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Standard enthalpy of formation

A standard enthalpy of formation Δ H f ° is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess’s law.

The standard enthalpy of formation of CO 2 ( g ) is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:

C ( s ) + O 2 ( g ) CO 2 ( g ) Δ H f ° = Δ H 298 ° = −393.5 kJ

starting with the reactants at a pressure of 1 atm and 25 °C (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of CO 2 , also at 1 atm and 25 °C. For nitrogen dioxide, NO 2 ( g ), Δ H f ° is 33.2 kJ/mol. This is the enthalpy change for the reaction:

1 2 N 2 ( g ) + O 2 ( g ) NO 2 ( g ) Δ H f ° = Δ H 298 ° = +33.2 kJ

A reaction equation with 1 2 mole of N 2 and 1 mole of O 2 is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, NO 2 ( g ).

You will find a table of standard enthalpies of formation of many common substances in Appendix G . These values indicate that formation reactions range from highly exothermic (such as −2984 kJ/mol for the formation of P 4 O 10 ) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, C 2 H 2 ). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions.

Evaluating an enthalpy of formation

Ozone, O 3 ( g ), forms from oxygen, O 2 ( g ), by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation, Δ H f ° of ozone from the following information:

3 O 2 ( g ) 2 O 3 ( g ) Δ H 298 ° = +286 kJ


Δ H f ° is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus, Δ H f ° for O 3 ( g ) is the enthalpy change for the reaction:

3 2 O 2 ( g ) O 3 ( g )

For the formation of 2 mol of O 3 ( g ), Δ H 298 ° = +286 kJ. This ratio, ( 286 kJ 2 mol O 3 ) , can be used as a conversion factor to find the heat produced when 1 mole of O 3 ( g ) is formed, which is the enthalpy of formation for O 3 ( g ):

Δ H ° for 1 mole of O 3 ( g ) = 1 mol O 3 × 286 kJ 2 mol O 3 = 143 kJ

Therefore, Δ H f ° [ O 3 ( g ) ] = +143 kJ/mol .

Check your learning

Hydrogen gas, H 2 , reacts explosively with gaseous chlorine, Cl 2 , to form hydrogen chloride, HCl( g ). What is the enthalpy change for the reaction of 1 mole of H 2 ( g ) with 1 mole of Cl 2 ( g ) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl( g ) is −92.3 kJ/mol.


For the reaction H 2 ( g ) + Cl 2 ( g ) 2 HCl ( g ) Δ H 298 ° = −184.6 kJ

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Questions & Answers

wat is electrolysis?
Mgbachi Reply
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Chelsea Reply
how do you ionise an atom
Rabeka Reply
many ways ,but one of them is when the atom becomes heated to a certain temperature the surface electron becomes too energetic and leaves the atom because the attraction between the nucleus and the electron becomes overpowered by the energetic eletron
also hitting of two atoms can cause transfer of surface electrons
and when this transfers occur the atom becomes ionised
who is doing Cape chemistry tomorrow?
caramel Reply
What is hybridization
edmondnti Reply
the mix between different breeds of species in one
it is the blending of orbitals.
the mixing of orbital
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patrick Reply
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Sanjay Reply
when an atom looses electron, what does it become?
Abdullahi Reply
it's oxidized and called an ion
Now, I get it
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a positive ion,become positively charged/a cation.
sodium plus one is simple cation is exmpl
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Bryan Reply
just an element
Why is water a single covalent bond?
Mohamed Reply
nitrogen is a gas whereas phosphorus is solid .Explain.
Jacky Reply
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cool nitrogen down to around negative 270 °F and it will be solid. now they are both solid
whats a base
Daksalma Reply
A base is a substance which will neutralize an acid to yield salt and water only
base is a substance that produces OH(aq) ions in aqueous solution. Strong soluable bases are in water and are completely dislocated. Therefore weak base ionize slightly...
a base is a substance that neutralise and acid to form salt and water
write electrolysis of bright solution using either carbon or platinum and write the reaction at the anode or at the cathode
Abdullah Reply
what is the H3O of a solution with the pH of 2.5
Sgt.Elliott_98 Reply
pH<7, therefore there are only H3O+HX3OX+particles in the solution. [H3O+]=10−pH=10−6.99=1.02⋅10−7[HX3OX+]=10−pH=10−6.99=1.02⋅10−7 When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking  and I'll give you another one if this is ki
if I'm answering and interpreting what you're asking correctly
When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking 
sorry I don't know why that sent again
We have [H3O+]=10−pH=10−6.99=1.02⋅10−7[HX3OX+]=10−pH=10−6.99=1.02⋅10−7 and [OH−]=10−pOH=10−7.01=9.77⋅10−8[OHX−]=10−pOH=10−7.01=9.77⋅10−8.  Because of H3O++OH−⟶2H2OHX3OX++OHX−⟶2HX2O we are left with [H3O+]=1.02⋅10−7−9.77⋅10−8=4.6⋅10−9
What is chemistry
Smile's Reply
chemistry is an experimental study of matter and substances which is concerned with the structure composition and the changes these substances undergo
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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