5 Energetics
Year 11 DP Chemistry
R Slider
5.1 Exothermic and
Endothermic
Chemical EnergyEnthalpy
Kinetic energy is the energy of motion
•Particles that make up matter are
constantly in motion
Potential energy is stored energy (no motion)
•Every bond between atoms and ions has
stored energy within the bonds
•This stored energy is known as chemical
potential energy
Total energy = Kinetic energy +Potential Energy (known as Enthalpy or heat content–
symbol H)
Change in Enthalpy (ΔH)
Because it is difficult to measure the amount of energy in individual
reactants or products, Chemists measure the change in enthalpy (ΔH)
that occurs when a reaction takes place.
ΔH is the difference in the enthalpy of the products and the enthalpy of
the reactants.
Enthalpy changes result from bonds breaking
and new bonds reforming.
This breaking and reforming of bonds results in
energy changes within the system which are
measured as temperature changes
Note: Standard enthalpy changes are
measured under standard temperature
(298K) and pressure (101.3kPa)
Endothermic or Exothermic?
Endothermic
Some chemical reactions absorb heat
from the surroundings so the products
of the reaction contain more heat.
The surroundings feel colder, ΔH = +
Examples: photosynthesis, cold packs
Exothermic
Some chemical reactions release heat
to the surroundings so the products
have less heat than the reactants
The surroundings feel hotter, ΔH = Examples: combustion, neutralisation
reactions
Heat exchange
Another way to visualise ΔH is to think about energy being exchanged
between the chemical system and the surroundings
Heat loss in a chemical system = Heat gain to the surroundings (Temperature increases)
Heat gain in a chemical system = Heat loss to the surroundings (Temperature decreases)
Source diagrams:
http://www.chemhume.co.uk/ASCHEM/Unit%203/13%20Enthalpy/13%20Enthalpyc.htm
Neutralisation is exothermic
Acid
Base
Salt
Water
Acid + base reactions are called neutralisation reactions because the
low pH (<7) of an acid and the high pH (>7) of a base, results in a pH
closer to neutral (7)
These reactions release heat and are exothermic. For example:
HCl + NaOH  NaCl + H20 + 58 kJ/mol
Combustion is exothermic
Combustion is the process of burning
Most often, the combustion of a
material involves its combination
with oxygen gas.
Because combustion reactions
release energy in the form of heat
and light, they are exothermic.
Example:
The combustion of methane
(natural gas) in oxygen:
CH4 + 2O2  CO2 + 2H2O
∆ Hrxn = -832 kJ/mol (exothermic)
Fire in the Penola Forest (SA), Ash Wednesday 1983
(Source: www.austehc.unimelb.edu.au/ fam/1611_image.html)
Combustion: why exothermic?
Energy required
to break bonds
• Energy is required to break bonds(endothermic)
• Energy is released when bonds form (exothermic)
In the previous example involving the combustion of methane:
CH4 + 2O2  CO2 + 2H2O + 832kJ
A thermochemical
equation
Energy released
when bonds formed
Forming bonds
Releases energy
Breaking bonds
Absorbs energy
H reactants > H products
In summary:
∆ Hrxn = H (products) – H (reactants)
This reaction releases more energy than it absorbs resulting in a negative value for ∆ Hrxn
Energy changes in chemical reactions –
Enthalpy diagrams
Exothermic reactions
The enthalpy of the reactants is
higher than the products.
Endothermic reactions
The enthalpy of the reactants is
lower than the products.
∆ H is -ve
Products
Heat released
Products
Reaction progression
Enthalpy (H)
Enthalpy (H)
Reactants
∆ H is +ve
Heat absorbed
Reactants
Reaction progression
Examples
The combustion of methane is
exothermic and releases
energy to the surroundings
The decomposition of calcium
carbonate is endothermic and
absorbs energy from the
surroundings
Source diagrams:
http://www.chemhume.co.uk/ASCHEM/Unit%203/13%20Enthalpy/13%20Enthalpyc.htm
Activation Energy
The activation energy Ea is the minimum
amount of energy that is required by the
reactants for bonds to be broken and the
reaction to proceed to the products.
