Hess’s Law
SECTION 5.3
Hess’s Law
 The
enthalpy change of a physical or
chemical process depends only on
the initial and final conditions of the
process.
 The
enthalpy change of a multistep
process is the sum of the enthalpy
changes of its individual steps.
Problem

Iron can be obtained from the following reaction:

Fe2O3(s) + 3CO(g) → 3CO2 + 2Fe(s)

Determine the enthalpy change of reaction, given the
following:
1.
CO(g) + ½ O2 → CO2 ∆H°=-283.0kJ
2.
2Fe(s) + 3/2 O2 → Fe2O3(s) ∆H°=- 824.3kJ
Solution

Manipulate equations to make them match the overall equation:

#1 equation proceeds in the correct direction but must be
multiplied by 3.

So:

CO(g) + ½ O2 → CO2 ∆H°=-283.0kJ
X3

3CO(g) + 3/2 O2 → 3CO2 ∆H°=-849.0 kJ

#2 equation needs to be reversed but the coefficients are ok.

So:

2Fe(s) + 3/2 O2 → Fe2O3(s) ∆H°= - 824.3kJ
becomes:

Fe2O3(s) → 2Fe(s) + 3/2 O2 ∆H°= 824.3kJ (note the sign is reversed)

To get the overall equation:

Since oxygen is the same on both sides of the equation,
cross it out and add the individual enthalpy changes:

Fe2O3(s) + 3CO(g) → 3CO2 + 2Fe(s) ∆H°= -24.7 kJ
Practice Problems – p. 316 – all
questions
Standard Molar Enthalpies of
Formation

Change in enthalpy when 1 mole of a compound is
formed directly from its elements in their most stable
state at SATP (25°C & 100kPa).

It’s a synthesis reaction when the compound is formed
directly from its elements and not from any other
compound.
 Eg.
C(s) + O2(g) → CO2 (g)

Coefficients are often fractions because there must be
only 1 mole of product formed.

See Table 5.5 and Appendix B
Thermal Stability

The ability of a substance to resist
decomposition when heated.

The greater the enthalpy change of a
decomposition reaction, the greater its thermal
stability.
Enthalpies of Formation and Hess’s
Law

∆H°r =∑(n∆H°f products) - ∑(n ∆H°f reactants)

n= coefficient in equation

∑ = “sum of”

∆H°r = enthalpy change of reaction

Use standard molar enthalpies of formation on Appendix
B
Determine the enthalpy of formation for the
following reaction:

CH4(g) + 2O2 → CO2 + 2H2O

∆H°r =∑(n∆H°f products) - ∑(n ∆H°f reactants)
**Use Appendix B

∆H°r =((1)(-393.2) + (2)(-241.9)) - ((1)(-74.6)(2)(0)) ** since
oxygen is an element in its most stable state, standard molar
enthalpy is 0.

= (-877.1kJ)-(-74.6kJ)

= -802.5 kJ
Videos/Resources to study at home

Hess’s Law:

http://www.ausetute.com.au/hesslaw.html

https://www.youtube.com/watch?v=u7aTBxA7sL8

Hess’s Law and Enthalpies of Formation

http://www.chemguide.co.uk/physical/energetics/sums.html

https://www.youtube.com/watch?v=iQuy2mgbV9o
Practice Problems – p. 323 – #51,
52, 54, 55, 56, 58.

Research Application Questions

Choose from:

P. 291: #4, 10, 12

P. 311: #8, 12

P. 335: #2, 5, 6, 7, 8

P. 349: #62, 63, 65, 66, 67, 69, 70

P. 351: #23

P. 408-409: #68, 69