Topic 5 Energetics Answers
5.1 Exercises
1. Define the following terms:
a) combustion
A chemical reaction in which a substance reacts with oxygen, producing just heat or heat and light.
b) neutralisation
The process by which an acid reacts with a base to form a salt and water.
c) exothermic reaction
A chemical reaction that releases heat into its surroundings.
d) endothermic reaction
A chemical reaction that absorbs heat from its surroundings.
e) standard state of a substance
Refers to a substance in its standard and normal physical state under standard conditions, ie at normal
atmospheric pressure (1 atm) and an ambient temperature of 25°C/298K.
f)
the º in ∆Hº
The º symbol indicates a standard value. That is ∆Hº means “the standard enthalpy change”. The standard state
of a system is a reference value in thermodynamic measurements, and denotes a pressure of 1 atm and
concentration 1 mol dm-3, and reactants and products in their standard states. Temperature should always be
indicated but is normally 298 K in order for comparisons to be made.
g) standard enthalpy change of reaction (∆Hº)
The enthalpy change for a reaction in which the reactants and products are all in their standard states, and the
reaction takes place under standard conditions, ie 1 atm pressure, 1 mol dm-3 concentration
h) ∆fHº (298 K)
Indicates the enthalpy change of formation of a compound in its standard state from its constituent elements in
their standard states, at 298 K
2. Give the meaning of the use of a – sign and of a + sign with ∆H, with reference to an enthalpy level
diagram.
A -∆H value indicates an exothermic reaction; energy is released (given out).
The negative value can be derived from an enthalpy level diagram for an exothermic reaction (see figure 2,
page 266) where products have less energy than the reactants; hence the negative enthalpy value, which
indicates that heat is released to the surroundings.
Similarly, a +∆H value indicates an endothermic reaction, heat is taken in (or absorbed) from surroundings.
The positive value can also be derived from an enthalpy level diagram for an endothermic reaction (see figure 2)
where products have more energy than the reactants; hence the positive enthalpy value, which indicates that
heat is absorbed from the surroundings.
1
3. What information does the following thermochemical equation provide?
0.5H2(g) + 0.5Br2(l) HBr(g)
∆fHº (298 K) = -36 kJ mol -1
The equation itself shows that 0.5 mole of hydrogen gas reacts with 0.5 mole of bromine liquid to form one mole
of HBr gas. The notation at the end of the equation indicates that when this reaction takes place under standard
conditions and involves products and reactants in their standard states (given by the º symbol) at a temperature
of 298 K. 36 kJ of energy per mole of HBr formed is released (deduced from negative ∆H value) to the
surroundings. This is known as the standard enthalpy of formation of HBr at 298 K, as one mole of HBr is
formed under standard conditions.
4. For any reaction what is the relationship between temperature change, enthalpy change and
whether the reaction is exothermic or endothermic?
If a reaction is accompanied by an increase in temperature, it has released heat to the surroundings and
therefore products have less energy than reactants, and there is a negative enthalpy change. This is known as
an exothermic reaction.
Likewise, if a reaction is accompanied by a decrease in temperature, it has absorbed heat from its surroundings
and therefore products have more energy than reactants, and there is a positive enthalpy change. This is known
as an endothermic reaction.
5. An exothermic reaction:
A: absorbs energy from the surroundings
B: has products with greater energy than reactants
C: has products that are more energetically stable than reactants
D: has a positive enthalpy change
Answer: C in an exothermic reaction, products have less energy than the reactants and are therefore more
energetically stable.
A incorrect as exothermic reactions release energy to the surroundings; B incorrect, the products of an
exothermic reaction have less energy than the reactants; D incorrect exothermic reactions have a negative
enthalpy change.
6. What is the difference between heat and temperature?
Heat (q) is the transfer of energy that results from a temperature difference. Temperature is a measure of how
hot or cold something is. Temperature determines the direction of heat flow between two objects in contact. Two
objects with the same temperature are said to be in thermal equilibrium as there is no net heat flow between
them. So heat is the method of energy transfer and temperature is a way of measuring the amount and direction
of that energy transfer.
7. What is the difference between heat and work?
From the answer to Q 6, heat is the transfer of energy that results from a temperature difference. Work is also a
transfer of energy, but it takes place when an object is moved against an opposing force. For example, gases
burning in the cylinder of a car engine transfer energy to the wheels of that car. Work is done when the pistons
in the cylinder pushes out against the opposing force of the atmospheric pressure, and makes the car move.
Work results from a useful transfer of energy.
2
8. What does the word stability mean in the context of chemistry? How do the relative stabilities of
reactants and products compare for an exothermic reaction?
When we talk about the stability of a molecule we are really talking about its energy. A high energy molecule is
said to be unstable. However chemical stability or “resistance to change” has two meanings. It must be
considered in both kinetic (movement) and thermodynamic (heat) terms. Just because a system or molecule is
thermodynamically unstable, (like O2(g) or H2(g)) does not mean it will react spontaneously. They are kinetically
stable. For the purposes of this course however, stability refers to thermodynamic energy. In an exothermic
reaction the products have less thermodynamic energy than the reactants, and thus the products are more
stable than the reactants.
9. Label the following equations as either endothermic or exothermic:
a) 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
This is combustion, and combustion is an exothermic process as it produces very the very stable (low energy)
products carbon dioxide and water.
b) H2O(s) H2O(l)
This is a phase change (melting ice) which required energy, so it is endothermic.
c) NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)
This is a neutralisation reaction, which are exothermic as it they produce low energy products (including water),
and remove high energy reactants (such as reactive acids and bases).
d) C2H5OH(l) + O2(g) 2CO2(g) + 3H2O(l)
This is combustion, and combustion is an exothermic process.
e) 2HNO3(aq) + Mg(OH)2(s) Mg(NO3)2(aq) + 2H2O(l)
This is a neutralisation reaction, which are exothermic.
