3.
12 Hess’s Law
Hess’s law states that total enthalpy change for a reaction is Hess’s law is a version of the first law
independent of the route by which the chemical change takes place of thermodynamics, which is that
energy is always conserved.
2H (g) + 2Cl(g)
On an energy level diagram the directions of the arrows can
show the different routes a reaction can proceed by
Energy
a The two routes must have the same starting point and the
same end point
H2 + Cl2 b
In this example one route is arrow ‘a’
ΔH
The second route is shown by arrows ΔH plus arrow ‘b’
2HCl (g)
So by applying Hess’s law
a = ΔH + b
And rearranged
ΔH = a - b
Interconnecting reactions can also be shown diagrammatically.
In this example one route is arrow ‘a’ plus ΔH
ΔH
H+ -
(g) + Br (g) H+ -
(aq) + Br (aq) The second route is shown by arrows ‘c’ plus arrow ‘d’
a So a+ ΔH = c + d
d
And rearranged
H (g) + Br (g) HBr (g) ΔH = c + d - a
Often Hess’s law cycles are used to measure the enthalpy change for a reaction that cannot be measured
directly by experiments. Instead alternative reactions are carried out that can be measured experimentally.
∆H reaction This Hess’s law is used to work out the
CuSO4 (s) + 5H2O (l) CuSO4.5H2O (s) enthalpy change to form a hydrated salt
from an anhydrous salt.
-66.1 kJmol-1 + aq + aq +11kJmol-1 This cannot be done experimentally
because it is impossible to add the exact
CuSO4 (aq) amount of water to hydrate the copper
salt and it is not easy to measure the
temperature change of a solid turning into
∆H reaction + 11kJmol-1 = -66.1 kJmol-1 another solid.
∆H reaction = -66.1 - 11 Instead both salts are dissolved in excess
water to form a solution of copper sulphate.
= -77.1 kJmol-1
The temperature changes can be measured
for these alternative reactions and by using
Hess’s law the ∆H for the original reaction
can be calculated.
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∆H reaction This Hess’s law is used to work out the
CaCO3 (s) CaO (s) + CO2 (g) enthalpy change for the thermal
decomposition of calcium carbonate.
ΔH1 ΔH2
(+2HCl) (+2HCl) This cannot be done experimentally
because it is impossible to add the add
CaCl2 (aq) + H2O (l) + CO2 (g) the heat required to decompose the
solid and to measure the temperature
change of a solid at the same time.
∆H reaction + ΔH2 = ΔH1
Instead both calcium carbonate and calcium
∆H reaction = ΔH1 - ΔH2 oxide are reacted with hydrochloric acids to
form a solution of calcium chloride. The
temperature changes can be measured for
these reactions.
Using Hess’s law to determine enthalpy changes from enthalpy changes of formation.
This is a very common Hess’s law cycle using
standard enthalpies of formation to work out the ∆H reaction
enthalpy change for any chemical reaction. Reactants Products
Looking at the cycle on the right we can see the
two routes shown with the different coloured Σ ∆fH reactants Σ ∆fH products
arrows, therefore the equation is
Elements in standard states
Σ ∆fH reactants + ∆H reaction = Σ ∆fH products
As this cycle is always the same we can just use
the equation below in questions.
Standard enthalpies of formation can be
looked up in data books for all
∆H reaction = Σ ∆fH products - Σ ∆fH reactants compounds
Example 1 . What is the enthalpy change for this reaction ?
