Standard Enthalpy Changes of Reaction
15.1
15.1.1 – Define and apply the terms
standard state, standard enthalpy
change of formation (ΔHf˚) and
standard enthalpy change of
combustion (ΔHc˚).
Q – What are
“standard” conditions?

A – 298K (25˚C)
 around room temp

1.00 x 105 Pa (101.3 kPa)
 around room pressure
standard enthalpy change of formation
(ΔHf˚)
 ΔHf˚=
the enthalpy change that occurs
when 1 mol of a substance is formed
from its elements in their standard
states. (see table 11 of IB data booklet)
standard enthalpy change of formation
(ΔHf˚)
 Enthaply of formation of any element in its standard
state is ZERO !!!
○ For a given state of matter – per standard conditions
○ For a given allotrope – usually the most stable one !!!
○ Superscript  may be used to indicate standard
conditions
 Gives a measure of the stability of a substance
relative to its elements
 Can be used to calculate the enthalpy changes of all
reactions, hypothetical or real
Sample problem 1

Q – the ΔHf˚ of ethanol is given in Table 11 of the
IB data booklet. Give the thermo chemical
equation which represents the standard enthalpy
of formation of ethanol.

A – Start with the chemical equation for the
formation of ethanol from its component
elements in their standard states.

_C(graphite) + _H2(g) + _O2(g)  _C2H5OH(l) ΔHf˚= -277 kJmol-1
 Continue by making the coefficient for ethanol 1 because ΔHf˚ is per mole

2 C(graphite) + 3 H2(g) + ½ O2(g)  C2H5OH(l) ΔHf˚= -277 kJmol-1
Sample problem 2

Q – Which of the following does NOT have a
standard heat of formation value of zero at
25˚C and 1.00E5 Pa?
 Cl2(g)
 I2(s)
 Br2(g)
 Na(s)

A – Elements in their STANDARD states have
a zero value. Bromine is a LIQUID in its
standard state, so bromine it its gas state
would have a ΔHf˚ not equal to zero (31
Sample problem 3

Q – Which of the following DOES have a
standard heat of formation value of zero at
25˚C and 1.00E5 Pa?
 H(g)
 Hg(s)
 C(diamond)
 Si(s)

A – Graphite is more stable (but not harder)
than diamond so Si is the only choice in its
standard state. [ΔHf˚ C (diamond) = 1.8 kJ/mol]
Using ΔHf˚
The following expression
is used to predict
the standard enthalpy change
for an entire reaction.
 ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants
 Why does this work?
○ Hess’s Law – see worked example in text pp 147148
ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants
Sample Problem 4
 Calculate the enthalpy change for the reaction
 C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
-105
zero
3(-394)
4(-286) kJ/mol
ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants
Sample Problem 4
 Calculate the enthalpy change for the reaction
 C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
 -105
 ΔH˚reaction
 ΔH˚reaction
zero
3(-394)
4(-286) kJ/mol
= (3(-394)+4(-286))-(-105)
= -2221 kJ/mol
ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants
Sample Problem 5
 Calculate the enthalpy change for the reaction
 C2H5OH (l) + CH3COOH (l) 
CH3COOC2H5 (l) + H2O(g)
-277
kJ/mol
-874
-2238
-286
ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants
Sample Problem 5
 Calculate the enthalpy change for the reaction
 C2H5OH (l) + CH3COOH (l) 
CH3COOC2H5 (l) + H2O(g)
 -277
-874
-2238
-286
kJ/mol
 ΔH˚reaction
 ΔH˚reaction
= [(-2238) + (-286)] – [(-277)+(-874)]
= -2524 – (-1151)= -1373 kJ/mol
ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants
Sample Problem 6
 Calculate the enthalpy change for the reaction
 NH4NO3(s) 
N2O (g) + 2 H2O(g)
 - 366
+ 82
 ΔH˚reaction
 ΔH˚reaction
- 285 kJ/mol
= [(+82) + 2(-285)] - (-366)]
= - 488 – 366 = -122 kJ/mol
Practice Problems with Heats of Formation
 Read
Section 15.1 pp 147-149
 Do Ex 15.1 # 1-3, 8, 10, 11
------------------------------------------------------More Practice – Try Talbot – HL Practice
Heat of Formation: MC 2, 14, 19, 21,
OR 3a