Time (mins)
Temp
5
64
10
85
10.3
70
11
60
11.3
56
12
55
12.3
54
13
54
13.3
54
14
53
14.3
53
15
50
15.3
46
Observations
solid

Thermochemical principles includes:
transfer of heat between the system and the
surroundings
the enthalpy change for any process is the
sum of the enthalpy changes for the steps into
which the process can be divided
definition of the following terms:  cH,  fH,
 rH,  vapH and  fusH.
Heat
Heat is the transfer of energy from regions of
high temperature to regions of low
temperature.
As temperature is a measure of the
movement of molecules (average kinetic
energy), heat moves from regions of high
thermal motion to a region where there is less
thermal motion.
Enthalpy
When considering the heat change in chemical
reactions we use a quantity called enthalpy (H)
which is a measure of the chemical potential
energy stored in the bonds of the substances
involved.
Enthalpy
A change in enthalpy of a system is equal to the
heat released or absorbed at constant pressure.
It is measured in kilojoules, kJ (or joules, J). The
change in enthalpy is given by rH, where
rH = Hfinal – H
= H
initial
products
- H reactants
Enthalpy
Exothermic and Endothermic processes
For an exothermic reaction, heat is released and
rH < 0
In this case the products are more stable because
they contain less enthalpy than they started with.
This means the temperature of all species and the
surroundings increases. All combustion reactions are
exothermic.
This is shown in an energy profile diagram
Enthalpy, H
 rH < 0
Products
Enthalpy
The chemical potential energy is not lost but is
converted into increased kinetic energy of all the
particles (both reactants and products) and also
transferred to the surroundings.
An endothermic reaction absorbs heat from the
surroundings because the enthalpy of the products is
more than reactants and rH > 0. This means the
temperature of the system gets colder.
Dissolving may be endothermic or exothermic
depending on the nature of the solute and solvent e.g.
dissolving NH4NO3(s) in water is endothermic while
dissolving conc H2SO4 is exothermic.
Exercise:
In the space below draw a labelled diagram for
an endothermic reaction, showing reactants,
products and rH.
PRODUCTS
Enthalpy, H
rH > 0
Products
REACTANTS
Enthalpy of physical changes
A heating curve of a substance shows the variation of temperature of a sample as it is heated.
gas
b.p.
Temperature
/ oC
liquid
m.p.
solid
Time
At both its melting point and boiling point, even though the
sample is being heated, temperature remains constant.
All heat being added is used to break the forces holding the
molecules in their solid or liquid states, rather than being
converted into kinetic energy.
Enthalpy of physical changes
A heating curve of a substance shows the variation of temperature of a sample as it is heated.
gas
b.p.
Temperature
/ oC
liquid
m.p.
solid
Time
Could you explain looking at the graph why a drink stays
cold while it contains ice cubes?
Molar heat of Fusion (∆fusH)
This is the energy required to change one mole of a substance
from a solid to a liquid at the melting point.
H2O (solid at 0 oC)
H2O (liquid at 0 oC) ∆fusH=6.0 kJmol-1
The ∆fusH gives an indication of the strength of forces
holding the particles together in the solid phase
The melting and boiling points of a
substance indicate the strength of the
forces between the particles in the solid
or liquid phases
Molar heat of vaporisation (∆vapH)
This is the energy required to change one mole of a substance
from a liquid to a gas at the boiling point.
H2O (liquid at 100 oC)
H2O (gas at 100 oC) ∆vapH = 41.0kJmol-1
The ∆vapH gives an indication of the strength of forces
holding the particles together in the liquid phase
Enthalpies
We can’t measure enthalpies of reactants and
products directly all we can measure are
enthalpy changes.
Data books list the standard enthalpies of
combustion and /or the standard enthalpy of
formation for compounds.
These figures can be used to calculate the
enthalpy change for other reactions
Enthalpy Definitions
(∆cH 0) – Standard enthalpy (heat) of combustion
The standard enthalpy of combustion is the enthalpy
change when one mole of the substance is
completely burnt with all reactants and products in
their standard states
Standard states are the states at room
temperature (25oC) and pressure - that is O2 is a
gas and water is H2O (l)
C2H6(g) + 3 ½ O2(g)
2CO2(g) + 3H2O(l)
Heats of combustion are always exothermic
∆cH
0
= -1557kJmol-1
(∆fH 0) – Standard enthalpy (heat) of formation
The standard enthalpy of formation of a compound is the
enthalpy change when one mole of the substance is formed
from its elements, with all the reactants and products in their
standard states.
2C(s) + 3 H2(g) + ½ O2 (g)
C2H5 OH(l) ∆fHo= -278kJmol-1
Do you think these enthalpies of formation look strange if so why?
Data tables list heats of formation for compounds not
elements - why? Think standard states
The standard enthalpy of formation for all elements is
zero
Remember fH0 is the standard reaction enthalpy for the
formation of one mole of substance from its elements in
their most stable form at standard state. The units of
fH0 are kJ mol-1 and the equation it refers to must have
only one mole of product.
