC. It is
measured in
, where g=mass, J=heat, and Cº= change in temperature!
The equation used is written
heat (J)/ (mass (g) x change in temperature ºC)
Joules or calories can be used to represent specific heat.
The amount of heat absorbed or released can be measured, as can the temperature change.
Metals tend to have low specific heats where non-metals tend to have high specific heats.
Substance
J/(g x ºC)
cal / (g x ºC)
Water
4.18
1.00
Alcohol
2.4
0.58
Ice
2.1
0.50
Wood
1.8
0.42
Steam
1.7
0.40
Chloroform
0.96
0.23
Aluminum
0.92
0.21
Glass
0.50
0.12
Iron
0.46
0.11
Silver
0.24
0.057
Mercury
0.14
0.0033
Exothermic and Endothermic Processes
Thermochemistry is the study of the flow of heat between a chemical reaction and its surroundings.
We can refer to the chemical reaction as a system — the thing you are focusing on. The surroundings include
everything outside the system.
In thermochemical calculations, the direction of the heat flow is given from the point of view of the system.
Endothermic - a process that absorbs heat from the surroundings.
The surroundings cool down in an endothermic reaction.
Exothermic — a process that loses heat to the surroundings.
Using the body as a system, an example of an exothermic process would be sweating. All
animals that maintain a constant body temperature must have a way for the system to lose heat.
How is heat energy measured?
Calorimetry — the accurate and precise measurement of the heat changes for chemical and physical
processes. A calorimeter is a device used to measure the amount of heat absorbed or released during chemical
and physical processes. Often styrofoam cups are used because of their strong insulating properties.
Most chemical reactions take place in the open air, at standard atmospheric pressure. For systems at constant
pressure, heat content = enthalpy (H) of the system.
Heat changes carried out at constant pressure are also enthalpy changes, therefore,
q = ΔH
ΔH is negative for exothermic reactions
ΔH is positive for endothermic reactions
Problems
1. How much heat energy, in joules, is released by 200 g of solid Al as it cools from 200ºC to
150ºC?
Rearranging the heat capacity equation,
q = ΔH = C • m • ΔT
C=0.92 J/g•Cº
m=200 g
ΔT = 150ºC-200ºC = -50ºC
9200 Joules of heat energy is released.
Thermochemical Equations
When a heat term is included in a chemical equation, it becomes a thermochemical equation.
CaO (s) + H2O (l) Ca(OH)2 (s) + 65.2kJ
The heat is given off during the chemical reaction. (notice the heat is on the product side)
The heat of reaction (ΔH) is the heat change for the equation exactly as it is written.
In this case, ΔH = -65.2kJ
ΔH is always negative for an exothermic reaction.
2NaHCO3 (s) + 129 kJ  Na2CO3 (s) + H2O (l) + CO2 (g)
The heat is absorbed during the chemical reaction. (notice the heat is on the reactant side)
 H = +129 kJ
For a typical endothermic reaction, ΔH is always positive:
The physical state of the products and the reactants must be stated in a thermochemcial reaction.
H2O (l)  H2 (g) + 1/2 O2 (g)  H = +285.8 kJ
H2O(g)  H2 (g) + 1/2 O2 (g)  H = +241.8 kJ
At 25 C, dissociation of water requires an extra 44.0 kJ of heat.
Heat of Combustion
The heat of combustion is the heat of reaction for the complete burning of a unit (mole) of a substance.
For methane,
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) + 890.4kJ
Can also be written
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
 H = -890.4 kJ
Heats of combustion are enthalpy changes at 1 atm with their reactants and products in their usual physical
states at 25 C.
Heat and Changes of State
Energy is absorbed or released when matter changes state. When ice melts, it absorbs energy from the
surroundings. As melting proceeds, the liquid and solid water remain at the same temperature until all of the ice
has melted. The heat absorbed by one unit (mole) of a substance when it melts from a solid to a liquid at
constant temperature is the molar heat of fusion,  Hfus. When the opposite occurs, (l)  (s) at constant
temperature, the heat lost is called the molar heat of solidification,  Hsolid.
Because of the law of conservation of energy, - Hsolid =  Hfus
For water, (s) (l)  Hfus = 6.01 kJ/mol
When liquids absorb heat at their boiling point, they become vapors. Heat flows from the surroundings to the
liquid, cooling the surroundings. For (l)  (g), an endothermic process, we have a molar heat of
vaporization,  Hvap. Similarly, for condensation, we have a molar heat of condensation.
For water,  Hvap = 40.7 kJ.
Heat of Solution
Heat changes can occur when a substance is dissolved in a solvent. When one unit (mole) of a substance is
dissolved in a solvent, the heat change is called the molar heat of solution,  Hsoln.
Example: dissolving NaOH in water.
NaOH (s)  Na+(aq) + OH- (aq)
 Hsoln = -445.1kJ/mol
This heat of solution is released when the sodium ions and hydroxide ions separate and interact with water.
Hot packs and cold packs work because of heat of solution.
Hot pack:
CaCl2 (s) + H2O  exothermic reaction
Cold pack:
NH4NO3 (s) + H2O (l)  NH4+ (aq) + NO3- (aq)
An endothermic reaction
 Hsoln=25.7 kJ/mol
Hess’s Law of Constant Heat Summation
In any series of reactions that start with the same reactants and products, the net change in energy must be the
same. In other words, when a reaction can be expressed as the algebraic sum of two or more other reactions, the
heat of reaction is the algebraic sum of the heats of reaction of these other reactions.
For example, for the reaction,
CuO (s) + H2 (g)  Cu (s) + H2O (g)
 H = ?? kcal/mol
This reaction can be expressed as the sum of the following reactions:
CuO (s)  Cu (s) + 1/2 O2 (g)
 H1 = 37.1 kcal
H2 (g) + 1/2 O2  H2O (g)
 H2 = -57.8 kcal
According to Hess’s law, the  H of the original reaction is the sum of the two other reactions.

 H =  H1 +  H2 = 37.1 kcal + (-57.8 kcal) = -20.7 kcal
What kind of reaction is this?
Problem
Apply Hess’ law of constant heat summation to calculate the heat of reaction for the burning of carbon
monoxide:
2CO (g) + O2 (g)  2CO2 (g)
 H = ??
There are two compounds in this equation. The standard heat of formation for each is given below:
C(s) + O2 (g)  CO2 (g)
C (s) + 1/2 O2 (g)  CO (g)
 H = -94.1 kcal/mol
 H = -26.4 kcal/mol
How do you reproduce the original equation?
Reverse the 2nd equation (sign on  H) and multiply by 2.
Multiply first equation by 2
2[C(s) + O2 (g)  CO2 (g)  H = -94.1 kcal/mol]
2[CO  C (s) + 1/2 O2 (g)  H = 26.4 kcal/mol]
2(-94.1 kcal/mol) + 2(26.4 kcal/mol) = -135.4 kcal/mol
The carbons on both sides cancel; one O atom is removed, leaving the original equation:
Standard Heats of Formation
Sometimes it is hard to measure heat changes for chemical reactions.
 materials may not be available
 reactions can be too slow for enthalpy to be measured
 chemicals may be too expensive
In these cases, enthalpy change can be calculated from standard heats of formation.
Standard Heat of Formation ( Hof) of a compound is the change in enthalpy that accompanies the formation of
one mole of a compound from its elements with all substances in their standard states at 25 C.
 Hof of the free element in its standard state is set at zero. This is true for diatomic molecules (HOFBrINCl) as
well as for the other elements.