10. Enthalpy level diagrams
a) Draw an enthalpy level diagram for the strongly exothermic reaction, A + B C. Now
superimpose on this diagram a diagram for the less strongly exothermic reaction: X + Y Z.
reactants:
least stable
energy
A+B
X+Y
−∆ H Z
−∆H
products:
most stable
reaction progress
3
C
b) Draw an enthalpy level diagram for the strongly endothermic reaction, C A + B. Now
superimpose on this diagram a diagram for the less strongly endothermic reaction: Z X + Y.
energy
products:
least stable
+∆H A + B
reactants:
most stable
C
+ ∆H X + Y
Z
reaction progress
c) On the diagrams you have drawn, indicate the relative stabilities of the reactants and products
and the sign of the enthalpy change for each reaction.
11. Supply the missing words: When a chemical bond forms energy is released When a chemical bond
is broken energy is absorbed. The overall difference in energy between reactants and products
determines whether or not the enthalpy change has a positive or a negative value.
12. Ethanol evaporates readily to form a gas: C2H5OH(l) C2H5OH(g)
a) The enthalpy change for this reaction is known as the enthalpy of vaporisation. What sign would
you give to this ∆H and would you expect the value to be large or small compared to the
enthalpy of vaporisation of water? Hint: how does the volatility of alcohol and water compare?
As this reaction produces molecules of greater energy (a gas) than the reactant (a liquid). Energy is required to
overcome the intermolecular forces between ethanol molecules in the liquid phase in order for molecules to
leave the surface of the liquid and evaporate (become a gas). Since energy is required, the reaction is
endothermic, and so it would have a positive value of ∆H. As ethanol evaporates more readily than water it must
require less energy to become a gas, so the value of ∆H must be smaller for ethanol than it is for water. There
are more hydrogen bonds per molecule of water than per molecule of ethanol overall.
13. Outline two main problems with the use of fossil fuels for energy considering what you have learnt
about their combustion.
Production of CO2 and CO emissions. CO2 is a greenhouse gas and is the main cause of global warming.
Carbon monoxide is a toxic air pollutant and also contributes to global warming as it reacts with other
compounds in the atmosphere to form methane, which is a potent greenhouse gas.
14. Calculate of the energy content of ethanol versus octane per gram, considering their standard heats
of combustion. What other factors should be taken into account when sourcing fuel?
o
C
The standard enthalpy of combustion of octane is ∆H
-1
= -5512 kJ mol .
The standard enthalpy of combustion of ethanol is ∆HoC = -1371 kJ mol-1.
4
Octane C8H16 Mr = 114.26
Number of moles in 1 g =
-1
-3
n = 1g/114.26 g mol = 8.75 x 10 moles
Energy content per gram =
8.75 x 10-3 moles x –5512 kJmol-1 = -48.24 kJ
Ethanol C2H5OH Mr = 46.08
Number of moles in 1g
n = 1g/46 g mol-1 = 0.0217 moles
Energy content per gram =
-1
0.0217 moles x –1371 kJmol = -29.75 kJ
As well as the energy content per gram of fuel the other factors that should be taken into account when sourcing
fuel include how much energy is needed to extract or mine the substance, how much energy is required to
transport the fuel, what are the by-products of combustion and how efficient is the fuel (i.e. a fuel may have a
high energy content but may be very inefficient, very little of the fuel may be converted to useful energy). In the
case of ethanol, which is often extracted from corn fields, we must also consider the cost of farming the land for
fuel, as in this case of many countries in South America, the growing of crops for ethanol has come at the great
price of deforestation, in particular the Amazon rainforest.
Geosequestration
Fossil fuels such as coal, oil and natural gas currently supply around 85 per cent of
the worlds energy needs. The International Energy Agency predicts that fossil fuels
will continue to be used heavily for many years. We can decrease our greenhouse
emissions such as CO2 gas by increasing energy efficiency, making greater use of
renewable energy, using low-carbon intensive fuels and through geosequestration.
Geosequestration is the long term storage of CO2 in the ground.
There are three main steps:
1. Capture the CO2 at the source, such as a power plant or industrial facility.
2. Transport the CO2, usually through a pipeline to the geological storage site.
3. Inject the CO2 deep underground into a geological reservoir, where it is stored.
5
5.2 Exercises
-1 -1
1. The specific heat capacity of water is given in the IB Chemistry data booklet as 4.18 J K g . What
does this mean?
4.18 joules of energy are required to raise the temperature of 1 gram of water by one Kelvin.
2. Define the following standard enthalpies of:
a) combustion
The standard enthalpy of combustion, ∆H°comb is the energy released when one mole of a substance undergoes
complete combustion in O2 under standard conditions.
b) neutralisation
The standard enthalpy of neutralization, ∆Hºneut is the energy released when an acid reacts with a base to form
a salt and water under standard conditions. The enthalpy of neutralisation is quoted in kJ mol-1 of water formed.
The standard enthalpy of neutralisation for a strong acid reacting with a strong base is
-1
– 57.1 kJ mol , no matter the acid or the base. Strong acids and strong bases are
completely ionised so the reaction is always the same:
H+(aq) + OH-(aq) H2O ∆Hºneut = -- 57.1 kJ mol-1
c) solution
The standard enthalpy of solution, ∆Hºsoln is the energy absorbed or released when a substance completely
dissolves into its constituent ions to form a solution under standard conditions. Completely dissolves means that
further dilution does not result in additional temperature change.
3. Calculate the heat needed to raise the temperature of 60.0 g of benzene from 21.2°°C to 36.2°°C given
the specific heat capacity of benzene is 1.05 J K-1 g-1.
