Title: Lesson 4 Enthalpy of Formation and
Combustion
Learning Objectives:
– Calculate change in enthalpy of reactions using enthalpy of formation or
combustion data
How to construct an enthalpy change for
the reaction



Reactants - the elements that make up the product in their standard state.
Theses need to be balanced (you can use fractions as the co-efficients)
Product – The compound formed from the elements.
Enthalpy value will be given in the IB booklet.
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Solutions
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Standard Enthalpy of Formation
You need to know the ΔHθf for all the reactants and products that are
compounds.
The ΔHθf for elements is zero – the element is being formed from the
element so there’s no change in enthalpy.
Note: add when going ‘with’ an arrow, subtract when going against an arrow…
(The questions will show arrows in a different directions within the cycle)
http://www.yout
ube.com/watch?v
=c8Adft3M8mg
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Using enthalpies of formation
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Enthalpies of formation calculations
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Energy of the system
Calculate the enthalpy change of the
combustion of ethanol
Compound
C2H5OH
CO2
H2O
ΔHf
-277
-394
-286
C2H5OH + 3O2
2CO2 + 3H2O
2C + 31/2O2 + 3H2
Reaction path
C2H5OH + 3O2 → 2CO2 + 3H2O
Energy of the system
Calculate the enthalpy change of the
combustion of ethanol
ΔHf
-277
-394
-286
Compound
C2H5OH
CO2
H2O
C2H5OH + 3O2
-277
2CO2 + 3H2O
2 x -394
2C + 31/2O2 + 3H2
Reaction path
3x -286
Energy of the system
Calculate the enthalpy change of the
combustion of ethanol
ΔHf
-277
-394
-286
Compound
C2H5OH
CO2
H2O
C2H5OH + 3O2
-1369KJmol-1
-277
2CO2 + 3H2O
2 x -394
2C + 31/2O2 + 3H2
3x -286
Reaction path
ΔHc = ∑ΔHofProducts - ∑ΔHofReactants
= ( 2 x 394 + 3x -286 ) – ( -277 )
Energy of the system
Calculate the enthalpy change of the
reaction below
CaO + H2O
Ca(OH)2
What are the intermediates?
Reaction path
CaO + H2O → Ca(OH)2
Energy of the system
Calculate the enthalpy change of the
reaction below
Compound
CaO
Ca(OH)2
H2O
ΔHf
-635
-986
-286
CaO + H2O
Ca(OH)2
Ca(s) + O2(g) + H2(g)
Reaction path
CaO
Now
Ca(OH)
∆Hf 2
+ Hwork
2O →out
Energy of the system
Calculate the enthalpy change of the
reaction below
ΔHf
-635
-986
-286
Compound
CaO
Ca(OH)2
H2O
CaO + H2O
-65KJmol-1
-635
Ca(OH)2
-286
-986
Ca(s) + O2(g) + H2(g)
This way is just following the direction of the arrows of the Hess cycle (elements
towards reactants and products). Using the equation will give you the same
answer!
Reaction path
-1
∆Hf = - ((-635)
CaO+ +(-286))
H2O →
+ (-986)
Ca(OH)
= -65KJmol
2
Calculate the enthalpy change of the
reaction below Compound ΔH
f
CaO
Ca(OH)2
H2O
-65KJmol-1
Ca(OH)2
CaO + H2O
-635
-286
-986
Ca(s) + O2(g) + H2(g)
This canCaO
be represented
Ca(OH)
a Hess2 cycle
+ H2O → as
-635
-986
-286
Calculate the enthalpy change of the
ΔH
reaction below Compound
CH NHNH
+54.0
f
3
N2O4
CO2
H2O
2
-20.0
-393
-286
∆H
4CH3NHNH2 + 5 N2O4
