Hess's law states that the total enthalpy change for a reaction is independent of the pathway taken. It allows calculation of enthalpy changes that cannot be measured directly by combining thermochemical equations. The standard heat of reaction (ΔH°) can be calculated from standard heats of formation (ΔHf°) of reactants and products. Hess's law and heats of formation are useful for determining enthalpy changes that cannot be measured in the laboratory.

1. Thermochemistry
Calculating Heats of
Reaction
HESS LAW

2. OBJECTIVES:
To understand thermochemical equations
To define Hess’s Law
To use Hess’s Law to find enthalpy changes that we cannot measure
directly
KEY WORDS:
ENTHALPY
HESS’S LAW
COMBUSTION
FORMATION

3. •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.

4. 1- If a certain process has an enthalpy
change Δ H, the reverse of that process
has an enthalpy change of - Δ H.
C(s) + O2(g) → CO2(g) ,ΔH = –393.5 kJ
CO2(g) → C(s) + O2(g) , ΔH = +393.5 kJ
Rules for Using Thermochemical
Equations

5. • Multiplying a thermochemical equation by a constant also
multiplies the thermodynamic quantity by that constant.
C(s) + O2(g) → CO2(g) ,ΔH = –393.5 kJ
2C(s) + 2O2(g) → 2CO2(g) ,
ΔH = 2*(–393.5 )kJ

6. •Hess’s Law states that the total enthalpy change is
independent of the route taken
2NO2(g)
2NO(g) + O2(g)
N2(g) +2O2(g)
ΔHr = -66.4kJmol-1
Route 1
Route 2
-180.8kJ+114.4kJ
Route 2
ΔHr = +114.4 + (-180.8) = -66.4kJmol-1
The total enthalpy change for route 1 is the same as
for route 2
This is a
thermo-
chemical
cycle

7. Hess’s Law
Hess’s Law: is it helpful?
How can you calculate the heat of
reaction when it cannot be directly
measured?
C (s, diamond) → C(s, graphite)

8. 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.

9. • A short video clip explaining Hess' Law -
• http://surfguppy.com/thermodynamics/hess-law/
• https://www.youtube.com/watch?v=QzhcRtf5A1w

10. 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

11. Hess’s Law
What if : we Write equation a in reverse to
give:
c. CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ
Do not forget : 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
b. C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ
C(s, diamond) → C(s, graphite)

12. 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
b. C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ
c. CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ
C(s, diamond) → C(s, graphite)
C (s, diamond) → C(s, graphite)

13. Hess’s Law
Hess’s Law ,
How can you calculate the heat of
reaction when it cannot be directly
measured?
C (s, diamond) → C(s, graphite)

14. 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) → C(s, graphite)
C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ
CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ
C(s, diamond) → C(s, graphite) ΔH = –1.9 kJ

15. Hess’s Law
C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ
CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ
C(s, diamond) → C(s, graphite) ΔH = –1.9 kJ

16. How can you determine ΔH for the conversion of
diamond to graphite without performing the
reaction?
CHEMISTRY & YOU

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

18. • 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 Law
Another 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

19. Hess’s Law
You 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

20. 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

21. •According to Hess’s law, it is possible to calculate an
unknown heat of reaction by using which of the
following?
A. heats of fusion for each of the compounds in
the reaction
B. two other reactions with known heats of
reaction
C. specific heat capacities for each compound in
the reaction
D. density for each compound in the reaction

22. •According to Hess’s law, it is possible to calculate an
unknown heat of reaction by using which of the
following?
A. heats of fusion for each of the compounds in
the reaction
B. two other reactions with known heats of
reaction
C. specific heat capacities for each compound in
the reaction
D. density for each compound in the reaction

23. Heats of
Formation
Standard Heats of Formation
How can you calculate the heat of
reaction when it cannot be directly
measured?

24. Heats of
Formation
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.

25. Heats of
Formation
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).

26. Interpret Data
Standard Heats of Formation (ΔHf°) at 25°C and 101.3 kPa
Substance
ΔHf°
(kJ/mol)
Substance
ΔHf°
(kJ/mol)
Substance
ΔHf°
(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

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

28. Heats of
Formation
• Such an enthalpy change is called
the standard heat of reaction (ΔH°).
For a reaction that occurs at standard
conditions, you can calculate the heat
of reaction by using standard heats of
formation.

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

30. Heats of
Formation
•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.

31. •What is the standard heat of reaction
(ΔH°) for the reaction of CO(g) with
O2(g) to form CO2(g)?
Sample Problem 17.8
Calculating the Standard Heat
of Reaction

32. 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
Sample Problem 17.8
•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.

33. First write the balanced equation.
Calculate Solve for the unknown.2
Sample Problem 17.8
2CO(g) + O2(g) → 2CO2(g)

34. Find and add ΔHf° of all the reactants.
Calculate Solve for the unknown.2
Remember to take into
account the number of moles
of each reactant and product.
Sample Problem 17.8
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 kJ
2 mol CO(g)
0 kJ
1 mol O2(g)

35. Find ΔHf° of the product in a similar way.
Calculate Solve for the unknown.2
Remember to take into
account the number of moles
of each reactant and product.
Sample Problem 17.8
ΔHf°(products) = 2 mol CO2(g) 
ΔHf°CO2(g)
= 2 mol CO2(g) 
= –787.0 kJ
–393.5 kJ
1 mol CO2(g)

36. Calculate ΔH° for the reaction.
Calculate Solve for the unknown.2
Sample Problem 17.8
ΔH° = ΔHf°(products) – ΔHf°(reactants)
= (–787.0 kJ) – (–221.0 kJ)
= –566.0 kJ

37. • 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
Sample Problem 17.8

38. Heats of
Formation
•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.

39. 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

40. 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
ΔHf°(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) = –

41. 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.
ΔH° = ΔHf°(products-ΔHf°(reactants)

42. 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

43. •The heat of reaction can be calculated by using the
known heats of reaction of two or more
thermochemical equations or by using standard
heats of formation.
BIG IDEA
Matter and Energy

44. 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.