thermochemistry heats of formation and calculating heats of reaction

Thermochemistry

Heats of Formation and Calculating Heats of Reaction

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•How much heat is released when a diamond changes into graphite?

Diamonds are gemstones composed of carbon. Over a time period of millions and millions of years, diamond will break down into graphite, which is another form of carbon.


Hess’s Law

Hess’s LawHow can you calculate the heat of reaction when it cannot be directly measured?


Hess’s Law

•Hess’s law of heat summation states that if you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.


Hess’s Law

Hess’s law allows you to determine the heat of reaction indirectly by using the known heats of reaction of two or more thermochemical equations.


C(s, diamond) → C(s, graphite)

Hess’s Law

Although the enthalpy change for this reaction cannot be measured directly, you can use Hess’s law to find the enthalpy change for the conversion of diamond to graphite by using the following combustion reactions.

a. C(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ

b. C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ


Hess’s Law

Write equation a in reverse to give:

c. CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ

When you reverse a reaction, you must also change the sign of ΔH.

a. C(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ




Hess’s Law

If you add equations b and c, you get the equation for the conversion of diamond to graphite.

C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ

CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ



c. CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ



Hess’s Law

If you also add the values of ΔH for equations b and c, you get the heat of reaction for this conversion.


C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ


C(s, diamond) → C(s, graphite) ΔH = –1.9 kJ


Hess’s LawC(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ


C(s, diamond) → C(s, graphite) ΔH = –1.9 kJ


•How can you determine ΔH for the conversion of diamond to graphite without performing the reaction?

CHEMISTRY & YOUCHEMISTRY & YOU


•How can you determine ΔH for the conversion of diamond to graphite without performing the reaction?

CHEMISTRY & YOUCHEMISTRY & YOU

You can use Hess’s law by adding thermochemical equations in which the enthalpy changes are known and whose sum will result in an equation for the conversion of diamond to graphite.


• Suppose you want to determine the enthalpy change for the formation of carbon monoxide from its elements.

• Carrying out the reaction in the laboratory as written is virtually impossible.

Hess’s LawAnother case where Hess’s law is useful is when reactions yield products in addition to the product of interest.

C(s, graphite)+ O2(g) → CO(g) ΔH = ? 1 2


Hess’s LawYou can calculate the desired enthalpy change by using Hess’s law and the following two reactions that can be carried out in the laboratory:

C(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ

CO2(g) → CO(g) + O2(g) ΔH = 283.0 kJ

C(s, graphite)+ O2(g) → CO(g) ΔH = –110.5 kJ

1 2

1 2


Hess’s LawC(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ

CO2(g) → CO(g) + O2(g) ΔH = 283.0 kJ

C(s, graphite)+ O2(g) → CO(g) ΔH = –110.5 kJ

1 2

1 2


Standard Heats of Formation

Standard Heats of FormationHow can you calculate the heat of reaction when it cannot be directly measured?



Enthalpy changes generally depend on the conditions of the process.

• Scientists specify a common set of conditions as a reference point.

• These conditions, called the standard state, refer to the stable form of a substance at 25°C and 101.3 kPa.



The standard heat of formation (ΔHf°) of a compound is the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.• The ΔHf° of a free element in its standard state

is arbitrarily set at zero.

• Thus, ΔHf° = 0 for the diatomic molecules H2(g), N2(g), O2(g), F2(g), Cl2(g), Br2(l), and I2(s).


Standard Heats of Formation (ΔHf°) at 25°C and 101.3 kPa

SubstanceΔHf° (kJ/mol)



Al2O3(s) –1676.0 F2(g) 0.0 NO(g) 90.37

Br2(g) 30.91 Fe(s) 0.0 NO2(g) 33.85

Br2(l) 0.0 Fe2O3(s) –822.1 NaCl(s) –411.2

C(s, diamond) 1.9 H2(g) 0.0 O2(g) 0.0

C(s, graphite) 0.0 H2O(g) –241.8 O3(g) 142.0

CH4(g) –74.86 H2O(l) –285.8 P(s, white) 0.0

CO(g) –110.5 H2O2(l) –187.8 P(s, red) –18.4

CO2(g) –393.5 I2(g) 62.4 S(s, rhombic) 0.0

CaCO3(s) –1207.0 I2(s) 0.0 S(s, monoclinic) 0.30

CaO(s) –635.1 N2(g) 0.0 SO2(g) –296.8

Cl2(g) 0.0 NH3(g) –46.19 SO3(g) –395.7



For a reaction that occurs at standard conditions, you can calculate the heat of reaction by using standard heats of formation.