Note:
If 4C(s) + 6H2(g) + O2(g)
2C2H5OH (l) rH0 = -555 kJ
Then f H0 (C2H5OH, l) = -555  2
= - 277.5 kJ mol-1
Exercise - Write equations for the reactions which
have the following enthalpies of combustion:
cH0 (H2, 25 oC) = -286 kJ mol-1
H2(g) + ½ O2(g)
H2O(l)
∆cH = -286kJmol-1
cH0 (CH3OH, 25 oC) = -726 kJ mol-1
CH3OH(g) + 1 ½ O2(g)
CO2(g) + 2H2O(l) ∆cH = -726kJmol-1
Exercise 1 - Write balanced equations for the
formation reactions from their elements
(a) C6H12O6(s)
6C(s) + 6H2(g) + 3O2(g)
C6H12O6 (s)
Note : We use mole
fractions where
necessary to give one
mole of the compound
(b) CO(g)
C(s) + ½ O2(g)
CO (g)
(c) HCl(g)
½ H2 (g) + ½ Cl2(g)
HCl (g)
Starter
cH0 means the e_______ of c__________
the 0 means s_______ c________
standard conditions - refers to
____ mole at ____degrees celsius and ___
atmosphere of p______
ENERGY CALCULATIONS
Calculations involving thermochemical principles may
include:
relating enthalpy changes to heat and mass, and use of
specific heat capacity of water
application of ΣrH = Σ fH (products) - Σ fH (reactants)
application of Hess’s Law
use of average bond energies in enthalpy change
calculations.
Measuring enthalpies of reaction – calorimetry
To measure enthalpy changes, the reaction is carried out in an
insulated container (such as a polystyrene cup) and the
temperature change is measured.
Using this temperature change, ΔT, and the value of the specific
heat capacity, c, the amount of energy transferred to the mass m
of substance can be calculated using the expression
E = m c ΔT
If the reaction takes place in an aqueous solution, and the
solutions are placed in an insulated container then we can
assume:
c =The specific heat capacity of the solutions is 4.18 J oC-1 g-1.
The densities of these aqueous solutions can be taken as 1.0 g
mL-1 so that 100 mL would have a mass of 100 g.
Calorimetry
Exercise
25.0 mL of 1.0 mol L-1 HCl is placed in a polystyrene cup and
its temperature is measured as 21.0 oC.
Following the addition of 25.0 mL of 1.0 mol L-1 NaOH solution,
the mixture is stirred and the final temperature recorded as
27.8 oC.
Calculate the molar enthalpy change, ΔrH in kJ mol-1, for this
reaction.
(Note: If the temperature increases then the reaction is
exothermic and ΔrH will be negative)
E = m c ΔT
= 50g X 4.18 J oC-1 g-1 x 6.8 oC
= 1421.2 J
If the reaction takes place in aqueous solution, and the
solutions are placed in an insulated container, the
specific heat capacity of the solutions is 4.18 J oC-1 g-1.
The densities of these aqueous solutions can be taken
as 1.0 g mL-1 so that 100 mL would have a mass of
100 g.
Energy per mole
E = m c ΔT
= 50g X 4.18 J oC-1 g-1 x 6.8 oC
= 1421.2 J
mol of NaOH (n) = 0.025L x 1molL-1 = 0.025 mol
E
H reaction 
n
1421.2J

0.025mol
 56848Jmol-1
ΔrH  56.848kJmol  1
Calculating Enthalpy From Heats of Formation
Given the enthalpies of formation of all the
compounds in a given reaction, we can calculate
the enthalpy of reaction using the following
formula
ΣrH = ΣnfH (products) - ΣnfH (reactants)
(the symbol Σ means sum of)
Calculating Enthalpy From Heats of Formation
ΣrH = ΣnΔ fH  (products) – ΣnΔ fH (reactants)
What does the “n” stand for?
These calculations are straight forward providing you
remember the following:
The formula only works if the heats of formation for all
the compounds are used
Set your work out very carefully, paying particular
attention to the + and - signs
Always check your answers!
Example
Using the standard heats of formation of CO2(g),
H2O(l), and C6H12O6(s), calculate the standard
enthalpy of combustion of glucose.
fHo(CO2, g)
fHo(H2O, l)
= -394 kJ mol-1
= -286 kJ mol-1
fHo(C6H12O6, s)
= -1268 kJ mol-1
fHo(CO2, g)
fHo(H2O, l)
= -394 kJ mol-1
= -286 kJ mol-1
fHo(C6H12O6, s)
= -1268 kJ mol-1
Note - Start by writing an equation for the combustion of 1
mole of glucose.
