Section 1: Introduction (background information)
Thermochemistry describes the changes in energy that take place in a chemical process. Ex. dissolving a solid, hydrocarbon combustion, melting a substance.
· the ability to do work
· sources: chemical, solar, nuclear, geothermal
· heat and electricity are forms of energy
· energy can be converted from one form to another: Ex. nuclear → heat → steam → electricity
· measured in Joules (SI unit) or in calories (old unit)
Two main types of energy:
· Kinetic energy - energy of motion
· Potiential energy - stored energy
Kinetic Molecular Theory (Main points)
· All matter is made up of tiny particles (atoms, molecules, ions).
· The particles are in constant motion (have kinetic energy).
· The particles of matter have spaces between them. The speed and spacing of the particles determine the state of matter.
· Adding heat (energy) increases the speed of the particles (increase in kinetic energy), as well as the space they occupy.
Temperature is a measure of the average kinetic energy of a substance. The higher the temperature, the higher the kinetic energy, and the larger space between the particles. A change in temperature is shown with the symbol ΔT.
ΔT = Tfinal - Tinitial
Heat is a form of energy that transfers from a system of higher kinetic energy (hot), to a system of lower kinetic energy (cold). Ex. A hot steel bar placed in cold water.
Law of Conservation of Energy (First Law of Thermodynamics)
Energy cannot be created or destroyed. However, it can be transferred from one thing to another, OR it can be converted into different forms.
When studying energy changes, scientists look at two things:
1. System: The part of the universe being studied
2. Surroundings: Everything else in the universe.
** Any energy change in the system, has an equal and opposite change in energy in the surroundings. **
ΔEsystem = -ΔEsurrondings
qsystem = -qsurroundings
There are three different ways to describe a system:
1. Open system: Matter and energy and be exchanged with the surroundings. Ex. An open flask with a reaction taking place.
2. Closed system: Matter cannot be exchanged with the surroundings, HOWEVER, energy can be exchanged. Ex. A stoppered flask with a reaction taking place.
3. Isolated system: Matter and energy cannot be exchanged with the surroundings. The system is isolated.
Section 2: Endothermic and Exothermic
Endothermic and Exothermic are two terms used to describe what happens to the energy of a system undergoing a change.
Endothermic processes require energy to take place. The system has to absorb energy from its surroundings. Ex. ice melting
Exothermic processes give off energy. The system releases energy to the surroundings. Ex. Water freezing
Potential Energy Diagrams
Section 3: Flow of Heat
The simplest situation in thermochemistry occurs when there is a flow of heat from one object to another. This causes a change in temperature without a change in state or chemicals. We use a formula to describe the energy flow:
q = mcΔT
q = quantity of heat flowing into or out of the system
m = mass of the substance
c = specific heat capacity of the substace
ΔT = change in temperature
Specific heat capacity is the amount of energy that is needed to increase the temperature of 1.0 grams of a substance by 1.0 °C.
Ex. H2O(l) = 4.18 J/g·°C meaning it takes 4.18 Joules of energy to heat 1.0 grams of water 1.0 °C.
Heat capacity is the amount of energy that is needed to increase the temperature of a substance 1.0 °C. Chemists are more interested in specific heat capacity and the two should not be confused.
Example 1. How much energy is needed to raise the temperature of a 250. g water sample from 20 °C to 85 °C?
Solution: q = mcΔT (ΔT = 85 °C - 20 °C)
q = 250g x 4.18 J/g·°C x 65 °C
q = 67925 J or 68 kJ
Example 2. A 2.50 L container of water at 22.3 °C is supplied with 20. kJ of energy. What is the final temperature?
Solution: q = mcΔT (m = 2.5 kg or 2500 g)
ΔT = q/mc → ΔT = 20000 J/(2500g x 4.18 J/g·°C)
ΔT = 1.9 °C
ΔT = Tf - Ti → Tf = ΔT + Ti
Tf = 1.9 °C + 22.3 °C = 24.2 °C
** Do questions 1-4 on page 634 and 5-10 on page 636 **
Section 4: Enthalpy Changes
All chemical bonds have stored or potential energy. Whenever the bonds in molecules or between molecules change, there is a change in the potential energy.
Enthalpy changes (q) describes changes in potential energy of a system during a chemical or physical change under constant pressure. Enthalpy can change during a chemical reaction, a phase change or in a nuclear reaction.
· Physical changes: A change in phase (melting, freezing, evaporating) or when something dissolves are physical changes. The bonds between molecules are broken or formed (intermolecular forces).