Use the equation q = mc ∆T
Where q = heat change in J
m = mass in g
c = specific heat capacity
∆T = temperature change in K
∆T = 15 K
q = mc ∆T
-1 -1
q = (60 g) x (1.05 JK g ) x (15 K)
q = 945 J required
4. A 10.0 g block of aluminium is supplied with 2510 J of heat. What is the temperature change of the
aluminium? The specific heat capacity of aluminium is 0.90 J K-1g-1.
q = mc ∆T
-1
-1
2510 J = (10 g) x (0.90 J K g ) x ∆T
∆T = 279 K
6
5. The specific heat capacity of stainless steel is 0.51 J K-1 g-1.
a) Calculate the heat that must be supplied to a 755 g stainless steel saucepan containing 536 g of
water in order to boil the water. The initial temperature of the water is 23.1°°C.
Energy required to heat the saucepan to 100°C.
Assume that the saucepan is in thermal equilibrium with the water (at 23.1°C.)
q = mc ∆T
-1
-1
q = (755 g) x (0.51 J K g ) x (76.9 K)
q = 29 610 J or 29.6 kJ required (to 3 sig. fig.)
Energy required to heat the water to 100°C.
q = mc ∆T
q = (536 g) x (4.18 J K-1 g-1) x (76.9 K)
q = 172 293 J or 172 kJ required (to 3 sig. fig.)
Total energy required
q =172 kJ + 29.6 kJ
q = 202 kJ (to 3 sig. fig.)
b) What percentage of the heat is used to raise the temperature of the water?
q Water 172 kJ
=
= 0.85 = 85%
q Total
202 kJ
c) What assumptions are made in the calculation of part b)?
This is assuming that no heat is lost to the surroundings. As heat is almost always lost to the surroundings,
more heat than necessary will be needed to boil the water, and a smaller percentage of all the heat supplied to
the system will actually be used to boil the water.
6. How much heat (in kJ) is given off when 1.26 x 1014 g of ammonia is produced according to the
equation at STP?
N2(g) + 3H2(g) 2NH3(g) ∆ H°° = -92.6 kJ
The energy released when one mole of ammonia, NH3, is produced is –92.6/2 = 46.3 kJ mol-1. In 1.26 x 1014 g
there are 1.26 x 1014 g/17.04 g mol-1 = 2.1 x 1015 moles, therefore energy released = 2.1 x 1015 moles x 46.3 kJ
mol-1 = 9.7 x 1016 kJ.
7. How much energy must be removed to freeze water in an ice cube rack containing 25 g of water?
The initial temperature of the water is 24.6°°C and the final temperature of the ice cubes in the freezer
-1
-1
is -18.0°°C. The specific heat capacity of ice is 2.03 J K g .
The heat energy removed to freeze the water and the heat removed to cool the ice must be considered
separately.
Heat removed to freeze water (from 24.6°C to 0°C). Use specific heat of water.
q = mc ∆T
q = (25 g) x (4.18 J K-1 g-1) x (24.6 K)
q = 2571 J removed
Heat removed to cool ice (from 0°C to -18°C). Use specific heat of ice.
q = mc ∆T
q = (25 g) x (2.03 J K-1 g-1) x (18 K)
7
q = 914 J removed
Total energy removed
q = 2571 J + 914 J
q = 3485 J or 3.49 kJ removed (3 sig. fig.)
8. Describe a simple laboratory experiment to measure the enthalpy of combustion of ethanol. Mention
apparatus required and the limitations and errors involved. Draw a simple diagram, if this helps with
your answer.
Need to know the mass of alcohol burnt and the temperature change in order to determine the heat transferred
by the burning of alcohol to a measured volume of water. As the ethanol burns, it will release energy equal to
the enthalpy of combustion, which will heat the water and the amount of heat transfer can be measured with the
thermometer. The following experiment can be used:
Limitations in the experiment include:
Lack of thermal insulation. Not all heat produced is transferred to the water.
This will not measure the standard enthalpy of combustion (ie heat released on a per mole basis) although this
may be calculated if the exact amount of ethanol burnt is known. The accuracy of the experiment will be limited
by the errors outlined below:
Errors in the experiment include:
Heat loss to the environment (surrounding air, beaker, matt, tripod, thermometer).
The measuring apparatus (such as the thermometer or balance) used have an associated error.
The amount of oxygen available to the ethanol, enough oxygen must be available to ensure complete
combustion, or by products such as carbon monoxide will be produced and this will affect the value obtained for
the heat of combustion.
8
9. The results from a student’s experiment to determine the enthalpy of solution of NaOH are as
follows:
mass NaOH = 4.20 g
dissolved in 100 cm3 of distilled water
initial temperature = 25.2oC
final temperature = 36.4oC
a) is the process endothermic or exothermic?
There was an increase in temperature of the water; meaning heat was released in order to heat the water,
therefore the process is exothermic.
b) calculate the ∆Hsoln of NaOH.
Use
q = mc∆T
= (100 g) x (4.18 J K -1g -1) x (11.2 K)
Note: ∆T = 36.5-25.2 = 11.2 ºC or, 309.5 – 298.2 = 11.2 K, i.e. ∆T is the same for oC or K.
= 4681.6 J
= 4680 J (to 3 sig. fig.)
This is the heat energy given off for the dissolution of 4.20 g of NaOH. In order to work out the ∆Hsoln of NaOH
per mole, the number of moles of NaOH in 4.20 g is required. Therefore, we need the Mr for NaOH.
Mr(NaOH)
= 22.99 + 16.00 + 0.105 = 40 g mol
n(NaOH)
= 4.20g/40gmol
-1
-1
= 0.105 mol (to 3 sig fig)
4680 J of heat are released when 0.105 moles of NaOH dissolve.
Therefore 4680/0.105 J of heat is released when 1.0 mole of NaOH is dissolved.
Heat liberated = 4680/0.105 J mol
= 44 571 J mol
-1
= 44 500 J mol
-1
-1
(to 3 sig. fig.)
∆Hsoln NaOH = –44.6 kJ mol-1
c) what is the term used for a well insulated reaction vessel used to measure ∆H of reaction?
calorimeter
d) List three assumptions or approximations made in this experiment.
•
That the reaction conditions were conducted in a well-insulated calorimeter where the amount of heat
lost or gained by the calorimeter itself or to the surroundings is negligible.
