1. The study of heat changes in chemical reactions
THERMOCHEMISTRY
The study of heat changes in chemical reactions 1

2. Exothermic and endothermic reactions.
What You Will Learn.
Definitions-heat of reaction, heat of combustion, kilogram calorific value, heat of neutralisation.
Definitions Honours Hess's law, heat of formation, law of conservation of energy, bond energy.
Exothermic and endothermic reactions.
Formula – heat given out = mass x specific heat capacity x change in temperature ( H = m c)
How heats of combustion are measured (bomb calorimeter) 2

3. Using kilogram calorific values to compare fuel.
What You Will Learn.
Using kilogram calorific values to compare fuel.
Honours only – how bond energy affects  H .
Explain and measure heat of neutralisation.
Honours only – explaining heat of formationn and writing heat of formation equations form the elements of a compound.
Honours only – using Hess's law and calculating heat of formation from other heats of reaction. 3

4. ENERGY TRANSFORMATIONS
Energy- defined as the ability to do work or supplying heat.
Energy is detected by observing the effects that it has.
It is weightless, odourless, and tasteless.
Chemical potential energy is the energy stored within the bonds of chemical substances.
Kinds and arrangements of the atoms in a substance determine the amount of energy stored.
Heat- represented by H, is energy that transfers from one substance to another. 4

5. IMPORTANT CONCEPT
When substances have a difference in temperature,heat always travels from warmer temperature to cooler.
The flow from a warmer object towards a cooler will occur until the temperature of both objects becomes the same.
Reactions can be either exothermic or endothermic 5

6. EXOTHERMIC AND ENDOTHERMIC PROCESSES
The law of conservation of energy states that in any chemical or physical process, energy is neither created nor destroyed.
Energy changes can be accounted for in terms of work, stored energy, or heat.
A process that absorbs heat from its surrounding is called an endothermic reaction and H is positive
A process that releases heat to its surroundings is an exothermic reaction and H is negative 6

7. Burning Mg metal in air is an example of an exothermic reaction.
In the lab. We can demonstrate exothermic and endothermic reactions easily.
Burning Mg metal in air is an example of an exothermic reaction.
Dissolving ammonium chloride or ammonium nitrate in water are examples of endothermic reactions. (how could we prove this?) 7

8. A Joule is symbolized with a capital “J”.
The amount of energy stored in the chemical bonds is thought of in terms of Joules.
A Joule is defined as the amount of work done by a force of one newton moving an object through a distance of one metre.
A Joule is symbolized with a capital “J”. 8

9. Heat Liberated = m x c x T
M = mass, c = specific heat capacity and T = change in temperature.
Specific heat capacity (c ) = the heat needed to the change the temperature of 1 kg of a substance by 1oC
Note for water and dilute solution 1 cm3 = 1g so 50cm3 of solution = 50 g and so on. 9

10. If you examine the table on slide 12 you can see that water has the highest specific heat compared to the other substances listed.
That means that water has a higher specific heat capacity (can absorb more heat) than any of those substances. 10

11. The table in the next slide shows some specific heat capacities.
Water has a specific heat 9 times greater than iron. This means it takes 9 times more heat to raist the temperature of water than Iron
The table in the next slide shows some specific heat capacities. 11

12. Table of specific heat capacities
Substance
Specific Heat capacity
(J kg-1 K-1)
Water
4180
Copper
390
Iron
451
glass
674
Aluminum
910
Alcohol
2500
Wood
1700
Ice
2100
Air
1000

13. The heat of reaction for the formation of 1 mole of water
Heat of reaction = heat change due to a complete chemical reaction.
The heat of reaction for the formation of 1 mole of water
H2(g) + ½ O2(g)  H2O(g) ∆H = -242 kJ mol-1
If the number of moles is changed then so does the heat of reaction. If two moles of hydrogen are burned completely in oxygen then the heat of reaction is
2H2(g) + O2(g)  2H2O(g) ∆H = -484 kJ mol-1