· Chemical reactions: Ionic and covalent bonds break and new ones form (intramolecular forces). The products have a different amount of potential energy.
· Nuclear reactions: Bonds between particles in the nucleus are broken.
Amount of energy given off in each enthalpy change is different:
Nuclear > Chemical > Physical (phase)
Enthalpy changes are energy changes and are measured in the units of energy, Joules or kilojoules.
Very often, the enthalpy change is known for common reactions and phase changes. The amount of the substance is important to the energy change. Chemists use moles to describe the amount of a substance. The energy change is given as molar enthalpy (ΔH).
Molar enthalpy (ΔH): The enthalpy change of 1.0 mol of a substance undergoing a change. The units are given as kilojoules per mole (kJ/mol).
Enthalpy and molar enthalpy are related using the formula:
q = nΔH
· q is the enthalpy change or energy change (also heat of reaction)
· n is the number of moles of the substance changing
· ΔH is the molar enthalpy
Phase Changes: A change in the state of matter (solid, liquid, gas) of matter without a change in the chemical composition of the system.
· There is a change in potential energy (Ep) of the substance.
· There is no change in temperature during the phase change. Therefore there is NO CHANGE in kinetic energy (Ek).
The molar enthalpy for the phase changes of water are well known:
H2O(s) → H2O(l) ΔHfus = 6.03 kJ/mol
H2O(l) → H2O(s) ΔHfre = - 6.03 kJ/mol
H2O(l) → H2O(g) ΔHvap = 40.8 kJ/mol
H2O(g) → H2O(l) ΔHcon = - 40.8 kJ/mol
Example: What is the enthalpy change when 50.0 grams of ice is converted at 0 °C is converted to water at 0 °C?
q = nΔH
First find the moles of water (molar mass = 18.02 g/mol):
nwater = 50.0 g/18.02 g/mol = 2.77 mol of water
q = 2.77 mol x 6.03 kJ/mol
q = 16.7 kJ
Do Questions 15-18 on Page 643, 19-23 in Page 645 and 24-29 on Page 648
Section 5: Heating and Cooling Curves (Total Energy Changes)
Heating curves show how the temperature of a substance changes as heat is added to it. Cooling curves show how the temperature of a substance changes as heat is removed from it.
Example: Draw a heating curve for 100.0 grams of ice at -10 °C that is warmed to 120 °C.
In the example heating curve above, there are 5 sections (A - E). Lets look at what is happening in each section.
A: Ice is warming from -10 °C to 0 °C: ↑T; ↑Ek; ↑q
B: The ice is melting: no ΔT; ↑Ep; ↑q
C: Water is warming from 0 °C to 100 °C: ↑T; ↑Ek; ↑q
D: Water is boiling (vaporizing): no ΔT; ↑Ep; ↑q
E: Water vapour is warming: ↑T; ↑Ek; ↑q
Question: How much energy is required to heat the 100.0 grams of ice as in the example above from -10 °C to 120 °C? (100.0 g H2O = 5.55 mol)
To find the answer, we must find the energy needed in each of the five sections of the heating curve:
ΔEtotal = qA + qB + qC + qD + qE
ΔEtotal = mcΔTice + nΔHfus + mcΔTwater + nΔHvap + mcΔTvapour
ΔEtotal = (100.0 g x 2.06 J/g·°C x 10 °C) + (5.55 mol x 6.03 kJ/mol) + (100.0 g x 4.18 J/g·°C x 100 °C) + 5.55 mol x 40.8 kJ/mol) + (100.0 g x 2.01 J/g·°C x 20 °C)
ΔEtotal = 2.06 kJ + 33.5 kJ + 41.8 kJ + 226 kJ + 4.02 kJ
ΔEtotal = 308 kJ
Example 2: How much energy is released when 155 grams of water at 35 °C is cooled to ice at -15 °C? (Draw cooling curve first)
ΔEtotal = qA + qB + qC
ΔEtotal = mcΔTwater + nΔHfre + mcΔTice
ΔEtotal = (155 g x 4.18 J/g·°C x -35 °C) + (8.60 mol x -6.03 kJ/mol) + (155 g x 2.06 J/g·°C x -15 °C)
qtotal = -26.0 kJ + -51.9 kJ + -4.79 kJ
qtotal = -82.7 kJ
Do Questions 30-34 on Page 655
Section 6: Enthalpy Changes in Reactions
Lets look at the reaction for the combustion of butane.