C6H12O6(s) + 6O2(g)

6CO2(g) + 6H2O(l)
rHo =  nfHoproducts -  nfHoreactants
rHo = ( 6 x -394 + 6 x -286) - ( -1268 + 0)
= - 2812 kJ mol-1
Ethanoic acid can be formed by the oxidation of ethanol.
C2H5OH(l) + O2(g) 
CH3COOH(l) + H2O(l)
This reaction occurs when wine goes sour. By calculating the
standard reaction enthalpy for this oxidation reaction, decide
whether the reaction is exothermic or endothermic given the
following data:
fHo(H2O, l)
= -286 kJ mol-1
fHo(CH3COOH, l) = -485 kJ mol-1
fHo(C2H5OH, l)
= -278 kJ mol-1
rHo =  nfHoproducts -  nfHoreactants
rHo = (- 485 + - 286) - (-278)
= - 493 kJ mol-1
Therefore exothermic
Exercises
1. a) Using the standard heats of formation given calculate the
enthalpy change for the following reaction.
2SO2(g) + O2(g) 
fHo(SO2, g)
= -297 kJ mol-1
fHo(SO3, g)
= -396 kJ mol-1
2SO3(g)
rHo =  nfHoproducts -  nfHoreactants
rHo = (2 x -396) - (2 x -297)
rHo = (2 x -396) - (2 x -297)
= - 198 kJ mol-1
b) What is the enthalpy change when 100 g of SO2 burn to form
SO3?
2SO2(g) + O2(g) 
fHo(SO2, g) = -297 kJ mol-1
2SO3(g)
fHo(SO3, g) = -396 kJ mol-1
100g
n(SO2 ) 
64gmol -1
 1.563mol
rHo =  nfHoproducts -  nfHoreactants
rHo = (1.563 x -396) - (1.563 x -297)
= - 154 .7 kJ
= - 155 kJ (3SF)
Calculate fHo(PCl5, s) from the following information.
fHo(PCl3, l)
= -320 kJ mol-1
PCl3(l) + Cl2(g)

PCl5(s)
rH = -124 kJ mol-1
rHo =  nfHoproducts -  nfHoreactants
This is the
unknown
(-124) = fHo - (-320)
(-124) = fHo + 320
fHo = - 444 kJ mol-1
Tricky enthalpy calculations
1. Calculate the final temperature when 50g of water at 70 C
is mixed with 150g of water at 40 C ?
Energy loss by hot water = Energy gained by cool water
E lost = E gained (conservation of energy)
∆H(loss) = ∆H (gained)
m c ∆T
= m c ∆T
m c ( 70 – T) = m c (T - 40)
Hot water
will cool
Cool water will
heat up
Tricky enthalpy calculations
m c ∆T
= m c ∆T
m c ( 70 – T) = m c (T - 40)
Hot water will cool
Cool water will heat up
Solve for T
50 x 4.18 x(70  T )  150 x 4.18 x(T  40)
70  T  3T  120
4T  120  70
T  47.5o C
Hess’s Law
If an overall reaction can be broken down into a series of two
or more steps, then the corresponding overall enthalpy of
reaction is the sum of the enthalpies of the individual reaction
steps.
None of the steps need to be a reaction that can be carried out
in the laboratory.
Another way of saying this is that the energy difference
depends only on the difference in energy between the
reactants and products, not on the reaction path taken.
The process of photosynthesis is an endothermic process in which
energy from the sun is trapped and stored in the bonds of glucose.
6CO2(g) + 6H2O(l)  C6H12O6(aq) + 6O2(g) rH0 = +2808kJ mol-1
It is however easier to measure the enthalpy change for the
reverse reaction, the combustion of glucose (i.e. the process of
respiration).
C6H12O6(aq) + 6O2(g)  6CO2(g) + 6H2O(l) rH0 = -2808kJ mol-1
Remember if you reverse the reaction
reverse the sign of rH0
reactants
Reaction 1, H1
Hess’s Law
illustrated
Reaction 2, H2
Enthalpy,
H
rH
the energy
difference
rH total
Reaction 3, H3 depends only on
the difference in
energy between
products
the reactants
and products,
total = rH1 - rH2 + rH3
not on the
reaction path
taken.
As a result of this principle of conservation of
energy we can say:
Chemical reactions and their
corresponding H values can be added ,
subtracted and multiplied as if they were
algebraic equations
Another illustration of Hess’s law is:
C(s)
+
½ O2 (g)
+ O2
CO (g)
H o3
H o1
H o2
+ ½ O2
H o 1= H o2 + H o3
CO2 (g)
In practice it is impossible to measure the value of H o3 (the
heat of formation of CO(g)) directly as some CO2 gas is always
formed as a by product.
However we can calculate it using Hess’s law.
Using algebra and the following info find H o3
H o1 = -393kJmol-1
and
H o2 = -282kJmol-1
C(s)
+
½ O2 (g)
+ O2
H o1
Answer
 H o 3 =  H o1 -  H o2
H o3
CO (g)
H o2
+ ½ O2
CO2 (g)
= -393 – (-282)
= -111kJ mol-1
How’d ya go?
