# 5.3 Enthalpy  (Page 6/25)

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## Standard enthalpy of formation

A standard enthalpy of formation $\text{Δ}{H}_{\text{f}}^{°}$ 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:

$\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}=\text{Δ}{H}_{298}^{°}=-393.5\phantom{\rule{0.2em}{0ex}}\text{kJ}$

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 ), $\text{Δ}{H}_{\text{f}}^{°}$ is 33.2 kJ/mol. This is the enthalpy change for the reaction:

$\frac{1}{2}{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}=\text{Δ}{H}_{298}^{°}=\text{+33.2 kJ}$

A reaction equation with $\frac{1}{2}$ 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, $\text{Δ}{H}_{\text{f}}^{°}$ of ozone from the following information:

$3{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{O}}_{3}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{298}^{°}=\text{+286 kJ}$

## Solution

$\text{Δ}{H}_{\text{f}}^{°}$ 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, $\text{Δ}{H}_{\text{f}}^{°}$ for O 3 ( g ) is the enthalpy change for the reaction:

$\frac{3}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}\left(g\right)$

For the formation of 2 mol of O 3 ( g ), $\text{Δ}{H}_{298}^{°}=\text{+286 kJ.}$ This ratio, $\left(\frac{286\phantom{\rule{0.2em}{0ex}}\text{kJ}}{2\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}\right),$ 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 ):

$\text{Δ}\text{H}\text{° for}\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}\text{mole of}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}\left(g\right)=1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{286\phantom{\rule{0.2em}{0ex}}\text{kJ}}{2\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}}\phantom{\rule{0.2em}{0ex}}=143\phantom{\rule{0.2em}{0ex}}\text{kJ}$

Therefore, $\text{Δ}{H}_{\text{f}}^{°}\left[{\text{O}}_{3}\left(g\right)\right]=\text{+143 kJ/mol}.$

## 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 ${\text{H}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2\text{HCl}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{298}^{°}=-184.6\phantom{\rule{0.2em}{0ex}}\text{kJ}$

#### Questions & Answers

wat is electrolysis?
list the side effect of chemical industries
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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
sunday
also hitting of two atoms can cause transfer of surface electrons
sunday
and when this transfers occur the atom becomes ionised
sunday
who is doing Cape chemistry tomorrow?
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the mix between different breeds of species in one
Jared
it is the blending of orbitals.
stanley
the mixing of orbital
caramel
are covalent bonds influenced by factors such as temperature and pressure?
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when an atom looses electron, what does it become?
it's oxidized and called an ion
Anora
thanks
Abdullahi
Now, I get it
Abdullahi
cation
Anora
can you give an example please, if you don't mind
Abdullahi
a positive ion,become positively charged/a cation.
Janis
sodium plus one is simple cation is exmpl
ajmal
what is copper
just an element
Power
Cu
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can you explain what you are needing it now better than maybe I'm just not interpreting it what you're needing to know
Alex
cool nitrogen down to around negative 270 °F and it will be solid. now they are both solid
daniel
whats a base
A base is a substance which will neutralize an acid to yield salt and water only
Zainab
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...
Roy
a base is a substance that neutralise and acid to form salt and water
Daksalma
write electrolysis of bright solution using either carbon or platinum and write the reaction at the anode or at the cathode
what is the H3O of a solution with the pH of 2.5
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
Alex
if I'm answering and interpreting what you're asking correctly
Alex
When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking
Alex
sorry I don't know why that sent again
Alex
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
Alex
What is chemistry
chemistry is an experimental study of matter and substances which is concerned with the structure composition and the changes these substances undergo
Mohamed
Chemistry=seemyhistory Life of the Past, Present and Future of a noun (person place or things) A compound of 3 words to create one word.
Debra
Chemistry=see my history che=see mi=my stry=story
Debra