2C4H10 + 13O2 → 8CO2 + 10H2O ΔHcomb = -5320 kJ
The equation is written so that 2 moles of butane are combusted with 13 moles of oxygen. 8 moles of carbon dioxide are produced along with 10 moles of water. When this happens, 5320 kJ of energy is given off (exothermic).
** In thermochemistry, the coefficients are taken to mean moles, not molecules. **
To find the molar enthalpy for butane, we would have to use our formula:
q = nΔH
This would rearrange to be ΔH = q/n
ΔH = -5320 kJ/2 moles
ΔH = -2660 kJ/mol of butane combusted
Section 7: Communicating Energy Changes
There are many different terms used to describe energy changes in physical changes or chemical reactions. Some of these terms will be a review of things already covered.
Heat of reaction: The enthalpy change in a chemical reaction (q) measured in kJ.
Molar heat of reaction: The molar enthalpy value for a specific reaction (ΔH) measured in kJ/mol.
Standard molar enthalpy (ΔH°): The enthalpy of a physical change or chemical reaction measured at SATP (standard ambient temperature and pressure - 25 °C and 100 kPa).
The following are symbols used to describe different processes.
Heat of combustion
Heat of formation
Heat of a reaction
Heat of fusion (melting)
Heat of vaporization
Heat of solidification (freezing)
Heat of condensation
Heat of solution (dissolving)
Standard Heats of Formation (ΔH°f): The energy change that occurs when 1 mole of a compound is produced from its elements at SATP. NOTE: The heat of formation (kJ) and the molar heat of formation (kJ/mol) have the same value for true formation reactions, because by definition, only 1 mole of product is made.
Ex. C + 2H2 → CH4 formation reaction
Ex. 1/2 O2 + H2 → H2O formation reaction
Ex. CH4 + 2O2 → CO2 + 2H2O Not formation rxn (more than one product)
Ex. 4Fe + 3O2 → 2Fe2O3 Not true formation rxn (2 moles of product)
Thermochemical Equations: A chemical equation that includes the energy value (ΔH or H). The location of the energy term tells us if the reaction is endothermic or exothermic. If the energy term is on the left of the equation, the reaction is endothermic. If the energy term in on the right of the equation, the reaction is exothermic.
Ex. H2O (l) + 285.8 kJ → H2 (g) + O2 (g) endothermic
Ex. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) exothermic
The energy term can also be written next to the chemical equation. The sign on the energy term tells us if the reaction is endothermic (+) or exothermic (-).
Ex. 2SO2 (g) + O2 (g) → 2SO3 (g) ΔHcomb = -197.8 kJ
Ex. H2SO4 (l) → SO3 (g) + H2O (l) ΔH = 133 kJ
Note: In thermochemical equations, the phase of the compound (s, l, g, aq) must be included.
Potential Energy (Enthalpy) Diagrams:
Section 8: Calorimetry
Calorimetry is used to measure to enthalpy of a reaction (also called heat of a reaction). It measures energy changes using an isolated system, a thermometer and a known mass of water. The simplest calorimeter is just two Styrofoam cups with a lid.
The isolated system does not allow energy to pass in or out. Therefore, all of the energy from the physical or chemical change, gets absorbed by the water (or the water releases energy if the process is endothermic).
Several assumptions are made when using calorimeters:
1. No heat is transferred between the calorimeter and the outside environment
2. Any heat absorbed or releases by the calorimeter materials is negligible
3. A dilute aqueous solution has the same density and specific heat capacity as pure water
A simple Calorimeter
The law of conservation of energy is used to explain how a calorimeter works.
A physical change or chemical reaction takes place causing a change in potential energy. This causes a change in kinetic energy (temperature) of the water.
Epotential = - Ekinetic
qsystem = - qcalorimeter (water)
Calorimetry problems can be solved in three steps.
Step 1: Finding qrxn (system) or qwater based on the information given. The equations qsystem = nΔH or qwater = mcΔT will have to be used.
Step 2: Stating that qsystem = - qcalorimeter (water). If qsystem was calculated, then qcalorimeter is also known.
Step 3: Finding the requested information with the q value.
