IB Chemistry home > Syllabus 2016 > Energetics > Reaction enthalpy

Syllabus ref: 5.2

Chemical reactions proceed with changes in energy. This energy may either be released or absorbed.

Definition

The enthalpy of reaction is the energy change when a reaction proceeds to completion in the stoichiometric amounts as expressed by the equation.

This definition may lead to confusion when the equation coefficients are greater than one. For example, in the reaction of nitrogen and hydrogen making ammonia the stoichiometric equation is:

N2(g) + 3H2(g) 2NH3(g)

However, the enthalpy of formation of ammonia definition requires that only 1 mole of ammonia is formed

½N2(g) + H2(g) NH3(g)

Clearly, the energy change of formation of ammonia is half that of the enthalpy change of reaction for the balanced equation.

ΔH(reaction) = 2 x ΔHfarr(NH3)

enthalpy change
formation:
1/2N2(g) + 3/2H2(g) NH3(g)
-46 kJ mol-1
reaction:
N2(g) + 3H2(g) 2NH3(g)
-92 kJ mol-1

Reaction enthapy may be determined directly by experiment, or by using other thermodynamic data such as enthalpies of formation, or bond enthalpies.


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Reaction enthalpy from enthalpy of formation values

The enthalpy of formation is: "the energy change when one mole of a substance is made from its constituent elements in their standard states"

A chemical reaction may be expressed in the following way:

Reactants Products

In theory, if the reactants were to be changed to their elements in their standard states this would be the reverse of the enthalpy of formation, i.e. -ΔHf. The values are the same numerically but with the opposite sign.

The products are formed from these same elements, this time the energy changes are their enthapies of formation. The overall reaction is the sum of these two parts.

Reaction enthalpy = - (enthalpy of formation of the reactants) + (enthalpy of formation of the products)

∴ Reaction enthalpy = (enthalpy of formation of the products) - (enthalpy of formation of the reactants)

∑ ΔH(formation) products - ∑ ΔH(formation) reactants

Example: Find the enthalpy of reaction for the decomposition of calcium carbonate, given the following enthalpies of formation:

1     ΔHf (CaCO3) = -1207 kJ mol-1

2     ΔHf2 (CaO) = -635 kJ mol-1

3     ΔHf3 (CO2) = -394 kJ mol-1


To change CaCO3 back into its elements is the reverse of the enthalpy of formation = - (-1207 kJmol-1) = +1207 kJmol-1

2 Formation of CaO, ΔHf2 = -635 kJmol-1, and 3 ΔHf3 of CO2 = -394 kJmol-1

Therefore the enthalpy change for the decomposition = +1207 + (-635 + -394) = +178 kJ

When using this method to calculate the reaction energy, ensure that the states of the compounds given are their standard states. If this is not the case then there will be a further energy step needed.

For example, in the combustion of methane:

CH4(g) + O2(g) CO2(g) + H2O(g)

As the equation as written, you produce water NOT in its standard state, which would be liquid at 25ºC. The equation:

Reaction energy = ∑ ΔH(formation) products - ∑ ΔH(formation) reactants

refers to the formation of water(l). To obtain water(g) from water(l) requires another endothermic process, the enthalpy of vaporisation of water.

H2O(l) H2O(g) ΔH(vap) = +45 kJ mol-1

Example: The standard enthalpy change of formation values of two oxides of phosphorus are:

1 P4(s) + 3O2(g) P4O6(s)   ΔHf = -1600 kJ mo-1
2 P4(s) + 5O2(g) P4O10(s) ΔHf = -3000 kJ mol-1

What is the enthalpy change in kJ mol-1 for the reaction shown below?

P4O6(s) + 2O2(g) P4O10(s)

Equation 1 shows the enthalpy of formation of phosphorus(III) oxide (the reactant)

Equation 2 shows the enthalpy of formation of phosphorus(V) oxide (the product)

REM Oxygen is already an element, so its enthalpy of formation is zero by definition.

ΔH(reaction) = ΔHf(products) - ΔHf(reactants)

∴ ΔH(reaction) = -3000 - (-1600) = -1400 kJ


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Reaction enthalpy from enthalpy of combustion data

Organic hydrocarbons, and compounds containing carbon hydrogen and oxygen, form the same products on complete combustion in excess air, or oxygen. The components of organic molecules, i.e. carbon and hydrogen, have known combustion values. This allows us to perform a similar treatment to that of formation enthalpies and use combustion enthalpy values to find ethalpies of reaction between two organic compounds.

For example if we wish to find the enthalpy change of the reaction:

C2H2(g) + H2(g) C2H4(g) ΔH = ?

It would be impossible to cary out the experiment in practice as the reaction could never be made to stop at ethene; a mixture of products would always ensue. However, combustion of the reactants gives two moles of carbon dioxide and two moles of water:

C2H2(g) + O2(g) CO2 + H2O(g) ΔH =
H2(g) + ½O2(g) H2O(l) ΔH = -268 kJ

This is exactly the same result as complete combustion of the products:

C2H4(g) + 1½O2(g) CO2 + 2H2O(g) ΔH = ?

Hence if we imagine that the combustion products are a kind of stepping stone to the reactants, we have the combustion of the reactants to get to carbon dioxide and water, followed by the reverse of the combustion of the products to go from carbon dioxide and water to the actual products themselves.

The enthalpy change is given by:

Reaction enthalpy = combustion enthalpy of the reactants - combustion enthalpy of the products

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Reaction enthalpy from bond enthalpy data

This is covered in detail under bond enthalpies in section 4.42


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Significance of the ΔH sign

If the sign of the enthalpy change of reaction is positive then it is an endothermic reaction. In other words a reaction that absorbs, or needs energy to proceed. In the example above, the decomposition of calcium carbonate would be expected to be endothermic, as the compound has to be broken apart. This required energy (bond breaking).

A negative sign, on the other hand, indicates that energy is released, i.e. it is an exothermic reaction. In the example below, ammonia formation, -46 kJ of energy is released for every mole of ammonia formed.

½N2(g) + 1½H2(g) NH3(g)ΔH = -46 kJ

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