enthalpy
Activation
energy
High Ea means strong
bonds in the reactants
Reactants
∆H
Products
Reaction progress
Chemical stability
Exothermic reactions
•Products are more stable than
reactants due to lower enthalpy
content
•The lower the enthalpy the more
stable
•The lower the enthalpy, the
stronger the bonds
Endothermic reactions
•Products are less stable than
reactants due to a higher enthalpy
content
•The higher the enthalpy the less
stable
•The higher the enthalpy, the
weaker the bonds
Source: http://www.teachmetuition.co.uk/Energetics/chemical_energetics1.htm
Summary
Thermochemical Reaction
Exothermic
Endothermic
Enthalpy Change
(ΔH = HP – HR)
HP < HR
HP > HR
Sign ΔH
Negative
Positive
ΔT of surroundings
Increases (warmer)
Decreases (colder)
Enthalpy Diagram
Source diagrams:
http://www.chemhume.co.uk/ASCHEM/Unit%203/13%20Enthalpy/13%20Enthalpyc.htm
Exercises
1. Classify each of the reactions below as endothermic or exothermic
a) C6H12O6(aq) + 6O2(g)  6CO2(g) + 6H2O(l) ΔH = -2801 kJ mol-1
b) 6CO2(g) + 6H2O(l)  C6H12O6(aq) + 6O2(g) ΔH = +2801 kJ mol-1
c) H2(g) + 1/2 O2(g)  H2O(l) + 286 kJmol-1
2. When a spark is introduced into a vessel containing a mixture of hydrogen and oxygen gases,
they react explosively.
a) What is the role of the spark in this reaction?
b) What bonds are broken when this reaction is initiated?
c) What bonds are generated when the products are formed?
d) Is the reaction endothermic or exothermic? Justify your answer.
3. 10.0g of ammonium nitrate is dissolved in 100cm3 of water and the temperature of the solution
decreases from 19.00C to 10.500C.
a) Which is greater, the enthalpy of the reactants or the products?
b) Is the reaction endothermic or exothermic?
c) What chemical bonds were broken when the ammonium nitrate dissolved in water?
4. Consider the reaction represented by the following equation: A + 2B  C
In this reaction, the total enthalpy of the reactants is 80kJmol-1, the total enthalpy for the products
is -90kJ.mol-1and the activation energy for the forward reaction is 120kJ.mol-1.
a) Draw a diagram for the energy profile for this reaction. Label the diagram clearly to show
ΔH and Ea.
b) State whether the forward reaction is endothermic or exothermic
c) Calculate the change in enthalpy
d) Calculate the amount of energy released as the products are formed
e) Calculate the activation energy of the reverse reaction: C  A 2B
Exercises - Answers
1. a) exothermic b) endothermic c) exothermic
2. a)spark is to overcome the activation energy
b) H-H and O=O covalent bonds broken
c) O-H bonds are made during the reaction
d) exothermic as energy is released
3. a) products>reactants
b) endothermic
c) ionic bonds between NH4+ and NO34. a) diagram should show exothermic forward rxn
b) exothermic
c) ΔH = -90 – 80 kJ mol-1 = -170 kJ mol-1
d) 290 kJ mol -1
e) 290 kJ mol-1
5.2 Calculating Enthalpy
Measuring Enthalpy Changes
Recall that changes in temperature are related to changes in the amount of
heat energy that is being absorbed or released in a system. This
quantity of heat energy, q in joules (J), can be calculated using the
following equation and often involves water being heated by reactions:
q = m C ∆T
Heat energy released or
absorbed in Joules (J)
Mass (g)
Change in
Temperature
(final – initial)
Specific heat
Capacity (4.18 J g-1K-1 for water)
Example:
What quantity of energy is required to raise the temperature of 0.5L of water from
200C to 1000C?
q = m C ∆T
= 500g X 4.18 J g-1K-1 X 800C or K
= 167,200 J
= 167 kJ
NB: 0C = K - 273
Heat Capacity Values
Heat capacity is a measure of how much energy a substance can absorb without
changing temperature. These values vary greatly among substances.
Substance
Specific Heat
Capacity
J K-1g-1
Substance
Specific Heat
Capacity
J K-1g-1
Water
4.18
Ethanol
2.41
Ethylene glycol
2.39
Hexane
2.26
50/50 water/E.
glycol
2.86
Chloroform
0.96
Acetone
2.17
Aluminium
0.90
Iron
0.448
Copper
0.386
Specific heat capacity (C), is the amount of heat energy in Joules (J),
required to raise the temperature of 1g of a substance by 1 Kelvin (K).
Practice Questions
1. A Bunsen burner was used to heat 100 cm3 of water for 10 min.
The temperature of the water increased from 15.00C to 800C.
Determine the heat energy change of the water. (note: the density
of water is 1.0 g cm-3)
2. The same Bunsen burner was used to heat a 500g block of copper
for 10 minutes. Assuming the same amount of energy is transferred
from the Bunsen burner to the block as in question 1, determine the
highest temperature that the block could reach if the starting
temperature was 15.00C.
