ENERGETICSThe study of energy changes that take place in a chemical

reaction

OBJECTIVES

At the end of the lesson you should be able to:1. describe energy changes in bond formation and bond breaking2. distinguish between exothermic and endothermic reactions in terms of

energy 3. explain energy diagrams representing endothermic and exothermic

reactions4. 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

ENDOTHERMIC PROCESS

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

EXOTHERMIC PROCESS

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

ENDOTHERMIC REACTION

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

EXOTHERMIC REACTION

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

COMPARING ENERGY PROFILE DIAGRAMS

Compare both energy profile diagrams. State the differences you observe

Energy Profile diagram for: An endothermic reaction An exothermic

reaction

DIFFERENCES BETWEEN ENERGY PROFILES

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)

Measuring the heat (enthalpy) change, ∆H

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

HEAT OF SOLUTION

• 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.

HEAT OF SOLUTION: PROBLEM

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)

ANSWER TO QUESTION

ACTIVITY

Try these using the same programme!

• 2.5g NH4Cl in 20 cm3 of water.

• 4.0 g Na2CO3 in 40 cm3 of water

HEAT OF NEUTRALISATION: PROBLEM

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.

Answer to Problem

Initial Temperature = 25 oC (average of both solutions)

Final Temperature = 36 oCDensity of solution = mass/volumeMass = volume x density

= 100 cm3 x 1 g cm-3

= 100 g

Answer to Problem (continued)

Calculation of the heat released:

∆H = m c ∆T

∆H = 100 x 4.2 x (36 – 25)

= 4620 J

Answer to Problem (continued)

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

Answer to Problem (continued)

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