4CO2 + 12 H2O + 9N2
∆H1
∆H2
4C(s) + 9N2(g) + 12H2(g) + 10O2(g)
∆H = ∆H2 -∆H1
Calculate the enthalpy change of the
ΔH
reaction below Compound
CH NHNH
+54.0
f
3
2
N2O4
CO2
H2O
-20.0
-393
-286
∆H
4CO2 + 12 H2O + 9N2
4CH3NHNH2 + 5 N2O4
4 x 54
4 x -393
5 x -20
12 x -286
4C(s) + 9N2(g) + 12H2(g) + 10O2(g)
∆H1 = (5 x -20) + (4 x 54) = +116KJmol-1
∆H2 = (12 x -286) + (4 x -393) = -5004KJmol-1
∆H = ∆H2 -∆H1
∆H = -5004 – (-116)
CaO
-1
2O → Ca(OH)
2
∆H+=H-4888KJmol
STANDARD ENTHALPY OF FORMATION
REACTANTS
SO2(g) + 2H2S(g)
ΔHθr
Route 1
PRODUCTS
3S(s) + 2H2O(l)
ΔHθf(products)
ΔHθf(reactants)
3S(s) + 2H2(g) + O2(g)
ELEMENTS
Step 1: Write the balanced equation for the reaction.
This will be ΔHθr
Step 2: Under the equation write a list of the elements
present. This must be balanced
STANDARD ENTHALPY OF FORMATION
REACTANTS
SO2(g) + 2H2S(g)
ΔHθr
Route 1
PRODUCTS
3S(s) + 2H2O(l)
ΔHθf(products)
ΔHθf(reactants)
3S(s) + 2H2(g) + O2(g)
Step 3:
ELEMENTS
ΔHθf[SO2(g)] = -297 kJmol-1
ΔHθf[H2S(g)] = -20.2 kJmol-1
ΔHθf[H2O(l)] = -286kJmol-1
ΔHθf values give
the enthalpy
change going from
the element to the
compound
Using Hess’ Law; Route 1 = Route 2
STANDARD ENTHALPY OF FORMATION
REACTANTS
SO2(g) + 2H2S(g)
ΔHθr
Route 1
ΔHθf(SO2) +
2 x ΔHθf(H2S)
PRODUCTS
3S(s) + 2H2O(l)
3 x ΔHθf(S) +
2 x ΔHθf(H2O)
3S(s) + 2H2(g) + O2(g)
Step 3:
ELEMENTS
ΔHθf[SO2(g)] = -297 kJmol-1
ΔHθf[H2S(g)] = -20.2 kJmol-1
ΔHθf[H2O(l)] = -286kJmol-1
ΔHθf values give
the enthalpy
change going from
the element to the
compound
Using Hess’ Law; Route 1 = Route 2
STANDARD ENTHALPY OF FORMATION
REACTANTS
SO2(g) + 2H2S(g)
ΔHθr
Route 1
-297 +
(2 x -20.2)
PRODUCTS
3S(s) + 2H2O(l)
(3 x 0) +
(2 x -286)
3S(s) + 2H2(g) + O2(g)
Step 3:
ELEMENTS
ΔHθf[SO2(g)] = -297 kJmol-1
ΔHθf[H2S(g)] = -20.2 kJmol-1
ΔHθf[H2O(l)] = -286kJmol-1
ΔHθf(s) is zero
because its an
element
Using Hess’ Law; Route 1 = Route 2
STANDARD ENTHALPY OF FORMATION
SO2(g) + 2H2S(g)
ΔHθr
Route 1
-297 +
(2 x -20.2)
Step 4:
3S(s) + 2H2O(l)
(3 x 0) +
(2 x -286)
3S(s) + 2H2(g) + O2(g)
ΔHθf(SO2) + ΔHθf(H2S) + ΔHθr = 3ΔHθf(S) + 2ΔHθf(H20)
-297 + (2 x -20.2) + ΔHθr = (3 x 0) + (2 x -286)
ΔHθr = (3 x 0) + (2 x -286) – (-297 + (2 x 20.2))
= -234.6 kJmol-1
Using enthalpies of combustion
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Enthalpies of combustion calculations
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HESS’S LAW AND ENTHALPY OF FORMATION
TASKS:
1.
Complete the ‘Enthalpy of Formation’ worksheet
2.
Complete the Hess’s Law cut and stick exercise
Terry’s Cutting tip: Keep the left column and top row
in one piece
1
One mole of carbon burns to give one mole of carbon dioxide,
releasing 393.5 kJ. One mole of carbon burns to give one mole of
carbon monoxide, releasing 110.5 kJ. Calculate the energy from
burning one mole of carbon monoxide.
 Hf[CO2(g)] = -393.5 kJ mole-1 and Hf[CO(g)] = -110.5 kJ mole-1
CO(g) + ½O2(g)
Hf[CO(g)]
H
CO2(g)
Hf[CO2(g)]
Alternative route
= -110.5
= -393.5
Elements
here
because
Hf given
C(s) + O2(g)
By Hess’s Law:
What info do you have?
Which equation to use?
H = (SUM)H PRODUCTS - (SUM)H
= (-393.5)-(-100.5)