Solving Problems Using Hess’s Law
1. Write the data in the form of equations
2. Rewrite the equations to give the desired
species on the correct side of the equation. If
the reaction must be reversed (perhaps because
we require a species to be a reactant and not a
product) then the sign of the H must also be
reversed.
3. Add the equations, and the H together
4. Check your working!!
Hang on to your hats!
Here we go………
Hess’s Law Calculation
Calculate the heat of formation of CS2(l) given that the heats
combustion of carbon, sulfur and carbon disulfide. Sulfur burns
to SO2 in oxygen
-393, -297 and -1007 kJ mol
-1
respectively
Write the data given in the form of equations
(1) C (s) + O2(g)
CO2 (g)
H = -393 kJ mol
-1
(2) S (s) + O2 (g)
SO2 (g)
H = -297 kJ mol
-1
(3) CS2(s) + 3O2 (g)
CO2 (g) + 2SO2 (g) H = -1007 kJ mol
The equation we are looking for is C(s) + 2S(s)
-1
CS2 (l)
Now rewrite the equations to give the desired species on
the correct side of the equation
Hess’s Law Calculation
The equation we are looking for is C(s) + 2S(s)
CS2 (l)
Rewrite the equations to give the desired species on the correct side
of the equation
(1) C (s) + O2(g)
CO2 (g)
H = -393 kJ mol -1
(2) 2S (s) + 2O2 (g)
(3) CO2 (g) + 2SO2 (g)
2SO2 (g)
H = ( 2 x -297) = - 594 kJ mol
CS2(s) + 3O2 (g)
H = +1007 kJ mol
-1
Now add equations and the H values together
cancelling where appropriate
(1) C (s) + O2(g)
(2) 2S (s) + 2O2 (g)
(3) CO2 (g) + 2SO2 (g)
Now
add
C(s) + 2S(s)
CO2 (g)
2SO2 (g)
CS2(s) + 3O2 (g)
CS2 (l)
H = -393 kJ mol
-1
H = -594 kJ mol
-1
H = +1007 kJ mol
H = +20 kJ mol
Therefore the fH o =(CS2(l)) is +20 kJ mol
-1
-1
-1
-1
Important
• All booklets are to be handed in
worksheets 1 (8),2 (9) ,3 (10) , 4 (11) and
5 (12) Change
• All worksheets must be completed and
marked before next Monday as all
booklets are handed in on day of test
(Monday)
• Read pathfinder thermochemistry page
54 to help understanding and bestchoice
for extra practice
Hess’s Law calculation
Calculate the enthalpy change for this reaction:
2
3 2(g) + ½O2(g) →
2C(s)
+ 3H
given:
C2H5OH(l)
C(s)
+
O2(g)
→
CO2(g)
∆H = –393 kJ mol–1
H2(g)
+
½O2(g)
→
H2O(l)
∆H = –285 kJ mol–1
C2H5OH(l)
+ 3O2(g)
→ 2CO2(g) + 3H2O(l)
∆H = –1364
1364 kJ
kJ mol–1
mol–1
2 C atoms are required as a reactant.
3 H2 molecules are required as a reactant.
C2H5OH is a product.
2
1
2
×2
3
×3
+
Remember to change the sign of the
∆H when reversing the equation.
2C(s) + + 2O2(g)
→
2CO2(g)
∆H = –393 kJ mol–1 × 2
3H2(g) + + 1½O
½O2(g)
→
3H2O(l)
∆H = –285 kJ mol–1 × 3
2CO2(g) + 3H2O(l)
→
C2H5OH(l) + 3O2(g)
∆H = +1364 kJ mol–1
→
C2H5OH(l)
∆H = –277 kJ mol–1
2C(s) + 3H2(g) + ½O2(g)
Before adding these three equations, cancel out the terms
which appear on both sides of the arrows.
Use the bracket keys on your calculator to add up
the ∆H values.
It’s very easy to make errors in the exam, so do
each sum twice.
You may use either of the two techniques to work
out the enthalpy of a reaction
ie
Hess’s law
Or
rHo=  nfHoproducts -  nfHoreactants
Very important homework
Read chapter 13 page 49
Complete questions 2 and 3 page 52
(relating ∆fusH to intermolecular forces is a favourite of
examiners!)
Complete All 3 worksheets ie 10,11 & 12 in preparation for
test
Bond Energies
Bond energy is a measure of the intramolecular bond
strength in a covalent bond
Note that these
AB (g)
A(g)
+ B(g)
energy equations
have their
species as gases
not as species in
their standard
states
Bond energy is the energy required to break one mole of a
particular bond when the reactants and products are in
the gaseous state
Bond Energies
The exact value of the C
the environment it is in:
H
H
C
H
H bond depends on
H
H
Cl
C
H
H
H
H
C
H
CH3
So the bond energies given in data books are
average bond energies
Bond Energy Calculations
A chemical reaction is a series of bond breaking
processes (∆H positive)
And bond making processes (∆H is negative)
We can estimate the enthalpy of a chemical
reaction by adding the positive bond energies
for those bonds which are broken to the
negative bond energies for those bonds which
are made.