14. Heat of combustion = heat change which occurs when one mole
Heat of combustion = heat change which occurs when one mole of a substance is burnt in excess oxygen.
Measuring the heat of combustion.
The principal is to find the rise in temperature of a measured quantity of water for a given quantity of burnt substance. The heat gained by the water is assumed = the heat given out by the substance and the formula E= m c  is used to calculate the heat for the known mass of substance. This is then adjusted for one mole.
Example
C4H10(g) + 6½O2 (g)CO2 (g)+ 5H2O(g) H = kJ mol-1
Why the equation with 6½O2? Shouldn't the equation have whole numbers?
To do this accurately a bomb calorimeter can be used as shown in the next slide.

15. Endothermic stirrer Exothermic Steel bomb System water
12/10/09
BOMB CALORIMETER- in a bomb calorimeter the system is closed in order to prevent mass loss. A sample is burned in a constant-volume chamber in the presence of oxygen at high-pressure.
Oxygen intake valve
Electrical leads with firing element
thermometer
Endothermic
stirrer
Exothermic
Steel bomb
System
Sample to be burned
water 15

16. System with can as boundary
12/10/09
A basic calorimeter
Endothermic = absorbing energy
Surroundings everything else
System with can as boundary
Exothermic = releasing energy 16

17. Heat of neutralisation = heat change when one mole of H+ from
Heat of neutralisation = heat change when one mole of H+ from an acid reacts with one mole of OH- ions from an alkali.
Measuring heat of neutralisation
Equivalent amounts of acid and alkali are mixed and the rise in temperature is recorded.
Remember the number of H atoms in the acid must be taken account off.
The formula E= m c  is used to calculate the heat where m is the mass of acid + the mass of akali.
The answer is then adjusted for one mole of H+ ions.
Note: The heat of neutralisation for a strong acid and a strong base is always the same as they are fully dissociated in solution, effectively being H+ + OH-  H2O H=-57.2kJ/mol.
With a weak acid or base it is less as some of the heat is used up in the dissociation of the acid or base which is endothemic.

18. Measuring the Heat of Neutralisation
1. Take the initial temperatures of known volumes of acid (HCl) and alkali (NaOH).
2.Place the two solutions into the calorimeter and mix them together.
3. After the reaction is complete, take the final temperature.
4. Because you know the initial and final temperatures and the s.h.c of the solution, you can calculate the amount of heat released(∆H) or absorbed in the reaction using the equation ∆H = m x c x T
No heat enters or leaves! 18

19. Calculating Heat of Neutralisation
The formula for heat liberated ∆H is :
∆H = m X c X ∆ T
mass(kg) X specific heat capacity X change in temperature
Now Find the heat liberated when 100 cm3 of 1.0M HCl is neutralised by 100 cm3 of NaOH when the temperature rose from 15.6oC to 22. 4oC. The specific heat capacity of the solution is J kg-1 K-1 and then calculate the heat for 1 mole of acid neutralised. 19

20. The temperature change ∆ T = (22.4⁰C – 15.6⁰ C) = 6.8⁰C
Answer
The temperature change
∆ T = (22.4⁰C – 15.6⁰ C) = 6.8⁰C
C = J kg-1 K-1
Mass of the solution is 200 g which is 0.2 kg. (remember 1 cm3 = 1 g so 200 cm3 = 200 g)
∆H = m X c X ∆ T
= 0.2 x 4080 x 6.8
= J 20

21. No of moles of acid = vol. (L) x molarity = 0.1 x 1.0 = 0.1 moles.
Answer continued.
No of moles of acid = vol. (L) x molarity
= 0.1 x 1.0
= 0.1 moles.
Now 0.1 moles  J
 1 mole = J or 55.4kJ 21

22. Kilogram calorific value
The kilogram calorific value of a fuel is the heat energy
Produced when 1 kg of a fuel is burned completely in
excess oxygen
Fuel
Formula
Kilogram calorific value in KJ kg-1
Methane
CH4
55,625
propane
C3H8
50,454
butane
C4H10
49,603
carbon C
32.750
hydrogen H2
143,000

23. Bond Energy (honours)       
Bond energy is the energy needed to break one mole of
covalent bonds and to separate the neutral atoms from
each other. H  O H
O=O
O=O 
H C H → O=C=O +     
O=O O H H
In the combustion of methane 4 C-H bonds and 2 O=O are broken
Then two C=O bonds and 4 O-H bonds are made.

