IB syllabus > bonding (sl) > 6.3

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

6.3 - Hess' law

6.3.1: Calculate the enthalpy change of a reaction which is the sum of two or more reactions with known enthalpy changes. Use examples of simple two- and three- step processes. Students should be able to construct simple enthalpy cycles, but will not be required to state Hess' law.

This topic is best addressed using examples

Example 1

Monoclinic sulphur is formed in volcanic regions by reaction between sulphur dioxide and hydrogen sulphide according to the equation:

SO2 + 2H2S --> 2H2O + 3S

Draw an enthalpy diagram or cycle and calculate the standard enthalpy change for this reaction.

Here are some values.

Standard enthalpy of formation (all in kJ)

Standard enthalpy of combustion of S (monoclinic) = -297.2kJ

equation 1: H2 + 1/2O2 --> H2O     -286
equation 2: H2 + S --> H2S     -20.2
equation 3: S + O2 --> SO2     -297.2

multiply E2 by 2
2H2 + 22 --> 2H2S     -40.4

add E3
equation 4: 2H2 + 3S + O2 --> SO2 + 2H2S     -337.6

multiply E1 by 2
equation 5: 2H2 + O2 --> 2H2O     -572 kJ

subtract 4 from 5
- 3S ---> 2H2O - SO2 - 2H2S     -234.4 kJ

SO2 + 2H2S --> 2H2O + 3S     -234.4 kJ


Example 2:

Calculate the enthalpy change in the reaction:

2KHCO3 (s) --------> K2CO3 (s) + CO2 (g) + H2O (l)

Hess' law by manipulation of the equations...

2KHCO3 --> K2CO3 + CO2 + H2O

2K + H2 + 2C + 3O2 --> 2KHCO3 ΔH = 2(-959) = -1918
2K + C + 3/2O2 --> K2CO3 ΔH = -1146


H2 + C + 3/2O2 --> 2KHCO3 - K2CO3 ΔH = -772
H2 + 1/2O2 ---> H2O ΔH = -286


C + O2 --> 2KHCO3 - K2CO3 - H20 ΔH = -486
C + O2 --> CO2 ΔH = -394


ZERO ---> 2KHCO3 - K2CO3 - H20 -CO2 ΔH = -92


K2CO3 + H20 + CO2 --> 2KHCO3 ΔH = -92


2KHCO3 --> K2CO3 + H20 + CO2 ΔH = +92

As all the elements in standard states are assigned an energy value of zero, then to break apart the compound on the left hand side is the opposite of the enthalpy of formation (must be broken apart) Formation of the products molecules is the actual value of the enthalpies of formation.

This makes the enthalpy change

ΔH = sum  ΔHf (Products) - sumΔHf (Reactants)

note that we use the actual values of the sum ΔHf (Products) and the negative of the sumΔHf (Reactants)

does that make it clearer?

you can think of any reactant as having to be broken down into its elements which then must be reformed as the products...

Hence sum of the ΔHf of the left hand side (reactants) but with the opposite sign (because they're being broken up)

Then they are reformed into the products --> hence sum of the ΔHf of the products but with the correct sign this time

Simply add up the two sets of enthalpy sums to give you the reaction enthalpy

- sum ΔHf (Reactants) + sum ΔHf (Products) = Reaction enthalpy

of course the books usually just state the rearranged form

Reaction enthalpy = sum ΔHf (Products) - sum ΔHf (Reactants)

Which does not particularly help in the understanding of WHY this is the case...

Don't worry about the exam just remember that you have to break the reactants down to the elements in the first stage (sum the enthalpies of formation and change sign)and then build them back up to the products in the second stage (sum the enthalpies of formation)

Then add it all up!


Bond breaking = endothermic as you would expect

I made a mistake when I did the original calculation and got an exothermic value. I knew that it could not be the answer as simple inspection tells me that it is a decomposition and is overwhelmingly likely to be endothermic.

Always useful to consider the likelihood of ANY answer.


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