Al2O3 + 3 Mg 3 MgO + 2 Al ∆H reaction
Al2O3 (s) + 3 Mg (s) 3 MgO (s) + 2 Al (s)
∆fH (MgO)= -601.7 kJ mol-1
∆fH (Al2O3) = -1675.7 kJ mol-1
∆fH (Al2O3) 3 x ∆fH (MgO)
Remember elements have ∆fH = 0
2Al (s) + 3 Mg (s) + 1.5O2 (g)
∆H = Σ ∆fH products – Σ ∆fH reactants
By application of Hess’s Law
∆H = 3 x ∆fH (MgO) – ∆fH (Al2O3)
∆fH(Al2O3) + ∆H reaction = 3 x ∆fH (MgO)
∆H = (3 x –601.7) – –1675.7
= -129.4 kJ mol-1
2Al (s) + 3 Mg (s) + 1.5O2 (g)
Energy
Note because there are –1675.7
3 MgO in balanced 3 x –601.7
equation then we The calculation can also be
multiply the ∆Hf (MgO) shown diagrammatically
Al2O3 (s) + 3 Mg (s)
by 3 using an energy level
diagram, -129.4 3 MgO (s) + 2 Al (s)
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Example 2. Using the following data to calculate the enthalpy of combustion of propene
∆fH C3H6(g) = +20 kJ mol-1 ∆fH CO2(g)= –394 kJ mol-1 ∆fH H2O(g)= –242 kJ mol-1
C3H6 + 4.5 O2 3CO2 + 3H2O
This question is identical to example 1 in how to approach it. The only difference is the enthalpy change of
reaction has been defined as an enthalpy of combustion. Many students get confused because they see the
words enthalpy of combustion and enthalpy of formation in the same question and do not know what cycle to
use. Draw it out to check.
∆cH
∆cH = Σ ∆fH products – Σ ∆fH reactants C3H6 (g) + 4.5 O2 (g) 3 CO2 (g) + 3 H2O (g)
∆cH = [3 x ∆fH (CO2) + 3 x ∆fH (H2O)] - ∆fH (C3H6) 3 x ∆fH (CO2)
∆fH (C3H6)
∆cH = [(3 x –394) + (3 x –242)] – 20 3 x ∆fH (H2O)
= -1928 kJ mol-1
3C (s) + 3 H2 (g) + 4.5 O2 (g)
Using Hess’s law to determine enthalpy changes from enthalpy changes of combustion.
This is another common cycle were the alternative ∆H reaction
routes involves combusting all the reactants and Reactants Products
products in oxygen and using the enthalpies of
combustion to calculate the overall enthalpy change of Σ∆cH reactants Σ∆cH products
reaction. This cycle can only be used if all the reactants
and products can be combusted in oxygen.
Combustion Products
∆H reaction = Σ ∆cH reactants - Σ∆
∆c Hproducts
Example 3. Using the following enthalpy combustion data to calculate the enthalpy of reaction for the
following reaction
CO (g) + 2H2 (g) CH3OH (g)
∆cH CO(g) = -283 kJ mol-1 ∆cH H2 (g)= –286 kJ mol-1 ∆cH CH3OH(g)= –671 kJ mol-1
∆H reaction
∆H reaction = Σ∆cH reactants - Σ∆cH products CO (g) + 2H2(g) CH3OH(g)
∆H = ∆CH (CO) + 2 x ∆CH (H2) - ∆CH (CH3OH) [+ O2] [+ O2]
∆CH (CO) + ∆CH (CH3OH)
∆H = -283+ 2x –286 - -671 2x ∆CH (H2)
CO2 (g) + 2 H2O (l)
= -184 kJ mol-1 Be really careful
with your minus
signs
CO (g) + 2H2(g)
Energy
-184
Note: unlike enthalpies of CH3OH(g)
formation, elements do
-283+ 2x –286
have an enthalpy of The calculation can also be
combustion value. shown diagrammatically -671
Burning hydrogen in using an energy level CO2 (g) + 2 H2O (l)
oxygen is a very diagram,
exothermic reaction!
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Example 4. Using the following combustion data to calculate the heat of formation of propene
3C (s) + 3H2 (g) C3H6 (g)
∆cH C (s) = -393kJ mol-1 ∆cH H2 (g)= –286 kJ mol-1 ∆cH C3H6(g)= –2058 kJ mol-1
This another example where both enthalpies of combustion and formation appear in the same
question. When constructing the original equation remember to only have one mole of the
compound on the right hand side of the equation because by definition the enthalpy of formation
is for 1 mole of the compound formed.