•
That the reaction conditions were those of constant pressure, in doing so, heat change (q) per mole =
∆Hsoln
•
That the heat capacity of the solution to that of pure water. This is approximately valid as only 4.02 g of
NaOH is dissolved in 100 g water (the solution is so dilute).
•
The mixing was consistent and complete and did not add further energy or heat to the system.
•
The mass of the solution was 100 grams, that is we are assuming that 100 cm3 of the solution = 100
grams and the effect of the dissolved NaOH or the density of the water at the given temperature will
have little effect on its mass.
e) Give a thermochemical equation for the reaction.
NaOH(s) Na+(aq) + OH-(aq) ∆Hsoln = -44.6 kJ mol
-1
9
10. In a calorimetry experiment 50.0 cm3 of 2.0 mol dm3 HNO3 and 50.0 cm3 of 2.0 mol dm3 NaOH, both
with initial temperatures 21.4oC were mixed in a calorimeter. The temperature of the mixture rose to
o
25.9 C.
a) Calculate the enthalpy of neutralization.
Use
q = mc∆T
The specific heat of water is 4.18 J K -1g -1
= (100 g) x (4.18 J K -1g -1) x (4.5 K) (∆T = 25.9 - 21.4 = 4.5 ºC, or 298.9 – 294.4 = 4.5 K)
= 1881 J
= 1880 J (to 3 sig. fig.)
Since the temperature rose, this is the heat given off for the reaction of 50.0 cm3 of 2.0 mol dm-3 HNO3 and 50.0
3
-3
cm of 2.0 mol dm NaOH.
Assumptions: see those from question 9 above, which all apply for this question too.
b) Write the thermochemical equation for the reaction.
To write the thermochemical equation, we need to know the amount of heat produced in accordance with the
moles presented in the chemical equation. The chemical equation is:
HNO3 + NaOH NaNO3 + H2O
How many moles of HNO3 and NaOH are in the solutions used?
n = cv
n = 2 x .05 dm3
n = 0.1 moles (of both HNO3 and NaOH)
Therefore the amount of energy released for the reaction involving 0.1 moles is 1880 J. For 1 mole of reactants,
the total energy released will be 1/0.1 x 1880J = 18 800J or 18.8 kJ.
The thermochemical equation is therefore:
HNO3 + NaOH NaNO3 + H2O
∆Hneut = -18.8 kJ mol
-1
c) What is the change in enthalpy for the neutralization per mole of HNO3?
The thermochemical equation above shows the reaction of one mole of HNO3 with 1 mole of HNO3 so the
-1
change in enthalpy for the neutralization of one mole of HNO3 is ∆Hrxn = -18.8 kJ mol .
11. In terms of heat changes bond making is exothermic and bond breaking is endothermic.
12. The heat of formation for water is – 286 kJ mol
-1
a) write the thermochemical equation for this reaction.
H2 (g) + ½O2 (g) H2O (l)
∆H = – 286 kJ
b) Is this reaction exothermic or endothermic?
Exothermic, as the heat of formation is negative in sign, signifying that energy is released.
c) Is the heat content of the products equal to, less than, or greater than the heat content of the
reactants?
The heat content (energy) of the products is less than the heat content (energy) of the reactants as this is an
exothermic reaction.
10
13. 50.0 cm3 of 2.0 M NaOH(aq) is place in an insulated cup, and a thermometer is used as a stirrer as
5.0 cm3 portions of HCl(aq), of unknown concentration, is added from a burette. The temperature
change results observed are:
3
Volume of HCl(aq) cm
o
Temperature C
0.0
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
45.0
50.0
22.2
24.4
26.4
28.4
30.1
31.1
30.4
29.2
29.9
28.8
28.2
a) Plot a graph showing temperature against volume of acid (the independent variable is on X-axis,
dependent variable on Y axis)
Heat of neutralization of NaOH by addition of HCl
35
Temperature ºC
30
25
20
15
10
5
0
0
5
10
15
20
25
30
35
40
45
50
55
3
Volume HCl solution added (cm )
b) Account for the shape of this graph. Does the shape suggest all readings are correct?
The graph shows a steady increase in temperature of the NaOH solution with increasing increments of HCl
solution added until 25 cm3 has been added. At this point the temperature reaches a maximum. Further
additions do not cause any further temperature rise and the temperature of the resultant solution begins
3
decrease with further increments. One may reasonably assume that at 25 cm of HCl added, the neutralization
of the NaOH is complete and no further reaction occurs, as no further heating is observed and the resultant
solution begins to cool. There seems to be an inconsistency with the reading recorded upon the addition volume
reaching 40 cm3, suggesting a random error such as an incorrect thermometer reading present, or the solution
was not stirred thoroughly.
c) Calculate the enthalpy of neutralization of sodium hydroxide with hydrochloric acid
Use
∆T
q = mc∆T
= 31.1ºC – 22.2ºC = 8.9 ºC)
The specific heat of water is 4.18 J K -1g –1
3
3
v = 50 cm + 25 cm = 75 cm
3
m = 75 g
-1
-1
= (75 g) x (4.18 J K g ) x (8.9 K)
11
= 2790.15 J
= 2790 J (to 3 sig. fig.)
Since the temperature rose, this is the heat given off for the reaction of 50.0 cm3 of 2.0 mol dm3 NaOH and 25.0
3
cm of HCl solution. Therefore the enthalpy of neutralisation is –2790 J mol
-1
d) Calculate the molarity of the hydrochloric acid used.
Write the balanced equation for the reaction:
NaOH + HCl NaCl + H2O
From the equation it can be seen that one mole of HCl is required to neutralise one mole of NaOH. If we
assume that when the temperature reaches its maximum the reaction is complete, then it is at this point that the
number of moles of acid and base are equal, ie nbase = nacid
nbase = cv = 2 mol dm-3 x 0.05 dm3 = 0.1 moles.