Enthalpy calculation using bond energies
Calculate the heat of reaction for the following
C2H6(g) + Cl2(g)
C2H5Cl(g)
+
HCl (g)
Given the following bond energies:
C
C
H 413 kJ mol-1
Cl
339 kJ mol-1
Cl
H
Cl 242 kJ mol-1
Cl
431 kJ mol-1
Calculate the heat of reaction for the following
C2H6(g) + Cl2(g)
C2H5Cl(g)
+
HCl (g)
Given the following bond energies:
C
C
H 413 kJ mol-1
Cl
Cl
H
339 kJ mol-1
Cl 242 kJ mol-1
Cl
431 kJ mol-1
This kind of problem is simple once you write out the equation
showing structural formula with every bond shown
H H
H
C C H + Cl
H H
H H
Cl
H
C C Cl + H
H H
Cl
H H
H
H H
C C H + Cl
Cl
H
H H
C C Cl + H
Cl
H H
Bond breaking (∆H +)
Bond making (∆H -)
C
H +413 kJ mol-1
C
Cl
- 339 kJ mol-1
Cl
Cl + 242 kJ mol-1
H
Cl
- 431 kJ mol-1
∆H = +413 + 242 - 339 - 431
= -115kJmol-1
If you use all positive
values you can
rHo = Ebonds broken - E
use
bonds formed
Note
In multiple bonds such as
O O we do not double the bond
energy for a single bond (O O).
The double bond is a different kind
of bond from a single bond
Enthalpy calculation using bond energies
Calculate the heat of reaction for the following
CH4(g) + 2O2(g)
CO2 (g)
+
Given the following bond energies:
C
H = 413 kJ mol-1
O
O
= 498 kJ mol-1
C
O
=805 kJ mol-1
H
O = 463 kJ mol-1
2H2O (g)
Enthalpy calculation using bond energies
Write out
structural
2H2O (g) formula
Calculate the heat of reaction for the following
CH4(g) + 2O2(g)
H
H
C
O
H +
O
CO2 (g)
O
O
O
+
C
H
O + H
O
O
H
H
H
Bonds Broken
4xC
H 4 x 413 kJ mol-1
2xO
O 2 x 498 kJ mol-1
+2648 kJ mol-1
Bonds Formed
2xC
O 2 x -805 kJ mol-1
4xH
O 4 x - 463 kJ mol-1
-3466 kJ mol-1
A slightly tricky example
E
E
E
C-H
C - Br
= 413 kJ mol-1
= 285 kJ mol-1
H - Br
= 366 kJ mol-1
5 b. Bromine reacts with methane as shown below
CH4 (g) + Br2 (g)
CH3Br (g) + HBr (g)
rH0 = - 45kJ mol-1
Given this equation and the bond enthalpies given above to
calculate the bond enthalpy for Br-Br
Bonds Broken
4xC
2xO
Bonds Formed
H 4 x 413 kJ mol-1
2xC
O 2 x 498 kJ mol-1
4xH
+2648 kJ mol-1
O
2 x -805 kJ mol-1
O 4 x - 463 kJ mol-1
-3462 kJ mol-1
rHo =   Hbonds broken +   H bonds formed
rHo = (+2648 kJ mol-1) + (-3462 kJ mol-1)
= - 814 kJmol -1
Therefore Reaction is exothermic
CH4 (g) + Br2 (g)
CH3Br (g) + HBr (g)
H
H
H
C
H + Br
Br
H
H
Br
C
Br + H
Br
H
Bond breaking (∆H +)
C
rH0 = - 45kJ mol-1
H + 413 kJ mol-1
Br
?
Bond making (∆H -)
C
Br
- 285 kJ mol-1
H
Br
- 366 kJ mol-1
∆H = +413 + Br-Br - 285 - 366
-45 = +413 + Br-Br - 651
-458 =
Br-Br - 651
193 kJmol-1= Br-Br
Hesses Law of Summation page 150
NaOH(s)
H o3
H o1
+ HCl (aq)
NaOH(aq)
H o2
+ HCl (l)
NaCl (aq) + H2O(l)
It can be seen that:
 H o 1 =  H o 2 +  H o3
What kind of question can you expect in the
exam?
Don’t worry if the question isn’t in a form that you
recognize immediatelyRemember always keep cool and think carefully!
Are you ready for a toughie?
