Standard Enthalpy Change of Reaction
15.1.1
15.1.2
15.1.1 – Define and apply the terms
standard enthalpy change of
formation (ΔHf˚) and standard
enthalpy of combustion (ΔHc˚).
Q – What are the “standard” conditions?
 A – 298K (25˚C) , 1.00E5 Pa (100 kPa) or
around room temp and pressure

15.1.1 – Define and apply the terms
standard enthalpy change of
formation (ΔHf˚) and standard
enthalpy of combustion (ΔHc˚).
 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)
 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

2C(graphite) + 3H2(g) + 1/2O2(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.
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.
Using ΔHf˚

The following expression is used to predict
the standard enthalpy change for an entire
reaction.
 ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants
 Sample Problem
 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
 ΔH˚reaction
= (3(-394)+4(-286))-(-105)
= -2221 kJ/mol
15.1.1 – Define and apply the terms
standard enthalpy change of
formation (ΔHf˚) and standard
enthalpy of combustion (ΔHc˚).
 ΔHc˚=
the enthalpy change that occurs
when 1 mol of a substance burns
completely under standard conditions (see
table 12 of IB data booklet)
 Balance these equations for one mole of the
reactant instead of the product.
15.1.2

Determine the enthalpy change of a
reaction using standard enthalpy
changes of formation and combustion.
 ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants
 ΔH˚reaction= ΔHc˚reactants -  ΔHc˚products

They are “flipped” because formation
focuses on the energy used in creation
and combustion focuses on the energy
used in destruction.
Summing up

So now we have several ways to predict
the enthalpy changes of a chemical
reaction
 Using Hess’s Law
 Average Bond Enthalpies
○ ΔH˚= bonds broken-  bonds formed
 Standard Enthalpies of formation
○ ΔH˚reaction= ΔHf˚products -  ΔHf˚reactants
 Standard Enthalpies of combustion
○ ΔH˚reaction= ΔHc˚reactants -  ΔHc˚products
HW (Due Monday)

Do # 6 from you Hess Lab post lab
questions and turn the whole thing in
tomorrow.