Therefore,
nacid = 0.1 moles
c = n/v = 0.1 moles /0.025 dm3
c= 4 mol dm-3
14. Carbon burns in a limited supply of oxygen according to this equation:
∆ H = -110 kJ mol-1
C(s) + ½O2(g) CO(g)
Explain why subscripts showing the physical state of chemicals, and coefficients, are vital in any
equation where a ∆ H is quoted.
A molecule in a different state can have a different energy. For example, water in the gas form is higher in
energy than water in the liquid form, as water must absorb energy to become a gas. The amount of a chemical
also affects enthalpy. In the equation above, when one mole of carbon monoxide is produced 110 kJ of energy
are released. If two moles of carbon monoxide were produced, then 220 kJ of energy would be released as
there would be twice the quantity of reactants.
15. Write thermochemical equations for the following:
-1
a) ∆ H(sol) = - 44.4 kJ mol for sodium hydroxide
+
-
NaOH(s) Na (aq) + OH (aq)
∆Hsol = - 44.4 kJ mol
-1
b) ∆ Hneut = -57 kJ mol-1 for potassium hydroxide and HCl(aq)
KOH(aq) + HCl(aq) KCl(aq) + H2O(l)
∆Hneut = -57 kJ mol
-1
c) ∆ Hcomb = -2220 kJ mol-1 for propane, C3H8
C3H8(g) + 6O2(g) 3CO2 + 4H2O
∆Hcomb = -2220 kJ mol
-1
16. In a laboratory experiment to determine the enthalpy of combustion of ethanol the mass of ethanol
burnt was measured, as was the temperature rise in the water being heated.
a) Name the apparatus used to contain the water.
Calorimeter
b) Write the thermochemical equation for the complete combustion of ethanol.
CH3CH2OH + 3½O2 2CO2 + 3H2O
∆Hcomb = -1371 kJ mol-1 (∆H combustion value from data book)
12
c) An accurate value for ∆ H of combustion for ethanol is –1371 kJ mol-1. However, the results from
the laboratory experiment consistently give results that are lower than this. List any
assumptions made and the sources of error associated with this experiment.
•
Assuming that heat transfer from burning ethanol was complete and there were no heat losses.
•
The heat absorbed by the apparatus used is not taken into account
•
Mass loss due to evaporation of ethanol throughout the course of the experiment is assumed to be
negligible
5.3 Exercises
1. Hess’s Law
a) What is Hess’s Law?
The energy released or absorbed in the process of converting reactants to products is constant and
independent of the pathway used for the conversion. If reactants can be converted into products by a series of
reactions, the sum of the heats of these reactions (taking into account their sign) is equal to the overall heat of
reaction for the direct conversion of reactants to products. It is sometimes called the law of constant heat
summation, is an extension of the law of conservation of energy.
b) Explain how Hess’s law and the law of conservation of energy are intrinsically linked.
The law of conservation of energy states that, in a closed system, energy cannot be created or destroyed – it
can only change form. If we were to create the same product by two different pathways, either of these
pathways cannot use either more or less energy than the other, as the law of conservation of energy must be
obeyed.
Defying the Laws of Thermodynamics
If we could somehow figure out how to defy the first law of thermodynamics, we could
effectively “create” energy and this would solve all the world's energy problems!
c) What are the advantages of Hess’s law in relation to reactions for which enthalpy changes
cannot be measured directly?
Hess’s law allows us to use known enthalpy values to calculate enthalpy changes for reactions that cannot be
directly measured. For a reaction where we cannot measure the heat change by experiment, we measure a
different route that uses the same reactant and product molecules, and then combine the enthalpy values for
these known reactions.
2. Why is the standard enthalpy change of formation of O2 zero?
O2 is the naturally occurring state of oxygen, and is not formed from any constituent elements, no energy is
released or required to “make” it!
3. Write the thermochemical equations that give the values of the standard enthalpies of formation of
a) KClO3(s) (∆
∆ H°f = -391 kJ mol-1)
The standard enthalpy of formation creates one mole of products formed from constituent elements in their
standard states.
∆H°f = -391 kJ mol-1
K(s) + ½Cl2(g) + 1½O2(g) KClO3(s)
-1
b) CH3COOH(l) (∆
∆ H°f = -487 kJ mol )
13
2C(s) + O2(g) + 2H2(g) CH3COOH(l)
∆H°f = -487 kJ mol
-1
-1
c) SO3(g) (∆
∆ H°f = -396 kJ mol )
∆H°f = -396 kJ mol-1
⅛S8(s) + 1½O2(g) SO3(g)
S8 is a common allotrope of sulfur.
4. The enthalpy of combustion of methanol, carbon (graphite) and hydrogen can be used to calculate
the enthalpy of formation of methanol.
a) Write thermochemical equations for each combustion reaction (Hint: use your data booklet to
find ∆H°° values)
The enthalpy of combustion refers to the complete combustion of one mole of substance.
CH3OH(l) + 1½O2(g) CO2(g) + 2H2O(l)
∆H = -715 kJ mol
-1
-1
C(s) + O2(g) CO2(g)
∆H = -394 kJ mol
H2(g) + ½O2(g) H2O(l)
∆H = -286 kJ mol-1
b) Calculate ∆H°° for the reaction: 2C(s) + 4H2(g) + O2(g) 2CH3OH(l). Use both enthalpy cycles and
equation manipulation methods; the answer obtained using each method should be the same.