Example: 14.50 g of ammonium nitrate (NH4NO3) is placed in 500.0 mL of water. When stirred, the temperature fell from 23.5 °C to 20.0 °C. What is the molar enthalpy of solution (Hsol) of NH4NO3? (Molar mass of NH4NO3 = 80.06 g/mol)
Step 1: We can calculate the change in energy of the water; qwater = mcΔT
qwater = 500.0g ∙ 4.18 J/g°C ∙ 3.5°C
qwater = -7.3 kJ
Step 2: qsol = - qwater
qsol = +7.3 kJ
Step 3: qsol = nΔHsol
ΔHsol = qsol/n nNH4NO3 = 14.50 g/80.06g/mol = 0.181 mol
ΔHsol = 7.3 kJ/0.181 mol
ΔHsol = +40.4 kJ/mol
Section 9: Bomb Calorimetry
Bomb Calorimeters are designed to find the molar enthalpy of substances involved in combustion reactions of fuels or in finding the energy value in food (Calories). The calorimeter consists of an inner chamber (bomb) where the substance is burned, a water filled outer chamber, a stirrer, a thermometer, and wires for ignition.
The total energy change of the calorimeter is the sum of the energy changes of all the components of the bomb calorimeter (water, container, stirrer & thermometer). The manufacturer finds the heat capacity of the whole bomb calorimeter. The heat capacity (C) is the amount of energy needed to increase the temperature 1.0 °C and is measured in the units kJ/°C. Therefore, the amount of heat involved is given as:
q = CΔT
The three step method can be used again in bomb calorimetry.
Example: In an experiment, 1.50 g of sucrose (C12H22O11) is burned and causes the temperature of the bomb calorimeter to rise from 25.00 °C to 27.88 °C. If the heat capacity of the bomb calorimeter is 8.57 kJ/°C, calculate the enthalpy of combustion.
Step 1: qcalorimeter = CΔTcalorimeter ΔT = Tf - Ti = 27.88 °C - 25.00 °C = 2.88 °C
= 8.57 kJ/°C ∙ 2.88 °C
= 24.7 kJ
Step 2: qcomb = - qcalorimeter
qcomb = -24.7 kJ
Step 3: qcomb = nΔHcomb
ΔHcomb = qcomb/n nsucrose = 1.50 g/342.34g/mol = 0.00438 mol
ΔHcomb = -24.7 kJ/0.00438 mol
ΔHcomb = -5.64 MJ
Section 10: Hess's Law
Hess's Law can be used to find the enthalpy of a reaction for which there is no experimental method to use. If the enthalpies of related chemical reactions are known, then these reactions can be manipulated so that they add up to give the chemical equation that is desired. The enthalpies of each reaction can then be added together to enthalpy of the desired reaction.
Hess's Law: If a reaction is the sum of two or more other reactions, the enthalpy change for the overall process is the sum of the enthalpy changes for the constituent reactions.
Example: What is the ΔH°f for the formation of nitorgen dioxide (NO2)?
1/2 N2 (g) + O2 (g) → NO2 (g) ΔH°f = ??
We can use the following information from equations with known enthalpies:
1/2 N2 (g) + 1/2 O2 (g) → NO (g) ΔH° = +180 kJ
NO2 (g) → NO (g) + 1/2 O2 (g) ΔH° = +112 kJ
We must manipulate the 2 known equations so that they add up to the unknown equation. There are 3 rules to follow:
1. If you switch an equation around (reactants become products and products become reactants), the ΔH changes sign (becomes negative if positive).
2. If you multiply or divide a reaction by a number to change the balancing, you must multiply or divide the ΔH by the same number.
3. When adding the reactions, any compound on both sides of the equation will cancel out.
Looking at these two reactions, we must switch the second reaction around to give:
1/2 N2 (g) + 1/2 O2 (g) → NO (g) ΔH° = +180 kJ
NO (g) + 1/2 O2 (g) → NO2 (g) ΔH° = -112 kJ
When the reactions are added together, the NO (g) on each side of the equation will cancel and the two 1/2 O2 (g) will add to make one O2 (g). The enthalpies are added together and we end up with:
1/2 N2 (g) + O2 (g) → NO2 (g) ΔH°f = +68 kJ
Example 2: Calculate the ΔH° for the following reaction:
2 N2 (g) + 5 O2 (g) → 2 N2O5 (g) ΔH° = ??
Use the following reactions:
1. H2 (g) + 1/2 O2 (g) → H2O (l) ΔH° = -285.8 kJ
2. N2O5 (g) + H2O (l) → 2 HNO3 (l) ΔH° = -76.6 kJ
3. 1/2 N2 (g) + 3/2 O2 (g) + 1/2 H2 (g) → HNO3 (l) ΔH° = -174.1 kJ
Reaction 1 is switched around and multiplied by 2.