3. Explain the difference in the temperatures considering the same
amount of energy was put into the water and the copper
Solutions:
1)25.9kJ 2) 1490C 3) The water has a much higher heat capacity than the copper which
means the water can absorb much more heat energy before changing temperature.
Calorimetry
What is calorimetry
Water can be used to measure the change in heat
energy in a chemical reaction due to its ability
to absorb heat. This is known as calorimetry.
A calorimeter is a device that is used to measure
the enthalpy change that occurs during a
chemical reaction.
Measuring heat in the laboratory
If a known quantity of water is placed in the
calorimeter and a reaction is carried out, the
change in temperature due to the reaction is
transferred to the water and is measured using
a thermometer placed in a hole in the lid.
The heat energy change is:
q = m C ∆T
The amount of heat lost or gained in the reaction is
equal in size but opposite in sign to the amount
of heat lost or gained by the water.
Coffee cup calorimeter
The simplest calorimeter makes use of two
polystyrene (Styrofoam) cups, one inside the other
with a lid on top. This minimises the amount of
heat lost to the surroundings. This is the most
significant source of error.
Calorimetry example
When a student added excess Mg is to 100cm3 of 2.00 mol dm-3 CuSO4, the temperature rose
from 20.0oC to 65oC.
i) The energy change:
Q = m x c x ΔT Q = 100 x 4.18 x (65 - 20) Q = 100 x 4.18 x 45 Q = 18810 J
Q = 18.81 kJ Convert the energy calculated into KJ (divide by 1000)
ii) Calculate the number of moles used:
No moles = c x V
1000
No moles = 2 x 100
1000
No moles = 0.2
iii) Calculate the amount of energy exchanged per mole , this is the enthalpy:
Enthalpy = Energy
moles
Enthalpy = - 18.81
0.2
Enthalpy = - 94.05 kJ mol-1
iv) Finally write the equation with the enthalpy change:
Mg(s) + CuSO4(aq)  MgSO4(aq) + Cu(s) ΔH = - 94.05 kjmol-1 It is exothermic, therefore negative
Heat of solution
In order for a solute to dissolve in a
solvent, the solute particles and the
solvent particles must separate, which
involves energy changes.
In addition, solute particles interact with
solvent particles, which also can result
in energy changes.
Overall:
ΔHsoln = ΔH1 + ΔH2 + ΔH3
The energy profile diagrams on the right
show net exothermic and endothermic
dissolutions
Question: How are these processes
similar to a chemical reaction? How are
they different?
Diagrams Source:
http://wps.prenhall.com/wps/media/objects/3312/3391718/blb1301.html
Heat of Combustion (ΔHc)
• Heat of Combustion of a substance is the heat
liberated when 1 mole of the substance undergoes
complete combustion with oxygen at constant
pressure.
• Combustion is always exothermic, ΔH is negative.
• By definition, the heat of combustion is minus the
enthalpy change for the combustion reaction.
• By definition, the heat of combustion is a positive
value.
• Heat of Combustion can be measured experimentally.
Heat of Combustion (ΔHc)
The diagram to the right shows a typical school
laboratory setup to test the heat of combustion of
a fuel such as ethanol.
1.
2.
3.
4.
5.
6.
7.
A known quantity of water is placed in a flask or
beaker (calorimeter)
A thermometer is positioned with bulb near the
middle of the volume of water
A known quantity of fuel, such as an alcohol
(alkanol), is placed in the spirit burner
The initial temperature of the water is measured
and recorded (Ti)
The wick on the spirit burner is lit, burning the
fuel, and heating the water
When the temperature has risen an appreciable
amount, the spirit burner is extinguished and the
final temperature recorded (Tf)
The final quantity of fuel is measured and
recorded
Source: http://www.ausetute.com.au/heatcomb.html
Heat of Combustion Example
A student used the apparatus on the previous slide to determine the
heat of combustion of ethanol. Their results for one trial are below:
initial water temperature (Ti) = 20oC
intial mass burner + ethanol = 37.25 g
final water temperature (Tf)= 75oC
final mass burner + ethanol = 35.50 g
change in temperature = Tf - Ti = 55oC
mass ethanol used = 1.75g
Use the results above to determine the heat of combustion for
ethanol in kJ/mol. (Solution on the next slide)
Heat of Combustion Example
1. Calculate moles (n) of fuel used
molecular mass (MM) of ethanol = 46.1g/mol
mass ethanol used = 1.75g
n = mass ÷ MM = 1.75 ÷ 46.1 = 0.0380 mol
2. Calculate energy required to change temperature of water
energy = mass of water x specific heat capacity of water x change in water
temp (mCΔT)
energy = 200g x 4.184 JK-1g-1 x 55oC = 46024 J = 46.024 kJ
3. Calculate the heat of combustion of ethanol
Assume all the heat produced from burning ethanol has gone into heating the
water, ie, no heat has been wasted.