Equation Method:
To get 2 carbons on the left, we need to multiply the combustion of graphite equation by 2:
∆H = -788 kJ mol-1 (= -394x2)
2C(s) + 2O2(g) 2CO2(g)
To get 4 H2 on the left, we need to multiply the combustion of hydrogen gas (H2) by 4:
∆H = -1144 kJ mol-1 (= -286 x 4)
4H2(g) + 4O2(g) 4H2O(l)
On the right side of the equation we need two moles of methanol, so we use the combustion of methanol
equation reversed (remember when reversing equations the sign of the ∆H value must be changed) and
multiplied by two:
2CO2(g) + 4H2O(l) 2CH3OH(l) + 3O2(g)
∆H = 1430 kJ mol-1 (= -715x -2)
Now we can write them together and add the equations:
2C(s) + 2O2(g) 2CO2(g)
∆H = -788 kJ mol-1
4H2(g) + 2O2(g) 4H2O(l)
∆H = -1144 kJ mol
2CO2(g) + 4H2O(l) 2CH3OH(l) + 3O2(g)
∆H = 1430 kJ mol
Total: 2C(s) + 4H2(g) + O2(g) 2CH3OH(l)
∆H = -502 kJ mol-1
-1
-1
Enthalpy Cycle Method:
2C(s)
+
2O2(g)
4H2(g)
∆Hx
2CH2OH(l)
2O2(g)
3O2(g)
2CO2(g)
+
4H2O(l)
14
In accordance with Hess’s Law, the energy required to form CO2 and H2O by the direct combustion of hydrogen
and oxygen or by the indirect route via ethanol is the same. This information can be used to construct the
relationship:
[(2x 384-) + (4 x –286)] = ∆H°x + (-1430)
∆Hx = [(2 x -394) + (4 x -286) – (-1430)
∆Hx = -502 kJ mol-1
c) Why is the value obtained in b) not the value for the enthalpy of formation of CH3OH?
The equation is for the formation of TWO moles of methanol, not one, as must be the case for an enthalpy of
formation value.
d) Calculate the value for the molar enthalpy of formation for methanol from your answer to b). It
should concur with the value given in the IB Chemistry data booklet.
C(s) + 2H2(g) + ½O2(g) CH3OH(l)
∆H°f = -251 kJ mol
-1
5. The partial oxidation of carbon to carbon monoxide, CO is not easy to control as CO2 also forms,
meaning it is hard to measure the reaction enthalpy. The complete oxidation of carbon and of CO to
form CO2 is a more straightforward reaction to induce, and we can use these known reactions to
determine the ∆H°° for the formation of CO.
a) Write the thermochemical equations for the enthalpy of formation of CO2 and combustion of CO
from their constituent elements. The enthalpy change for the formation of CO2 is ∆H°° = -394 kJ
mol-1 and the enthalpy of combustion for CO is ∆H°° = -283 kJ mol-1
C(s) + O2(g) CO2(g)
∆H = -394 kJ mol-1
CO(g) + ½O2(g) CO2(g)
∆H = -283 kJ mol-1
b) Using the equations you have written above, calculate the ∆H°° for the formation of CO.
Combine the equations as follows (reversing the combustion of carbon monoxide)
C(s) + O2(g) CO2(g)
∆H = -394 kJ mol-1
CO2(g) CO(g) + ½O2(g)
∆H = +283 kJ mol-1
Total: C(s) + ½O2(g) CO(g)
∆H = -111 kJ mol-1
6. Using the molar enthalpy of vaporisation of water (+ 44 kJ) and the molar enthalpy of formation for
water (see IB Chemistry data book) calculate the value of ∆H°° for the reaction:
H2(g) + ½O2(g) H2O(g)
The equation for the molar enthalpy of vaporisation of water is:
H2O(l) H2O(g)
∆H° = + 44 kJ mol-1
This states that 44 kJ of energy is required to convert 1 mol of water from the liquid to the gaseous phase.
The equation for the molar enthalpy of formation of water is:
H2(g) + ½O2(g) H2O(l)
∆H° = -286 kJ mol-1
This equation also represents the molar enthalpy of combustion of H2 and this is where the value can be found
in your IB Chemistry data booklet. Note that the value of -286 kJ mol-1 corresponds to the formation of liquid
water. The question asks for the enthalpy of formation of gaseous water, which of course will have a different
value.
15
To get 1 H2(g) on the left, we need 1 x the second equation and to get 1mole of gaseous water H2O(g) on the
right we need 1 x the first equation and add together.
H2(g) + ½O2(g) H2O(l)
∆H° = -286 kJ mol-1
H2O(l) H2O(g)
∆H° = + 44 kJ mol-1
Total : H2(g) + ½O2(g) H2O(g)
∆H° = -242 kJ mol
-1
The liquid water in each given equation cancels out to give the desired equation. Check that your answer makes
sense! The energy released in the formation of one mole of gaseous water should be less than the energy
released when one mole of liquid water is formed, as the gaseous form of water contains more energy.
7. Calculate the change in enthalpy for the graphite diamond transition. Use information given in the
data book.
C(graphite) + O2(g) CO2(g)
∆H = -394 kJ mol-1
C(diamond) + O2(g) CO2(g)
∆H = -395 kJ mol-1
So for the reaction C(graphite) C(diamond) the enthalpy will be: (-394 kJ + 395 kJ) ∆H = +1 kJ mol-1
8. Calculate the standard reaction enthalpy for the hydrogenation of ethyne to ethane. (Hint: use data
from the IB data booklet).
C2H2(g) + 2H2(g) C2H6(g)
C2H2(g) + 2½O2(g) 2CO2(g) + H2O(l)
∆H = -1299 kJ mol-1
H2(g) + ½O2(g) H2O(l)
∆H = -286 kJ mol-1
C2H6(g) + 3½O2(g) 2CO2(g) + 3H2O(l)
∆H = -1560 kJ mol-1
Add the equations as follows:
C2H2(g) + 2½O2(g) 2CO2(g) + H2O(l)
∆H = -1299 kJ mol-1
2H2(g) + O2(g) 2H2O(l)
∆H = -572 kJ mol-1
2CO2(g) + 3H2O(l) C2H6(g) + 3½O2(g)
∆H = +1560 kJ mol-1
Total: C2H2(g) + 2H2(g) C2H6(g)
∆H = -311 kJ mol-1
9. Calculate the standard enthalpy of formation of SO3 at 25 oC given:
S8(s) + 8O2(g) 8SO2(g)
∆H°° = -2375 kJ mol
2SO2(g) + O2(g) 2SO3(g)
∆H°° = -198 kJ mol
-1
-1
The reaction for the standard enthalpy of formation of SO3 is:
S8(s) + 1½O2(g) SO3(g) ∆H = ?