Reaction 2 is switched around and multiplied by 2.
Reaction 3 is multiplied by 4.
1. 2 H2O (l) → 2 H2 (g) + O2 (g) ΔH° = +571.6 kJ
2. 4 HNO3 (l) → 2 N2O5 (g) + 2 H2O (l) ΔH° = +153.2 kJ
3. 2 N2 (g) + 6 O2 (g) + 2 H2 (g) → 4 HNO3 (l) ΔH° = -696.4 kJ
Everything cancels out to give our final equation and the enthalpies are added.
2 N2 (g) + 5 O2 (g) → 2 N2O5 (g) ΔH° = +28.4 kJ
Do questions from Hess's Law worksheet.
Section 11: Heats of Formation
Another way to predict the enthalpy of a reaction for which there is no experimental method to find it is the use the Standard Heats of Formation (ΔH°f). Remember that a standard heat of formation is the energy change that occurs when 1 mole of a compound is produced from its elements at SATP.
Note: The ΔH°f of an element on its own in its standard state is set at 0 kJ.
We can find the ΔH°rxn for a reaction by using the heats of formation of the reactants and products and the following equation:
ΔH°rxn = Σ nΔH°f (products) - Σ nΔH°f (reactants)
Σ means "sum of" n = # of moles from the balanced equation
ΔH°f = molar enthalpy of formation
Example 1: Using a table of standard heats of formation, find the ΔH°rxn for the following reaction:
2 CO (g) + O2 (g) → 2 CO2 (g)
We know the following information: ΔH°f O2 = 0 kJ/mol
ΔH°f CO = -110.5 kJ/mol ΔH°f CO2 = -393.5 kj/mol
Using our equation: ΔH°rxn = Σ nΔH°f (products) - Σ nΔH°f (reactants)
ΔH°rxn = 2 mol x -393.5 kJ/mol - (2 mol x -110.5 kJ/mol + 1 mol x 0 kJ/mol)
ΔH°rxn = -787 kJ - (-221 kJ)
ΔH°rxn = -787 kJ + 221 kJ = -566 kJ
Example 2: Using a table of standard heats of formation, find the ΔH°r for the following reaction:
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g)
We know the following information:
ΔH°f C2H4 = 52.5 kJ/mol ΔH°f O2 = 0 kJ/mol
ΔH°f CO2 = -393.5 kj/mol ΔH°f H2O (g) = -241.8 kJ/mol
Using our equation: ΔH°r = Σ nΔH°f (products) - Σ nΔH°f (reactants)
ΔH°rxn = (2 mol x -393.5 kJ/mol + 2 mol x 241.8 kJ/mol) - (1 mol x 52.5 kJ/mol + 3 mol x 0 kJ/mol)
ΔH°rxn = (-787 kJ + (-483.6 kJ)) - (52.5 kJ)
ΔH°rxn = -1270.6 - 52.5 kJ = -1323.1 kJ
Section 12: Bond Energy Method
Another way to find the enthalpy of a reaction (ΔHrxn) is to use the bond energies of all the bonds broken and formed in the reaction.
Bond dissociation energy or just bond energy is the energy required to break a chemical bond between two atoms. The values are measured in kJ/mol, which signifies the number of kilojoules of energy required to break one mole of bonds. (See table below)
Energy is required to break bonds and is released when new bonds form.
If the energy required to break the bonds is greater than the energy released when new bonds form, the reaction is endothermic. If the energy required to break the bonds is less than the energy released when new bonds form, the reaction is exothermic.
The following equation is used to help find the enthalpy of a reaction (ΔHrxn):
ΔH°rxn = Σ (bond energies reactants) - Σ (bond energies products)
Example 1: Find the ΔH°rxn for the following reaction:
H2 (g) + Cl2 (g) → 2 HCl (g) ΔH°r = ??
We break 1 mol of H-H and Cl-Cl bonds. We form 2 mol of H-Cl bonds.
We know that: H-H bond = 436 kJ/mol Cl-Cl bond = 243 kJ/mol
H-Cl bond = 432 kJ/mol
Using our equation: ΔH°rxn = Σ (bond energies reactants) - Σ (bond energies products)
ΔH°rxn = (1 x 436 kJ/mol + 1 x 243 kJ/mol) - (2 x 432 kJ/mol)
ΔH°rxn = 679 kJ - 864 kJ = -185 kJ