Here’s a question from the 2000 exam
Q5a. Shorter covalent bonds are stronger than
longer covalent bonds as illustrated by the
following bond enthalpies:
E
C-H
E
C - Br
E
H - Br
= 413 kJ mol-1
= 285 kJ mol-1
= 366 kJ mol-1
Circle the value below that is more likely to be
the bond enthalpy for C- Cl
346 kJ mol-1
Justify your choice
243 kJ mol-1
Hint: compare the radius of the Cl atom
with the Br atom by looking back on the
handout chart on atom radius
E
E
E
C-H
C - Br
H - Br
= 413 kJ mol-1
= 285 kJ mol-1
= 366 kJ mol-1
Circle the value below that is more likely to be
the bond enthalpy for C- Cl
346 kJ mol-1
243 kJ mol-1
Justify your choice
Ans: C- Cl bond is stronger because the Cl atom has a smaller
radius than Br so there is stronger attraction between the
nucleus and the bonded electrons / C-Cl bond is shorter than
the C-Br bond
(Note : discussions involving electronegativity difference or
bond polarity were considered to be irrelevant)
Note
Correct answer must include units for
full credit
QUESTION FOUR: ENTHALPY OF VAPORISATION
Use the following information to answer the question below.
ethanal
∆vapH / kJ
mol–1
26
propanal butanal
30
34
ethanoic
acid
52
Discuss the trend in vapH of the compounds in the table
above in terms of the attractive forces between the particles
and the factors affecting those forces.
Evidence
Ethanal, propanal and butanal are all aldehydes. They
all have weak intermolecular or permanent dipole forces
attracting molecules together.
vapH increases from ethanal to butanal due to increasing
M / electron numbers. The greater the M / electron numbers
the greater the strength of intermolecular or temporary
dipole forces.
Ethanoic acid has an H atom bonded to an O atom so can
form H-bonding between molecules.
H bonding is a stronger intermolecular attraction
compared to temporary and permanent dipole forces
between the aldehydes.
Achieved
Identifying H-bonding with ethanoic acid and any
weak intermolecular attraction involved.
OR
Identifying of H-bonding in ethanoic acid and
explaining that there is a difference in
electronegativity between O and H.
OR
Relating increase in MW to increase in vapH.
Merit
Identifying H-bonding with ethanoic acid and any weak
intermolecular attraction involved.
AND
Explaining presence of
H-bonding in ethanoic acid
using a difference in electronegativity.
AND ONE OF
H-bond stronger than other intermolecular forces of
attraction.
OR
Explaining trend in vapH of aldehydes by relating molecular
mass to strength of intermolecular forces.
Excellence
Comprehensive analysis and explanation of trends
in the table.
Explain temporary dipoles and the effect of
increasing mass.
Explain H bonding and how it occurs in ethanoic
acid.
Compare the strength of H bonding as weak
intermolecular attractions.
Explain how these attractions affect vapH
Q7 (2000)
The contact process for the preparation of sulfuric acid begins
with the conversion of sulfur to sulfur dioxide, followed by
oxidation of the sulfur dioxide to sulfur trioxide
(1) S (s) + O2(g)
(2) 2 SO2 (g) + O2 (g)
SO2 (g)
fHo (SO2) = -297 kJ mol
2SO3 (g)
rHo = -191 kJ mol
a. Is reaction 2 exothermic or endothermic?
exothermic
b. Use the information above to calculate fHo (SO3,g)
rHo =  nfHoproducts -  nfHoreactants
This is what we
want to find out
-1
-1
(1) 2 S (s) + 2 O2(g)
(2) 2 SO2 (s) + O2 (g)
2 SO2 (g)
2SO3 (g)
2 x -297 = -594 kJ mol -1
rHo = -191 kJ mol
rHo =  nfHoproducts -  nfHoreactants
-191 = 2fHoproducts - (-594)
-191 = 2fHoproducts + 594
(- 594 to bs)
-785 = 2fHoproducts
(divide by 2) -392.5 = fHoproducts
Therefore fHo (SO3 g) = - 392.5kJ mol-1
-1
Or you could have used Hesses law adjusting each
reaction to give 1 mole of product where necessary :
(1)
SO2 (s) + ½ O2 (g)
SO3 (g)
(2)
S (s) + O2(g)
SO2 (g)
S (s) + 1½ O2 (g)
SO3 (g)
rHo = -95.5 kJ mol
-1
rHo = -297 kJ mol
-1
fHo = -392.5 kJ mol
You will know you are right when you end up with the
balanced equation for the formation of SO3
Quite satisfying really don’t you think?
