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Hess’s cycle, Enthalpy change of formation and Enthalpy change of combustion: 

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  • Hess’s Law states that the enthalpy change accompanying chemical change is independent of the route taken. Therefore the enthalpy change for the overall process will be identical no matter how many steps are taken.

 

  • Hess cycle uses the concepts of Hess’sss’s law to find the unknown enthalpy value needed in a chemical reaction.

 

  • The units for enthalpy change are kJ/mol. 

 

  • There are many types of enthalpy which you can use the Hess cycle to work out and two of the major ones which come up in OCR exams are enthalpy change of formation and enthalpy change of combustion

Enthalpy change of formation:

 

  • Using Hess's cycle we can work out the enthalpy change of formation which is when 1 mole of a substance is formed from its element under standard conditions. 

  •  This can be represented by the symbol: ΔH f

Using a worked exam of a past exam question we can understand how to find the unknown enthalpy using enthalpy of formation: 

 

Use the standard enthalpy change of formation data from the table and the equation for the extraction of manganese to calculate a value for the standard enthalpy change of the extraction of manganese from Mn2O3 by reduction with carbon monoxide. 

 

 

Step 1: 

  • Work out ΔH f. Of the reactants using the equation

  • Mn2O3 = -971 , 3CO = 3 x (-111) Overall ΔH f = -1304 

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Step 2: 

  • Work out ΔH f. Of the products using the equation

  • 2Mn = 2(0) , 3CO2 = 3(-394)  Overall ΔH f = -1182 

 

Step 3: Draw a Hess cycle 

 

  • Because it's enthalpy change of formation (when 1 mole of a substance is formed from its elements) the arrows will go from the elements to the substances as shown below

  • The arrow between the reactants and products represents the value of the unknown ΔH 

Mn2O3(s) + 3CO → 2Mn(s) + 3CO2(g)

Step 4: 

Add  the values for ΔH f to the cycle 

Step 5: 

 

  • Follow the path of the arrows and rearrange the equation to find ΔH?

  • -1304 + ΔH? = -1182 

 

Therefore ΔH? = -1182 + 1304 = 122

ΔH = 112 kJ/mol (because it’s a positive value the reaction is endothermic energy stored is positive) 

Enthalpy change of combustion: 

 

  • Enthalpy change of combustion is when 1 mole of a substance burns/ reacts completely with oxygen under standard conditions.

  • Enthalpy change of combustion can be represented using the symbol:  ΔH∘C

We can use a worked example of a past exam question to work out the enthalpy change using enthalpy of combustion values:

 

 The equation below shows the formation of buta-1,3-diene, C4H6 

 

4C(s) + 3H2(g) → C4H6 (g) 

 

Using the following data workout the standard enthalpy of formation of buta-1,3-diene. 

For this type of question you are given the enthalpy change of combustion of some reactions and with this information you need to construct a Hess cycle and work out the enthalpy of formation of buta-1,3-diene. 

 

 

 

Step 1: Work out what will be made when burning the reaction completely in oxygen 

 

4C(s) + 3H2(g) → C4H6 (g) 

 

  • Burning this will make the same thing however you burning it with oxygen therefore  oxygen will be included as a result you would make CO2 and H2O 

 

  • Then you will have to balance what you're making with the equation you can use either of the reactant or product values and you should get 4CO2 + 3H2O.


 

Step 2: Draw a diagram with arrows coming from the reaction to what you are making from the reaction as shown below. This step helps you understand what the reaction goes to. For combustion enthalpy the arrows will always go down. 

Step 3: Add the enthalpy changes of combustion to the diagram from the question 

  • 4C = 4(-394) = -1576

  • 3H2 = 3(-286) = -858

  • C4H6 = -2542 

Step 4: 

Swap the arrow of the C4H6 so there's a direct route going from the reactants to the products via the carbon dioxide and water and add the unknown enthalpy sign that you're finding.  

 

IMPORTANT: Remember when swapping arrows around the sign also has to change therefore in this case the -2542 will become +2542.

Step 5: 

Follow the path of arrows and construct an equation to find ΔH. 

 

-1576 + -858 + 2542 = ΔH

So ΔH = 108 kJ/mol (This reaction is endothermic because ΔH is positive)

Author: Natasha fuller

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