• Such an enthalpy change is called the standard heat of reaction (ΔH°).




• The standard heat of reaction is the difference between the standard heats of formation of all the reactants and products.

ΔH° = ΔHf°(products) – ΔHf°(reactants)




•This enthalpy diagram shows the standard heat of formation of water.

• The enthalpy difference between the reactants and products, –285.8 kJ/mol, is the standard heat of formation of liquid water from the gases hydrogen and oxygen.

• Notice that water has a lower enthalpy than the elements from which it is formed.


•What is the standard heat of reaction (ΔH°) for the reaction of CO(g) with O2(g) to form CO2(g)?

Calculating the Standard Heat of Reaction


Analyze List the knowns and the unknown.1

KNOWNS UNKNOWN

ΔHf°CO(g) = –110.5 kJ/mol

ΔHf°O2(g) = 0 kJ/mol (free element)

ΔHf°CO2(g) = –393.5 kJ/mol

ΔH° = ? kJ

•Balance the equation of the reaction of CO(g) with O2(g) to form CO2(g). Then determine ΔH° using the standard heats of formation of the reactants and products.


First write the balanced equation.

Calculate Solve for the unknown.2

2CO(g) + O2(g) → 2CO2(g)


Find and add ΔHf° of all the reactants.


Remember to take into account the number of moles of each reactant and product.

ΔHf°(reactants) = 2 mol CO(g) ΔHf°CO(g) + 1 mol O2(g) ΔHf°O2(g)

= 2 mol CO(g) + 1 mol O2(g)

= –221.0 kJ

–110.5 kJ2 mol CO(g)

0 kJ1 mol O2(g)


Find ΔHf° of the product in a similar way.


Remember to take into account the number of moles of each reactant and product.

ΔHf°(products) = 2 mol CO2(g) ΔHf°CO2(g)

= 2 mol CO2(g)

= –787.0 kJ

–393.5 kJ

1 mol CO2(g)


Calculate ΔH° for the reaction.



= (–787.0 kJ) – (–221.0 kJ)

= –566.0 kJ


• The ΔH° is negative, so the reaction is exothermic.

• This outcome makes sense because combustion reactions always release heat.

Evaluate Does the result make sense?3



•Standard heats of formation are used to calculate the enthalpy change for the reaction of carbon monoxide and oxygen.

• 2CO(g) + O2(g) → 2CO2(g)

• The diagram shows the difference between ΔHf°(product) and ΔHf°(reactants) after taking into account the number of moles of each.


Calculate the standard heat of reaction for the following:

CH4(g) + Cl2(g) → C(s, diamond) + 4HCl(g)

ΔHf°(CH4(g)) = –74.86 kJ/mol

ΔHf°(C(s, diamond)) = 1.9 kJ/mol

ΔHf°(HCl(g)) = –92.3 kJ/mol


Calculate the standard heat of reaction for the following:

CH4(g) + Cl2(g) → C(s, diamond) + 4HCl(g)

ΔHf°(CH4(g)) = –74.86 kJ/mol

ΔHf°(C(s, diamond)) = 1.9 kJ/mol

ΔHf°(HCl(g)) = –92.3 kJ/molΔHf°(reactants) = [1 mol CH4(g) ΔHf°CH4(g)] + [1 mol Cl2 ΔHf°Cl2(g)]

= –74.86 kJ + 0.0 kJ = –74.86 kJ

ΔHf°(products) = [1 mol C(s) ΔHf°C(s, diamond)] + [4 mol HCl ΔHf°HCl(g)]

= 1.9 kJ + (4 –92.3 kJ) = –367.3 kJ

ΔH° = ΔHf°(products) – ΔHf°(reactants) = –367.3 kJ – (–74.86 kJ) = –292.4 k


Key Concepts & Key Equation

•Hess’s law allows you to determine the heat of reaction indirectly by using the known heats of reaction of two or more thermochemical equations.

•For a reaction that occurs at standard conditions, you can calculate the heat of reaction by using standard heats of formation.



Glossary Terms

• Hess’s law of heat summation: if you add two or more thermochemical equations to give a final equation, then you also add the heats of reaction to give the final heat of reaction

• standard heat of formation (ΔHf°): the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 25°C

thermochemistry heats of formation and calculating heats of reaction

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