To get 1/8 moles of sulfur on the left we must divide the first equation by 8, and to get 1 mole of SO3 on the right
hand side we need to divide equation 2 by 2:
⅛S8(s) + O2(g) SO2(g)
∆H° = -296.9 kJ mol-1 (= -2375/8)
SO2(g) + ½O2(g) SO3(g)
∆H° = -99 kJ mol-1 (=198/2)
Total: ⅛S8(s) +1½O2(g) SO3(g)
∆H° = -396 kJ mol-1
16
10. The intense heat given out by the oxyacetylene torch is used to cut and weld metals.
a) Given the molar enthalpy for the complete combustion of ethyne (acetylene), C2H2, is -1299 kJ
-1
mol , write the thermochemical equation for this reaction.
C2H2(g) + 2½O2(g) 2CO2(g) + H2O(l)
∆H = -1299 kJ mol
-1
b) What amount of heat is liberated when 0.260 kg of acetylene is burnt?
Convert mass into moles:
m = 260 g , M = (2 x 12.01 + 2 x 1.01) = 26.04
Therefore n = 9.98 moles
The enthalpy of combustion is quoted in kilojoules per mole. For 9.98 moles the enthalpy must be (-1299 kJ x
9.98 moles) = -12 970 kJ of heat is liberated.
11. Using the following equations:
1) 2NO(g) + O2(g) 2NO2(g)
∆H°° = -114.1 kJ mol-1
2) 4NO2(g) + O2(g) 2N2O5(g)
∆H°° = -110.2 kJ mol-1
3) N2(g) + O2(g) 2NO(g)
∆H°° = +90.25 kJ mol
-1
a) Calculate the standard enthalpy of formation of dinitrogen pentoxide
The reaction for the standard enthalpy of formation of 2N2O5 is
N2(g) +2½O2(g) N2O5(g)
We can rearrange the above equations in the following way to yield the equation. Note that the second equation
has been halved to make cancelling out easier.
N2(g) + O2(g) 2NO(g)
∆H° = +90.25 kJ mol-1
2NO(g) + O2(g) 2NO2(g)
∆H° = -114.1 kJ mol-1
2NO2(g) + ½O2(g) N2O5(g)
∆H° = -55.1 kJ mol-1
Total: N2(g) + 2½O2(g) N2O5(g)
∆H° = -78.95 kJ is the standard enthalpy of formation
b) Why is the standard enthalpy of formation of dinitrogen pentoxide not just equal to the value
given in equation 2)?
Equation 2 is not a reaction of the formation of dinitrogen pentoxide from its constituent elements, as nitrogen
dioxide is not a constituent element. Furthermore, equation 2 produces 2 moles of dinitrogen pentoxide, and the
enthalpy of formation is the production of only 1 mole.
17
12. The Apollo 11 Project landed the first man on the Moon on 21 July, 1969. The engines of the lunar
module used methylhydrazine (CH3-NH-NH2) and dinitrogen tetroxide (N2O4) as the propellant. The
reaction is:
4CH3-NH-NH2(l) + 5N2O4(l) 4CO2(g) + 12H2O(l) + 9N2(g)
a) Use the following information to calculate the enthalpy change for the above reaction:
1. C(s) + 3H2(g) + N2(g) CH3-NH-NH2(l) ∆H°° = +53 kJ mol-1
2. N2(g) + 2O2(g) N2O4(l)
∆H°° = -20 kJ mol-1
3. C(s) + ½O2(g) CO(g)
∆H°° = -110 kJ mol-1
4. C(s) + O2(g) CO2(g)
∆H°° = -393 kJ mol
5. H2(g) + ½O2(g) H2O(l)
∆H°° = -286 kJ mol-1
-1
Use the equation method, and rearrange the equations as follows so that the terms on the left hand side of the
equation match the desired equation, and the ones on the right match those on the right, and then cancel out
like terms.
4CH3-NH-NH2(l) 4C(s) + 12H2(g) + 4N2(g)
∆H° = -212 kJ mol-1 (Eq. 1 reversed and multiplied by 4)
5N2O4(l) 5N2(g) + 10O2(g)
∆H° = +100kJ mol-1 (Eq. 2 reversed and multiplied by 5)
4C(s) + 4O2(g) 4CO2(g)
∆H° = -1572kJ mol-1 (Eq. 4 multiplied by 4)
12H2(g) + 6O2(g) 12H2O(l)
∆H° = -3432 kJ mol-1 (Eq. 5 multiplied by 12)
Total: 4CH3-NH-NH2(l) + 5N2O4(l) 4CO2(g) + 12H2O(l) + 9N2(g) ∆H° = -5116 kJ mol-1
b) Is the reaction between methyl hydrazine (CH3-NH-NH2) and dinitrogen tetroxide (N2O4)
exothermic or endothermic? Explain.
This is a highly exothermic reaction, as the enthalpy of reaction is large and negative in sign.
13. Photosynthesis can be represented by the reaction:
6CO2(g) + 6H2O(l) C6H12O6(aq) + 6O2(g)
where C6H12O6 is glucose.
a) How would you determine experimentally the ∆H°° value for this reaction?
The easiest way would be to measure the reverse reaction, which is the combustion of glucose. Through using
a calorimeter, the heat produced by the combustion of glucose could be determined, and by Hess’s law it would
be the same energy (but opposite in sign) as the enthalpy for photosynthesis, taking into account heat lost to the
surroundings and other experimental errors.
b) About 7.0 x 1014 kg of glucose is produced by solar radiation per year on Earth. What is the ∆H°°
change for this amount of glucose? The ∆H°°(form) for glucose = -1274.5 kJ mol-1
From the data book, the enthalpy of combustion of glucose is -2816 kJ mol-1. From Hess’s law the ∆H° for the
-1
production of one glucose by photosynthesis is +2816 kJ mol (see part a.) as this simply the reverse reaction
of the combustion of glucose.