0.0380 mole ethanol produced 46.024 kJ of heat.
Therefore 1 mole of ethanol would produce 46.024 kJ ÷ 0.0380 mol = 1211
kJ/mol
The heat of combustion of ethanol is 1211 kJ/mol
The experimentally determined value for the heat of combustion of ethanol is usually less
than the accepted value of 1368 kJ/mol because some heat is always lost to the
atmosphere and in heating the vessel.
5.3 Hess’s Law
Hess’s Law
According to the Law of Conservation of Energy, the energy exchanges that take place in
reactions must be conserved.
Hess’s Law states that the enthalpy of a chemical process is the same, whether the process
takes place in one step or several steps. H is the same regardless of how the reactants
change into products
The diagram above shows an enthalpy cycle and reflects Hess’s Law:
ΔHx = ΔH1 + ΔH2
So, if we cannot measure the enthalpy for the overall reaction (x), but can measure the reactions (1) and (2), we
can determine the enthalpy of reaction mathematically.
Hess’s Law Analogy
Little Red Riding Hood wants to visit her
grandmother, who lives at the top of the
mountain.
Her grandmother’s house is 2000 meters
above Red’s house.
Regardless of how Red climbs the
mountain, her change in altitude will be the
same: 2000 meters.
Grandma’s House
Red’s House
The pathway
doesn’t matter;
the altitude
change is the
same.
Hess’s Law Problems
Helpful Hints
1. If you reverse a reaction, the sign of H
changes.
– 2H2O2(l)  2H2O(l) + O2(g)
• H = -196.4 kJ
– 2H2O(l) + O2(g)  2H2O2(l)
• H = +196.4 kJ
Hess’s Law Problems
Helpful Hints
2. If you multiply the coefficients of a
thermochemical equation by some number,
you must multiply H by the same number.
– 2H2O2(l)  2H2O(l) + O2(g)
• H = -196.4 kJ
– 4H2O2(l)  4H2O(l) + 2O2(g)
• H = -392.8 kJ
– H2O2(l)  H2O(l) + ½O2(g)
• H = -98.2 kJ
Hess’s Law Problems
Helpful Hints
3 Thermochemical equations can be combined
to give new thermochemical equations.
– N2 + 2O2  2NO2
• H = 67.8 kJ
– N2O4  N2 + 2O2
• H = -9.6 kJ
– N2 + 2O2 + N2O4  2NO2 + N2 + 2O2
– N2O4  2NO2
• H = (67.8 kJ) + (-9.6 kJ) = 58.2 kJ
Hess’s Law Problems summary
Below are the steps to follow when performing Hess’s Law problems:
1. Rearrange the balanced chemical equations so that the reactants and
products are on the correct sides. If you reverse a chemical equation, then
the sign of the enthalpy also changes.
2. Check that you have the correct states, enthalpy changes will be different
for species in the solid, liquid and gas states.
3. Assign each rearranged equation the correct ΔH value.
Remember, if you need to multiply each species in the chemical equation
by 2, then the enthalpy change must also be multiplied by 2.
4. Add the rearranged equations together to give the overall equation for the
reaction. Add the ΔH values for these equations to calculate ΔH for the
overall reaction.
Example 1
When carbon combusts in an excess of oxygen, carbon dioxide is formed and
393.5 kJ of heat is released per mole of carbon.
C(s) + O2(g) -----> CO2(g)
ΔH = -393.5 kJ
This overall reaction can also be produced as a two stage process:
Carbon combusts in limited oxygen producing carbon monoxide:
C(s) + ½O2(g)  CO(g) ΔH = -110.5 kJ
Carbon monoxide then combusts in additional oxygen:
CO(g) + ½O2(g)  CO2(g) ΔH = -283.0 kJ
These two equations can be added together to calculate ΔH for the overall
reaction:
C(s) + ½O2(g) ---> CO(g) ΔH = -110.5 kJ
CO(g) + ½O2(g) ---> CO2(g) ΔH = -283.0 kJ
C(s) + O2(g) ---> CO2(g) ΔH = -393.5 kJ
Example 1 - Graph
This shows a graphical
representation of the
previous example.
Using one step or two,
the overall enthalpy
change is the same.
