The document provides objectives and information about calculating enthalpies, solving problems in calorimetry, and Hess' Law. The objectives are to know how to calculate enthalpies, solve problems in calorimetry, and apply Hess' Law. It defines important terms like enthalpy, enthalpy of reaction, and discusses how to calculate enthalpies using heat of formation values from tables. Examples are provided to demonstrate calculating enthalpy of reaction. Important enthalpy changes like combustion, formation, atomization, fusion, and vaporization are defined. The document also discusses calorimetry, defining key terms, and outlines Hess' Law which states the heat of a reaction is the same whether it occurs

2. Objectives
To know how to calculate enthalpies
To know how to solve problems on
calorimetry
To know how to solve problems on Hess’ Law

3. How to Calculate
Enthalpies
How?

4. Let’s redefine some terms
• Enthalpy (H) is the sum of the internal energy of the
system plus the product of the pressure of the gas in
the system and its volume
Hsys = Esys + PV
If pressure is kept constant, we can arrive at:
H sys = q (at constant pressure)
Where: H = H final – H initial
q --- is heat

5. Let’s redefine some terms
• Enthalpy of Reaction ( H) is the difference
between the sum of the enthalpies of the products
and the sum of the enthalpies of the reactants
H = nH products – mH reactants
n
where and m are the coefficients of
the products and the reactants in the
balanced equation

6. Remember
• Enthalpies are usually computed at standard
conditions (25oC = 298K)
• Note that enthalpies of formation of elements at
standard conditions are equal to zero

7. Let’s Calculate
• Example 1: Calculate the enthalpy of the oxidation
reaction of benzene (C6H6) given with the chemical
equation: C6H6(l) + 4.5O2(g) = 6CO2(g) + 3H20(l)
1. Find the enthalpies of formation for all chemical
components of the reaction using a Table of Heat Formation
H C6H6(g) = 48.85 KJ/mol
H O2(g) = 0
H CO2(g) = -393.509 KJ/mol
H H2O (l) = -285.83 KJ/mol

8. C6H6(l) + 4.5O2(g) = 6CO2(g) + 3H20(l)
2. Multiply each enthalpy value on the corresponding reaction
coefficient and sum up the enthalpies of formation
C6H6(l) + 4.5O2(g) H C6H6(g) = 48.85 KJ/mol
H O2(g) = 0
H initial = 48.95 KJ/mol+ 4.5 x O H CO2(g) = -393.509 KJ/mol
H H2O (l) = -285.83 KJ/mol
H initial = 48.95 KJ/mol
3. Do the same to the final reagents
6CO2(g) + 3H20(l)
H final = 6 x (-393.509 KJ/mol) + 3 x (-285.83 KJ/mol)
H final = -3218.544 KJ/mol

9. H initial = 48.95 KJ/mol
H final = -3218.544 KJ/mol
4. Subtract the enthalpy of formation of the initial
reagents from the final reagent
H reaction = H final – H initial
H reaction = -3218.544 KJ/mol – 48.95 KJ/mol
H reaction = -3267.494 KJ/mol

10. Some Important Enthalpy Changes
1. Enthalpy Change of Combustion
- the enthalpy change which
occurs when one mole of the
substance is completely burnt
in oxygen under standard
conditions
Eg. C (graphite) + ½ O2 (g)  CO2 (g)
C (graphite) + O2 (g)  CO2 (g)
chemist’s shorthand:
H c,m [(graphite)] = -393.5 kJ/mol

11. Some Important Enthalpy Changes
2. Enthalpy Change of Formation
Eg. The SECF of methane, CH4, refersone mole of
- the enthalpy change when to the change:
the compound is formed from its elements
C (graphite) + 2 H2 (g) ---->CH4 (g) H = -74.8 kJ/mol
under standard conditions
H f,m [CH4(g)] = -74.8 kJ/mol
- may also be called Heats of Formation.
 3. Enthalpy Change of Atomisation
Eg. C (graphite  C of an element is kJ/mol
The SMECA (g) H = 716.7 the
enthalpy change when one mole of its
atoms in the gaseous 716.7is formed from
H at,m [(graphite)] = state kJ/mol
the element under standard conditions .
* Atomisation is always endothermic.

12. Some Important Enthalpy Changes
4. Enthalpy Change of Fusion
- The enthalpy change when 1 mole of solid is
converted to one mole of liquid at its melting point at
standard pressure
H fus,m [(H20)] = 716.7 kJ/mol H = 6.01 kJ/mol

13. Some Important Enthalpy Changes
5. Enthalpy Change of Vaporisation
- The enthalpy change when 1 mole of liquid is
converted to one mole of gas at its boiling point at
standard pressure
H vap,m [(H20)] = 716.7 kJ/mol H = 41.09 kJ/mol

14. Calorimetry

15. What is it?
Calor (Latin) + metry (Greek) = Calorimetry
Science of measuring
“heat” + “to measure” = the amount of heat
Two types of calorimetry
• 1. measurements based on constant pressure
• 2. measurement based on constant volume

16. Other terms
Calorimeter – the device used to measure heat of
reaction
Heat capacity – the amount of heat required to raise
its temperature by a given amount
– SI unit: J/K
Formula: q= C T
where: q - heat
C - heat capacity
T - change in temperature
= Tf-Ti

17. Other terms
• Specific heat capacity – gives the specific heat
capacity per unit mass of a particular substance
- SI unit: J/kgK
Formula: q= mc T
where: q - heat
m - mass
c - specific heat capacity
T - change in temperature
= Tf-Ti

18. Other terms
• Molar enthalpy of a substance
Formula: H= mc T
n
where: H - enthalpy change
m - mass
c - specific heat capacity
T - change in temperature
= Tf-Ti
n - moles of substance

19. Hess’s Law of Heat
Summation
By Germain Henri Hess

20. What is it?
• Hess Law of Heat Summation states that the heat
absorbed or released during a reaction is the same
whether the reaction occurs in one or several steps
• Rules
1. Make sure to rearrange the given equations so that the
reactants and products are on the appropriate sides of the
arrows
2. If you reverse equations, you must also reverse the sign of
H
3. If you multiply/divide equations to obtain a correct
coefficient, you must also multiply/divide the H by this
coefficient

21. Get ready
now for
some brain
exercises