Convert mass of glucose into moles:
17
m = 7.0 x 10 g , M = (6 x 12.01 + 6 x 16.00 + 12 x 1.01) = 180.18
15
Therefore n = 3.89 x 10 moles
The enthalpy of combustion is quoted in kilojoules per mole. For 3.89 x 1015 moles the enthalpy must be (+2816
kJ x 3.89 x 1015moles) = 1.09 x 1019 kJ of heat absorbed.
19
∆H° = +1.09 x 10 kJ
18
14. The enthalpy of formation of trinitrotoluene (TNT, C7H5N3O6) is –67 kJ mol -1. Although TNT is
dangerous as a fuel as it is sensitive to shock, it releases a large amount of gases on
decomposition, making it a powerful rocket fuel as it provides a large amount of thrust. Calculate
the enthalpy of the decomposition reaction:
4C7H5N3O6(s) + 21O2(g) 28CO2(g) + 10H2O(g) + 6N2(g)
(equation 1)
Enthalpy of formation reaction is given as follows:
7C(s) + 3O2(g) + 2½H2(g) + 1½N2(g) C7H5N3O6(s) ∆H° = –67 kJ mol-1
There are 2 other reactions that need to be used (they are numbered equation 2 and 3 with equation 1 being the
given reaction)
C(s) + O2(g) CO2(g) ∆H = -394 kJ
(equation 2)
H2(g) + ½O2(g) H2O(l) ∆H = -286 kJ
(equation 3)
These reactions can be combined as follows to produce the desired equation:
4C7H5N3O6(s) 28C(s) + 12O2(g) + 10H2(g) + 6N2(g)
∆H° = +268 kJ (Eq 1 reversed and multiplied by 4)
28C(s) + 28O2(g) + 21O2(g) 28CO2(g)
∆H = -11032 kJ (Eq 2 multiplied by 28)
10H2(g) + 5O2(g) 10H2O(l)
∆H = -2860 kJ (Eq 3 multiplied by 10)
Total: 4C7H5N3O6(s) + 21O2(g) 28CO2(g) + 10H2O(g) + 6N2(g) ∆H = -13606 kJ mol-1
5.4 Exercises
1. Supply the missing words. Breaking chemical bonds requires energy, so breaking bonds is an
endothermic reaction. When new bonds are made energy is released; the reaction is exothermic.
2. What is the difference between bond enthalpy and average bond enthalpy?
A bond enthalpy is known precisely; as it is only one of a kind (eg. hydrogen to hydrogen bonds can only occur
in H2 so it is known exactly.) An average bond enthalpy is a guide to the strength of a particular kind of bond, for
example carbon hydrogen bonds can occur in many different molecules, and each bond may have a slightly
different enthalpy depending on the rest of the molecule.
3. Why are bond enthalpies of covalent bonds always positive?
Bond enthalpies are the energy required to break bonds, therefore they are endothermic and so have a positive
value.
4. Using the average bond enthalpy for C-H, estimate the energy required to dissociate all of the C- H
bonds in 0.1 mol of CH4?
The average bond enthalpy for C-H is 412 kJ mol-1. There are 4 such bonds in CH4, and to dissociate all of them
-1
would take 1648 kJ mol . In this case there is only 0.1 mol, so the total energy required is 16.5 kJ (3 sig. fig.)
5. Estimate the standard enthalpy of the reaction and state whether it is exothermic or endothermic.
a) CH4(g) + 2F2(g) CH2F2(g) + 2HF(g)
Using the IB Chemistry data booklet the average bond enthalpies of the bonds in the above equations are given
and multiplied by the number of times they occur in the equation:
REACTANTS (bonds broken)
19
C-H = 412 kJ mol-1 x 4 = 1648 kJ mol-1
F-F = 158 kJ mol-1 x 2 = 316 kJ mol-1
Total: 1964 kJ mol-1
PRODUCTS (bonds formed)
C-H = 412 kJ mol-1 x 2 = 824 kJ mol-1
C-F = 484 kJ mol-1 x 2 = 968 kJ mol-1
H-F= 562 kJ mol-1 x 2 = 1124 kJ mol-1
Total: 2916 kJ mol
-1
Sum of bonds broken -sum of bonds formed = 1964-2916 = -952 kJ mol-1
The reaction is an exothermic reaction.
Products are more stable than reactants (more energy required to break bonds of the products) therefore
energy is released, exothermic.
b) CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)
Using the IB Chemistry data booklet the average bond enthalpies of the bonds in the above equations are given
and multiplied by the number of times they occur in the equation:
REACTANTS (bonds broken)
-1
C-H = 412 kJ mol x 4 = 1648 kJ mol
-1
Cl-Cl = 242 kJ mol-1
Total: 1890 kJ mol-1
PRODUCTS (bonds formed)
C-H = 412 kJ mol-1 x 3 = 1236 kJ mol-1
C-Cl = 338 kJ mol-1
H-Cl= 431 kJ mol
-1
Total: 2005 kJ mol-1
Sum of bonds broken -sum of bonds formed = 1890 -2005 = -115 kJ mol-1
The reaction is an exothermic reaction.
Products are more stable than reactants (more energy required to break bonds of the products) therefore
energy is released, exothermic.
6. What is the major limitation of the values determined for ∆ H reaction using average bond
enthalpies?
The bond enthalpies used are only an approximation . For example, a C-H bond in methane, CH4, will have a
different bond enthalpy value to a C-H bond in chloroform, CH3Cl, due to the electronic effects of differing atoms
present in the molecule. Cl is more electronegative than H, this will affect the bond strength of neighbouring
bonds in the molecule.
20
21