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) ∆ f Hº (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
Transcript
TTooppiicc 55 EEnneerrggeettiiccss AAnnsswweerrss
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.
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3. What information does the following thermochemical equation provide?
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.
X + Y
en
erg
y
reaction progress
Z
A + B
C−−−−∆∆∆∆H
reactants: least stable
products:most stable
−−−−∆∆∆∆H
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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. en
erg
y
reaction progress
X + Y
A + B
Z
C
reactants: most stable
products:least stable
++++∆∆∆∆H
++++∆∆∆∆H
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?
The standard enthalpy of combustion of octane is ∆HoC = -5512 kJ mol
-1.
The standard enthalpy of combustion of ethanol is ∆HoC = -1371 kJ mol
-1.
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Octane C8H16 Mr = 114.26
Number of moles in 1 g =
n = 1g/114.26 g mol-1
= 8.75 x 10-3
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 =
0.0217 moles x –1371 kJmol-1
= -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.
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5.2 Exercises
1. The specific heat capacity of water is given in the IB Chemistry data booklet as 4.18 J K-1
g-1
. 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 – 57.1 kJ mol
-1, 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
q = (60 g) x (1.05 JK-1
g-1
) 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-1
g-1
.
q = mc ∆T
2510 J = (10 g) x (0.90 J K-1
g-1
) x ∆T
∆T = 279 K
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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
q = (755 g) x (0.51 J K-1
g-1
) 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?
%8585.0kJ202
kJ172
q
q
Total
Water===
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
is -18.0°°°°C. The specific heat capacity of ice is 2.03 J K-1
g-1
.
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.
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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 -1
g -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 -1
n(NaOH) = 4.20g/40gmol -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 -1
= 44 571 J mol -1
= 44 500 J mol -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
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10. In a calorimetry experiment 50.0 cm3 of 2.0 mol dm
3 HNO3 and 50.0 cm
3 of 2.0 mol dm
3 NaOH, both
with initial temperatures 21.4oC were mixed in a calorimeter. The temperature of the mixture rose to
25.9 o
C.
a) Calculate the enthalpy of neutralization.
Use q = mc∆T
The specific heat of water is 4.18 J K -1
g -1
= (100 g) x (4.18 J K -1
g -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
cm3 of 2.0 mol dm
-3 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.
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