24. Bond Energy Bond Average Bond Energy kJ mol-1 C-H 212 C-C 348 C=C 612
837
O-H
463
O=O
496
C=O
743
The bond energies are averages as breaking the first C-H bond in methane does not take the same energy as the second bond and so on so we use the average of the four bonds

25. HONOURS
Heat of formation =heat change when one mole of a compound is formed from its elements in their standard state.
The heat of formation of water from it's elements is
H2(g) + ½ O2(g)  H2O(l) ∆H = -286 kJ mol-1
The heat of formation of Methane is
C(s) + 2H2(g) CH4(g) ∆H = kJ mol-1
Writing equations for heat of formation is easy just look at what elements are in the formula.
e.g write the equation for heat of formation of Ethanol C2H5OH.
2C(s) + 3H2(g) + ½ O2(g) Cl
Now write heat of formation equations for the following:
(a) NH (b) C6H12O (c) C2H (d) CH3COOH (e) H2S

26. Answers (a) ½N2 + 1½H2 → NH3 (b) 6C + 6H2 + 3O2 → C6H12O6
(c) 2C + 3H2 → C2H6
(d) 2C + 2H2 + O2 → CH3COOH
(e) H2 + S → H2S

27. It is important to state the physical states of the reactants and products because it can mean different ∆ H’s
H2O(l)→ H2(g) + ½ O2(g) ∆H = 285.8kJ H2O(g)→H2(g) + ½ O2(g) ∆H = 241.8kJ
A difference of 44.0kJ
It makes sense that it would require a greater amount of energy for water as a liquid to split into oxygen and hydrogen gas than it would for a water vapour molecule to split into oxygen and hydrogen gas. 27

28. Thermochemistry
Hess's Law states that if a chemical reaction takes place in a number of stages, the sum of the heat changes in the separate stages is equal to the heat change if the reaction was carried out in one stage.
According to Hess's law heat change for a single step reaction is the same as for the reaction done in two or more steps.
∆H=?
∆H1 = ∆H + ∆H2
= ∆H-283
∆H = kJ mol-1
∆H2
∆H1
Note ∆H2 is endothermic i.e. Positive as heat absorbed to make CO and ½O2 28

29. Calculations involving Hess's Law
Things to remember
If you multiply or divide an equation then you must multiply or divide the ∆H
If you reverse an equation must change the sign of the the ∆H
Find the heat of formation of Butane given the following information
The heat of combustion of Butane is H = kJ mol-1
The heat of combustion of C is H = KJ mol-1
The heat of combustion of H2 is H = KJ mol-1

30. Calculations involving Hess's Law (honours)
First we must write out the equations for the heats of combustion
❶ C4H10(g) + 6½O2(g) → 4CO2(g) + 5H2O H = kJ mol-1
❷ C(s) + O2(g) → CO2(g) H = KJ mol-1
❸ H2(g) + ½O2(g) → H2O(g) H = KJ mol-1
Now we must the equation for the heat of formation of butane
Need 4C(s) + 5H2(g) → C4H10(g) ∆H = ?
What must do the each equation so that they add up to what we
need?
Equation ❶ needs to be reversed
Equation ❷needs to be multiplied by 4
Equation ❸must be multiplied by 5

31. Calculations involving Hess's Law (Honours)
❶ x R
❷ x 4
❸ x 5
ADD
∆H = kJ mol-1
Hess's Law is just a restatement of the law of conservation of energy.
It allows to calculate heats of reaction that cannot be measured directly by experiment because the thermochemical equation can be added, subtracted, divided or multiplied.