-1
QUESTION ONE: SMELTING ZINC
(a) The smelting of zinc ores involves the reaction of Zn and ZnS with
oxygen gas according to the following equations:
Zn(s) + ½ O2(g)

ZnO(s)
rH ° = – 348 kJ mol–1
Reverse this reaction
ZnS(s) + 1 ½ O2(g)  ZnO(s) + SO2(g)
rH ° = – 441 kJ mol–1
Using the following information and the data above,
S(s)
+ O2(g) 
SO2(g) fH ° = – 297 kJ mol–1
calculate the value of the enthalpy change, r H °, for the reaction:
Zn(s)
+
S(s)  ZnS(s)
Evidence
fHo = –297 kJ mol–1
1(a) S + O2  SO2
SO2 + ZnO  ZnS + 1½ O2
fHo = +441 kJ mol–1
Zn + ½O2  ZnO
fHo = –348 kJ mol–1
By Hess’s law of heat summation – adding these equations
and the enthalpies gives
Zn(s)
+
S(s)  ZnS(s) :
rHo = –297 + 441 – 348 = –204 kJ mol–1
Achieved
Merit
Correct process
Correct value
with one error.
rHo with units.
Q 1 (b)
Using the result of the calculation in part (a) above,
describe, with a reason, whether the heat of
formation of ZnS is endothermic or exothermic.
rHo = –297 + 441 – 348 = –204 kJ mol–1
Evidence
1(b)
Reaction is exothermic
since rHo is negative.
Achieved
Explanation links answer to
sign of rHo calculated.
QUESTION TWO: FUEL CELLS
A fuel cell, such as that used on a space-craft, is
similar to a battery. An example is the fuel cell that
‘burns’ hydrogen and oxygen to produce water and
energy.
The overall equation for the reaction is
2H2(g)
+ O2(g)

rH ° = – 572 kJ mol–1
2H2O(l )
(a)If the water produced is in the gas phase, the equation for the reaction
is
2H2(g)
+
O2(g)

2H2O(g)
Use the following bond enthalpies to calculate rH ° for this reaction.
Bond
Bond Enthalpy kJ mol1
H–H
436
O=O
498
O–H
460
Evidence
2(a)
rHo = Ebonds broken - E bonds formed
= (2 × 436 + 498) – (4 × 460)
= +1370 – 1840
= – 470 kJ mol-1
Achieved
Merit
Correct process
Correct value
with one error.
rHo with units.
2(b)Write an equation for which the enthalpy change is
equal to vapH ° (H2O).
Answer For Achieved
H2O(l)  H2O(g)
Correct equation showing states and 1 mole
2(c)
By considering the nature of the reaction in
part (b), describe why it is an endothermic
change.
Evidence
2(c) Energy must be absorbed to break the attractions (hydrogen
bonds) holding the molecules together in the liquid state.
Achieved
Recognition that bond breaking is endothermic or that energy has
to be put in to overcome the intermolecular
attractive forces in liquid state.
(d) Using the information in parts (a) to (c) above, calculate
the value of vapH ° (H2O).
2(d)
2H2O( l )  2H2(g) + O2(g)
 rH = +572 kJ
2H2(g) + O2(g)  2H2O(g)
rH = - 470 kJ
2H2O(l)  2H2O(g)
Therefore H2O(l)  H2O(g)
vapHo(H2O) = +51.0 kJ mol–1
rH = +102 kJ
rH = +51 kJ
Some recognition of the fact that vapHo can be
related to the enthalpies of the two reactions at the top of the
page.
Achieved
Merit
Excellence
Some recognition of Appropriate
Correct answer
the fact that vapHo
calculation
with units of kJ
can be
with one error. mol–1.
related to the
enthalpies of the two
reactions at the top
of the page.
The diagram to
the right shows
a simple
calorimeter.
It can be used to measure the enthalpy of combustion of
ethanol, C2H5OH.
(a) If 1.00 g of ethanol is burned in the spirit burner, the
temperature of the 200 g of water is found to increase from
22°C to 40°C. Using these results, calculate the experimental
value of ∆cH (C2H5OH, ).
M(C2H5OH, ) = 46 g mol–1
Specific heat capacity of water = 4.18 J g–1 °C–1
Evidence
Achievement
Merit
Excellence
E = 200 × 4.18 × 18
= 15 048 joules
= 15.048 kJ released
One step
calculated
correctly.
One
error in
calculation
.
cH
1.00 g
n (ethanol ) 
46 gmol
1
= 0.0217 mol
15.048kJ
cH 
0.0217mol
 693kJmol
calculated
correctly
including
Negative
sign &
Correct
units of
kJ mol–1.
1
The experimental value of ∆cH(C2H5OH, ) calculated above,
is found to be only about half the ‘accepted’ value. Use the
following data to calculate ∆cH °(C2H5OH, ).
C2H5OH() + 3O2(g)  2CO2(g) + 3H2O()
∆fH °(C2H5OH, ) = –277 kJ mol–1
∆fH °(H2O, ) = –286 kJ mol–1
∆fH °(CO2, g) = –394 kJ mol–1
Evidence
rHo =  nfHoproducts -  nfHoreactants
= (2 × –394 + 3 × –286) – (–277)
= –1369 kJ mol–1
Achievement
Merit
One error in calculation.
Answer calculated
correctly including
Negative sign &
Correct units of
kJ mol–1.
