This document discusses energy changes that occur during chemical reactions. It defines endothermic and exothermic reactions, and explains how to interpret energy diagrams for these reactions. Key terms like activation energy, enthalpy change, heat of solution, and heat of neutralization are defined. The document provides examples of how to calculate heat of solution and heat of neutralization using experimental data and equations.

1. The study of energy changes that take
place in a chemical reaction

2. At the end of the lesson you should be able to:
1. describe energy changes in bond formation and bond breaking
2. distinguish between exothermic and endothermic reactions in
terms of energy
3. explain energy diagrams representing endothermic and
exothermic reactions
4. explain the following terms: activation energy, enthalpy change
5. define heat of solution
6. calculate the heat of solution from experiments or from
experimental data
7. define heat of neutralisation
8. calculate the heat of neutralisation from experiments or from
experimental data

3. To break a bond, energy (heat)
has to be absorbed.

4. When a bond is formed, energy
(heat) is released.

5. If the energy (heat) absorbed is greater than the energy (heat)
released, then the overall reaction is ENDOTHERMIC.
Energy Profile diagram for an endothermic reaction

6. If the energy (heat) released is greater than the energy (heat)
absorbed, then the overall reaction is EXOTHERMIC.
Energy Profile diagram for an exothermic reaction

7. Compare both energy profile diagrams. State the differences you observe
Energy Profile diagram for:
An endothermic reaction An exothermic reaction

8. ENDOTHERMIC REACTION EXOTHERMIC REACTION
Heat content of the
products is greater than
that of the reactants
Heat content of the
products is less than that
of the reactants
Heat (enthalpy) change, ∆H
is positive (+∆H)
Heat (enthalpy) change, ∆H
is negative (-∆H)

9. ∆H = m c ∆T
where:
• m is the mass of the solution
• c is the specific heat capacity
• ∆T is the change in temperature

10. • The heat absorbed or evolved when one mole of a substance
is dissolved in an infinite amount of solvent so that further
dilution causes no further heat change.
• Go to the link below and note the instructions before closing
as you will do this as a planning and designing lab.
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles
/thermochem/heat_soln.html
• We will now use this link to calculate heat of solution.

11. 1. Add 3.00 g of NaOH to 100 cm3 of water.
2. Note the initial temperature and the highest
or lowest temperature reached. (If you are
not sure, the change in temperature is given
on the graph.)
3. Calculate the heat of solution of NaOH.
(specific heat capacity of water is 4.2 J g-1 K-1
and the density of water is 1 g cm-3)

13. Try these using the same programme!
• 2.5g NH4Cl in 20 cm3 of water.
• 4.0 g Na2CO3 in 40 cm3 of water

14. The heat absorbed or evolved when one mole of water is
produced when an acid reacts with a base.
Problem:
• When 50 cm3 of 2 mol dm-3 hydrochloric acid at 25 oC
is reacted with 50 cm3 of 2 mol dm-3 sodium hydroxide
at 25 oC, the temperature increased to 36 oC.
Determine the heat of neutralisation.
• Assume that the density and the specific heat capacity
of the solution is the same as that for water.

15. Initial Temperature = 25 oC (average
of both solutions)
Final Temperature = 36 oC
Density of solution = mass/volume
Mass = volume x density
= 100 cm3 x 1 g cm-3
= 100 g

16. Calculation of the heat released:
∆H = m c ∆T
∆H = 100 x 4.2 x (36 – 25)
= 4620 J

17. Calculation of the number of moles of reactants:
Number of moles of HCl (and NaOH):
1000 cm3 contain 2 moles
50 cm3 contain x
x = 50 x 2
1000
= 0.1 moles
NaOH + HCl → NaCl + H2O
1 mol NaOH ≡ 1 mol HCl → 1 mol H2O
0.1 mol NaOH ≡ 0.1 mol HCl → 0.1 mol H2O

18. Calculation of the heat of neutralisation:
Formation of 0.1 mole of water produces 4620 J
Hence formation of 1 mole of water would produce 4620 J
0.1
= 46200 J
∆H(neut) = -46200 J mol-1
= -46.2 kJ mol-1