∆f H
∆fH = Σ ∆cH reactants - Σ ∆cH products
3C (s) + 3H2 (g) C3H6 (g)
∆fH = 3 x ∆cH(C) + 3 x ∆cH(H2 ) - ∆cH(C3H6)
3 x ∆cH(C) + ∆cH(C3H6)
∆fH = 3x -393+ 3x –286 - -2058 3 x ∆cH(H2 )
= +21 kJ mol-1 3 CO2 (g) + 3 H2O (l)
You may well notice I am using the notation ∆fH and ∆CH rather than ∆Hf and ∆HC. I have been
using the latter for many years and you will find it in almost every text book. It is actually incorrect
though. The reason is in the name. ∆fH is used for enthalpy change of formation. The definition
refers to change in enthalpy and ∆ is the sign for change so the f or c goes next to the ∆. H means
enthalpy and there is no such thing as Hf of Hc only H. On the new syllabus AQA are using ∆fH and
∆CH so I am changing all my notes to agree with this.
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Questions about Hess’s Law
1 A student used Hess’s Law to determine a value for the enthalpy change that occurs when anhydrous
copper(II) sulfate is hydrated. This enthalpy change was labelled ΔHreaction
∆H reaction
Anhydrous hydrated copper(II)
copper(II) sulphate sulphate
+ aq + aq
∆H1 ∆H2
copper(II)
sulphate solution
1 (a) State Hess’s Law.
1 (b) Write a mathematical expression to show how ΔHreaction, ΔH1 and ΔH2 are related to each
other by Hess’s Law.
1 (c) Using the values below for the two enthalpy changes ΔH1 and ΔH2, calculate a value for
ΔHreaction
ΔH1 = –156 kJ mol–1
ΔH2 = +12 kJmol–1
2) Using for following table of date and scheme of reactions, calculate a value for ΔHreaction
Reaction Enthalpy change ΔHreaction
kJ mol-1 H+ (g) +Cl - (g) H+ (aq) + Cl - (aq)
HCl (g) H+ (aq) + Cl-(aq) -75
H (g) + Cl (g) HCl (g) -432
H (g) + Cl (g) HCl (g)
H (g) + Cl (g) H+ (g) + Cl- (g) +963
3) On strong heating, zinc carbonate decomposes to zinc oxide and carbon dioxide:
ZnCO3(s) ZnO(s) + CO2(g)
Owing to the conditions under which the reaction occurs, it is not possible to measure the enthalpy change
directly. An indirect method uses the enthalpy changes when zinc carbonate and zinc oxide are
neutralized with hydrochloric acid.
(i) Write the equation for the reaction of zinc carbonate with hydrochloric acid. And label this ΔH1
(ii) The reaction of zinc oxide with hydrochloric acid is
ZnO(s) + 2HCl(aq) ZnCl2(aq) + H2O(l) ΔH2
Use the equations in parts (i) and (ii) to complete the Hess’s Law cycle below to show how you could
calculate the enthalpy change for the decomposition of ΔHreaction. Label the arrows in your cycle
∆H reaction
ZnCO3 (s) ZnO (s) + CO2 (g)
(iii) Write an expression for the ΔHreaction in terms of ΔH1 and ΔH2.
(b) Suggest two reasons why the value obtained by carrying out these two experiments and using the
equation gives a value different to the data booklet value for the decomposition reaction of zinc carbonate.
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Questions using Enthalpy Change of Formation Data
Enthalpy change Enthalpy change Enthalpy change
Compound of formation Substance of formation Compound of formation
/ Kj mol-1 / Kj mol-1 / Kj mol-1
Al2O3(s) -1669 Fe(l) +14 NH3 (g) -46
BaO(s) -558 Fe2O3(s) -822 NF3(g) -114
CH4(g) -75 H2O(l) -286 NH4NO3(s) -365
C3H7OH(l) -304 H2O(g) -242 NH4F(s) -467
C4H9OH(I) -327 H2O2(g) -133 N2H4(g) +74
CH3COCH3(l) -248 HCl (g) -92 NO2(g) +34
CO(g) -111 MgCl2(s) -642 N2O(g) +82
CO2(g) -394 MgO(s) -602 SO2(g) -297
For the following questions use the enthalpy change of formation data in the table above
1) Calculate the enthalpy change for the following reaction
NH4NO3(s) + ½ C(s) → N2(g) + 2H2O(g) + ½ CO2(g)
2) Calculate the enthalpy change for the following reaction
MgO(s) + 2HCl(g) → MgCl2(s) + H2O(l)
3) Calculate the enthalpy change for the following reaction
N2H4(g) + 2H2O2(g) → N2(g) + 4H2O(g)
4) Calculate the enthalpy change for the following reaction
4NH3(g) + 3F2(g) → NF3(g) + 3NH4F(s)
5) Use the data above and the following equation to calculate a value for the standard enthalpy of
formation for nitric acid.