(c) Give two reasons why the experimental value for
the enthalpy of combustion of ethanol calculated in
part (a) is so much less than the ‘accepted’ value
calculated in part (b).
Evidence
3(c)
1. Heat is lost to the surroundings/lack of insulation.
2. Some of the ethanol that is burned
undergoes incomplete combustion that
releases less energy.
3. Experiment not carried out under standard conditions.
Achievement
One correct
reason given.
Merit
A physical and
a chemical
reason given.
NCEA Level 3 (Chemistry) 2004 — page 5
Students were able to interpret data and explain it in terms of the nature
of the solid and the forces of attraction between the particles. At this
level, candidates were able to write a response which linked their
understanding of thermochemical principles to the question, rather than
simply write all they knew about the topic.
Many candidates who completed other parts of the paper to at least merit
level omitted Question Three
(a), suggesting that they were unfamiliar with the concept of specific heat
capacity. This is one of the thermochemical principles included in the
explanatory notes for this standard.
Candidates need to be reminded that in the case of fusH (H2O), it is
simply the melting of water and NOT the reaction to produce H2 and O2.
A very high proportion of the candidates in this paper gave answers that
conveyed the latter misconception.
1. Calculate the enthalpy change, ΔrH° for the decomposition of
sodium chlorate.
NaClO3(s)
NaCl(s) + 3/2O2(g)
ΔfH° (NaClO3(s)) = –359 kJ mol1.
ΔfH° (NaCl(s))
= –411 kJ mol-1.
Evidence
ΔrH° = [ΔfH°(NaCl) + ΔfH°(O2] – [ΔfH°(NaClO3)]
= – 411 + 0 – [–359]
= – 411 + 359
= – 52 kJ mol–1.
A
Correct
method
with one
error.
M
Correct
answer
and
units
2. Calculate the enthalpy change for the oxidation of ammonia.
4NH3(g) + 5O2(g)
6H2O(g) + 4NO(g)
given:
N2(g) + 3H2(g)
2NH3(g)
ΔH = –92 kJ mol–1
2H2(g) + O2(g)
2H2O(g)
ΔH = –484 kJ mol–1
N2(g) + O2(g)
2NO(g)
ΔH = +180 kJ mol–1
Evidence
4NH3
2N2 + 6H2
6H2 + 3O2
6H2O
2N2 + 2O2
4NO
4NH3(g) + 5O2(g)
ΔH = +184 kJ mol–1
ΔH = –1452 kJ mol–1
ΔH = +360 kJ mol–1
6H2O(g) + 4NO(g) ΔH= –908 kJ mol–1.
OR
ΔrH°
= [6 ΔfH°(H2O) + 4 ΔfH°(NO)] – [4 ΔfH°(NH3) + 5ΔfH°(O2)]
= (–1452 + 360) – (–184 + 0)
= –1092 + 184
This was an AME question
–1
= – 908 kJ mol .
Use the bond energy data provided below calculate the
energy change for the reaction:
CH4(g) + 2O2(g)
Bond
Average bond energies in
kJ mol–1
CO2(g) + 2H2O(g)
C–H
413
Evidence
Bonds broken: Bonds formed:
4  C–H = 4(413)
2  C=O = 2(– 745)
2  O=O = 2(498)
4  O–H = 4(– 463)
+2648
–3342
Overall –3342 + 2648 = – 694 kJ.mol–1
C=O
745
O=O
498
H–O
463
A
Correct
method
with one
error.
M
Correct
answer
and
units
Explain why the reaction
H2O(g)
H2O(l) is exothermic
Energy is released when bonds are formed (between molecules).
Explain in full what is meant by ΔcH° (S(s)) = –297 kJ mol–1.
When one mole of sulfur is burned (completely) to produce
SO2, 297 kJ of heat energy is released, when starting and
finishing conditions are 25 °C and 101.3 kPa (or at
standard conditions).
Explain why ΔcH° (S(s)) = ΔfH° (SO2(g)).
ΔfH°(SO2(g)) is the heat energy released or
absorbed when 1 mole of SO2 is formed from its
elements under standard condition. Burning
sulfur in oxygen produces sulfur dioxide from its
elements so the two energy changes represent
the same process.
Sulfuric acid is made by dissolving SO3 in water. Calculate the final
temperature of a solution of H2SO4 made by dissolving 0.03 moles
of SO3(g) in 250mL of water at 20 °C given:
ΔH = m  c  Δt
Where: m = mass, t = temperature
c = specific heat capacity of water, 4.18 J g–1 °C–1.
H2O(l) + SO3(g)
H2SO4(l) ΔrH = –814 kJ mol–1.
Molar masses H = 1, S = 32, O = 16 g mol–1
ΔH
= 0.03 x 814
= 24.42 kJ
= 24420 J
H
t 
mc
24420

250  4.18
23.36 °C
Final temperature = 23.36 + 20
AME
= 46.36 °C