Thermochemistry 2:
Energy & Enthalpy
Reminder
Enthalpy change, ΔH = ΔE+ PΔV
ΔH = Hproducts – Hreactants
W = -PΔV
Thermodynamic standard state
 The value of the enthalpy change, ΔH varies with temperature,
pressure, and concentration
 When we talk about ΔHo, we use a standard set of conditions
 We say ΔHo for a reaction is 699 kJ/mol, 1 atm pressure, and if in a
solution, at 1 M concentration. The temperature must be given
 Any change in these conditions would give a different value for ΔH
 These are standard conditions
 Std. enthalpy change is ΔHo
 Std energy change is ΔEo
 Std pressure, Po is 1 atm …
The thermodymanic std. state
The most stable form of a substance at 1
atm pressure (1 bar) and at a specified
temperature, usually 25oC; 1 M
concentration for all substances in solution
(page 307)
Si (s) + 2 Cl2 (g)
SiCl4 (l) ΔH = -687.0 kJ/mol
 Determine the value of ΔE, if 1.81 mole of Si react with a
stoichiometric amount of CL2 (g) when the reaction is carried out at
0.50 atm of pressure and with a volume change of -22.4 L.
 Step 1: Determine the ΔH for the reaction, taking into consideration
that 1.81 moles of Si reacted
 Step 2: Determine ΔE from the equation ΔH = ΔE+ PΔV
(watch your units!)
Problem 8.6
 The reaction between hydrogen and oxygen to yield
water vapour has ΔHo = -484 kJ
2 H2 (g) + O2 (g)
2 H2O (g)
ΔHo = -484 kJ
How much PV work is done, and what is the value of
ΔE (In kJ) for the reaction of 0.50 mol of H2 with
0.25 mol of O2 at atmospheric pressure if the
volume change is -5.6 L?
1) Find w = -PΔV
2) Making sure that w is expressed in kJ, use the eqn. ΔE=ΔH+PΔV
to find ΔE
(remember you are only using 0.5 mol H2)
Problem 8.7
 The explosion of 2.00 mol of solid
trinitrotoluene (TNT) with a volume of
approximately 274 mL produces
gases with volume 448 L at room
temperature.
 How much PV work (in kJ) is done
during the explosion?
2 C7H5N3O6 (s)
12 CO (g) + 5H2 (g) + 3 N2 (g) + 2 C (s)
Enthalpies of Physical Change
 Heat of fusion: The amount of heat energy needed for melting
(6.01 kJ/mol for water)
 Heat of vaporization: the amount of heat energy needed for
evaporation, breaking down the intermolecular bonds.
40.7 kJ/mol for water at 100oC
 Sublimation – the direct conversion of a solid to a vapour without
going through a liquid state
 Heat of sublimation = heat of fusion + heat of vaporization
Heat of sublimation
The enthalpy change is the same,
whichever route is taken.
It doesn’t matter that one route went
Via a liquid, and the other sublimed.
What matter is the enthalpy of the vapour
Minus the enthalpy of the solid:
ΔH = Hproducts – Hreactants
Enthalpies of Chemical Change
 Heats of reaction – enthalpies of chemical change
 Endothermic reactions.
 Hproducts > Hreactants
 Heat flow into the system from the surroundings
 ΔH is positive
 Exothermic reactions
 Hproducts < Hreactants
 Heat flow to the surroundings from the system
 ΔH is negative
 ΔHo values for a given equation




Eqn is balanced for the number of moles of reactants and products
All substances are in their std states
Physical state of each substance is given
Refer to the reaction going in the direction given
• Reverse the direction, change the sign of ΔH
Practice, practice,
practice, …
 Worked example 8.3
and problems 8.8, 8.9
on page 310
 Finish for homework
I want to see your
working clearly set
out