H2O(l) + 2NO2(g) + O2(g) → 2HNO3(l) ΔH = –128 kJ mol–1
6) Use the data above and the following equation to calculate a value for the standard enthalpy of
formation for carbon disulphide.
CS2(l) + 3O2(g) → CO2(g) + 2SO2(g) ΔH = –1076 kJ mol–1
7) Calculate the enthalpy change for the following reaction
3BaO(s) + 2Al(s) →3Ba(s) + Al2O3(s)
8) Calculate the enthalpy change for the following reaction
2NH3(g) + 2O2(g) N2O(g) + 3H2O(l)
9) Calculate the enthalpy change for the following reaction
Fe2O3(s) + 3CO(g) 2Fe(I) + 3CO2(g)
10) calculate the standard enthalpy change of combustion of propanone using equation below
CH3COCH3(l) + 4O2(g) → 3H2O(l) + 3CO2(g)
11) calculate the standard enthalpy change of combustion of an alcohol C3H7OH, as shown by the
equation:
C3H7OH(l) + 4 ½ O2(g) 3CO2(g) + 4H2O(l)
12) Write an equation for the combustion of methane (CH4) and use it to calculate the enthalpy change of
combustion of methane using the data in the above table
13) Write an equation for the combustion of butanol (C4H9OH(I)) and use it to calculate the enthalpy
change of combustion of butanol using the data in the above table
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Questions using Enthalpy Change of Combustion Data
Enthalpy change Enthalpy change Enthalpy change
Substance of combustion Substance of combustion Substance of combustion
/ Kj mol-1 / Kj mol-1 / Kj mol-1
C(s) -393.5 C4H6 (g) -2542 C2H5OH(l) -1370
H2(g) -285 C6H12 (l) -4003 C2H5SH(l) -1170
S(s) -297 C6H14 (l) -4163 CH3COOH(l) -870
CH4(g) -890 C7H8 (l) -3909 C4H4O4(s) -1356
CO(g) -283
For the following questions use the enthalpy change of combustion data in the table above
1) Calculate the enthalpy change for the following reaction
C6H12(l) + H2(g) C6H14(l)
2) Calculate the enthalpy change for the following reaction
CH4(g) + 1½O2(g) CO(g) + 2H2O(l)
3) Use the data above and the following equation to calculate a value for the standard
enthalpy of combustion for gaseous methanol.
CO(g) + 2H2(g) CH3OH(g) ΔH = –91 kJmol–1
4) Calculate the standard enthalpy change of formation of C4H6 using the equation below
and the data above.
4C(s) + 3H2(g) → C4H6(g)
5) Calculate the standard enthalpy change of formation of C7H8 using the equation below
and the data above.
7C(s) + 4H2(g) → C7H8(l)
6) Calculate the standard enthalpy change of formation of Maleic acid C4H4O4 using the
equation below and the data above.
4C(s) + 2H2(g) + 2O2(g) C4H4O4(s)
7) Calculate the standard enthalpy change of formation of ethanoic acid CH3COOH using
the equation below and the data above
2C(s) + 2H2(g) + O2(g) → CH3COOH(l)
8) Calculate the standard enthalpy change of formation of C2H5SH using the equation
below and the data above
2C(s) + 3H2(g) + S(s) → C2H5SH(l)
9) Write an equation for the formation of ethanol (C2H4OH(I)) from its elements and use it
to calculate the enthalpy change of formation of ethanol using the data in the above table
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