= - 283.0 kJ mole-1
REACTANTS
2
Use the Hf values given to calculate H of :
CH3COCH3(l) + 4O2(g)  3CO2(g) + 3H2O(l)
Hf[CO2(g)]
=
Hf[CH3COCH3(l)] =
Hf[H2O(l)]
=
CH3COCH3(l) + 4O2(g)
H
- 394 kJmol-1
- 248 kJmol-1
- 286 kJmol-1
3CO2(g) + 3H2O(l)
3Hf[CO2(g)]
+ 3Hf[H2O(l)]
Hf[CH3COCH3(l)]
= -248
= 3(-394) +
3(-286)
Elements
here
because
Hf given
3C(s) + 3H2(g) + 4.5O2(g)
 H =
(SUM)HPRODUCTS - (SUM)HREACTANTS
= (3(-394) + 3(-286)) – (-248)
= - 1792 kJ
mole-1
What info do you have?
Which equation to use?
3
Calculate Hf [CH4(g)], given
Hf [CO2(g)] = -393.5 kJ mole-1 also = HC [C(s)]
Hf [H2O(l)] = -285.8 kJ mole-1 also = HC [H2(g)]
Hc[CH4(g)] = -890.3 kJ mole-1
C(s) + 2H2(g)
CH4(g)
H
HC [C(s)]
+ 2HC [H2(g)]
Hc[CH4(g)]
= - 890.3
= (-393.5)
+ 2(-285.8)
CO2(g) + 2H2O(l)
= (SUM)HREACTANTS- (SUM)HPRODUCTS
 H = (-393.5 + 2(-285.8)) – (-890.3)
What info do you have?
Which equation to use?
Oxides
here
because
HC given
= - 74.8 kJ mole-1
4
Calculate HR for : C2H4(g) + H2(g)  C2H6(g), given
Hc [C2H4(g)] = - 1410.8 kJ mole -1
=
H c [H2(g)]
- 285.8 kJ mole -1
H c [C2H6(g)] = - 1559.7 kJ mole -1
C2H4(g) + H2(g) + 3.5O2(g)
C2H6(g) + 3.5O2(g)
H
Hc [C2H4(g)]
+ H c [H2(g)]
H c [C2H6(g)]
= -1559.7
= (-1410.8)
+ (-285.8)
2CO2(g) + 3H2O(l)
 H = (SUM)H REACTANTS- (SUM)H PRODUCTS
= ((-1410.8) + (-285.8)) -(-1559.7)

= - 136.9 kJ mole-1

Oxides
here
because
HC given
What info do you have?
Which equation to use?
NB
H values calculated from bond
energies are APPROXIMATE because:
1.
2.
Average values used
Gaseous state may not apply.
Also, you may care to remember :
Hr = Hf [PRODUCTS] - Hf [REACTANTS]
Hr = HC [REACTANTS] - HC [PRODUCTS]
Hr = E [REACTANTS] - E [PRODUCTS]
Key Points

∆Hor is the enthalpy change of a reaction for molar
quantities under standard conditions

∆Hor can be calculated from enthalpies of
combustion/formation using Hess Cycles


With formation, the arrows point up
With combustion, the arrows point down
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