This follows Hess’s
Law
Source: http://www.ausetute.com.au/hesslaw.html
Example 2 – you try
Calculate ΔH for the reaction:
NH3(g) + HCl(g)  NH4Cl(s)
Given that:
½N2(g) + 1½H2(g)  NH3(g) ΔH = -46.1 kJ
½H2(g) + ½Cl2(g)  HCl(g) ΔH = -92.3 kJ
½N2(g) + 2H2(g) + ½Cl2(g)  NH4Cl(s) ΔH = -314.4 kJ
Example 2 - Solution
1. Rearrange each balanced equation with reactants and products on the correct side, ie,
NH3(g) and HCl(g) on the left, NH4Cl(s) on the right hand side.
NH3(g) ½N2(g) + 1½H2(g)
HCl(g)  ½H2(g) + ½Cl2(g)
½N2(g) + 2H2(g) + ½Cl2(g)  NH4Cl(s)
2. Assign correct ΔH values to each equation.
NH3(g)  ½N2(g) + 1½H2(g) ΔH = +46.1 kJ
HCl(g)  ½H2(g) + ½Cl2(g) Δ H = +92.3 kJ
½N2(g) + 2H2(g) + ½Cl2(g)  NH4Cl(s) Δ H = -314.4 kJ
3. Add the rearranged equations together.
NH3(g)  ½N2(g) + 1½H2(g) ΔH = +46.1 kJ
HCl(g)  ½H2(g) + ½Cl2(g) Δ H = +92.3 kJ
½N2(g) + 2H2(g) + ½Cl2(g)  NH4Cl(s) Δ H = -314.4 kJ
NH3(g) + HCl(g)  NH4Cl(s)
4. Add the ΔH values.
NH3(g)  ½N2(g) + 1½H2(g) Δ H = +46.1 kJ
HCl(g)  ½H2(g) + ½Cl2(g) Δ H = +92.3 kJ
½N2(g) + 2H2(g) + ½Cl2(g)  NH4Cl(s) Δ H = -314.4 kJ
NH3(g) + HCl(g)  NH4Cl(s) Δ H = -176.0 kJ
Exercise
Calculate the enthalpy change (Δ H°), in kJ, of the
reaction:
N2H4 + 2 H2O2  N2 + 4 H2O
using the following enthalpy of combustion data:
a. N2H4 + O2  N2 + 2 H2O
ΔH° = -622 kJ
b. H2 + O2  H2O2
Δ H° = -188 kJ
c. H2 + ½ O2  H2O
Δ H° = -286 kJ
To solve this, leave equation (a) alone. Reverse equation (b) and double the
coefficients. Then, double
the coefficients for equation (c). When that is complete, equations a, b, and c
add up to the desired
equation and the overall DH° value equals -818 kJ.
Finding Enthalpy changes using
Hess’s Law cycles
Standard enthalpy change of
formation ΔH0f is the energy
change when 1 mole of a
compound forms from its
elements at under standard
conditions and constant
pressure.
If we want to determine ΔH0f the for a
substance such as benzene C6H6, we can
use the standard enthalpy changes of
combustion and Hess’s Law cycles to help
us. You can find the data below in your
data booklet.
Standard enthalpy changes of combustion
ΔH°c (kJ mol-1)
C6H6(l)
-3267
C(s)
-394
H2(g)
-286
Drawing the cycle
1. The formation of benzene is drawn
horizontally
2. The Hess’s Law cycle
is completed showing
the combustion
products. (O2 has been
left off for convenience
only - it would be added
in excess anyway)
3. There are two possible routes
to the same products. These
routes are shown in blue.
4. Construct and solve the equation:
ΔH - 3267 = 6(-394) + 3(-286)
ΔH = 3267 + 6(-394) + 3(-286)
ΔH = +45 kJ mol-1
Hess’s Law cycle – example 2
Calculate the enthalpy change of the reaction between ethene and hydrogen
chloride gases to form chloroethane gas using a Hess’s Law cycle.
Standard enthalpy changes of formation
ΔH°f (kJ mol-1)
C2H4(g)
+52
HCl(g)
-92
C2H5Cl(g)
-137
-137
These standard enthalpy changes of formation can
be found in your data booklet (except HCl)
ΔH for this reaction can easily be found by the same method as in the previous question
once the cycle has been drawn. Find the two pathways that lead to the same product and
equate them. Then solve for the unknown.
+52 - 92 + ΔH = -137
ΔH = -52 + 92 - 137
ΔH = -97 kJ mol-1
5.4 Bond Enthalpies
Bond enthalpy
Also known as bond energies, this is the enthalpy change required to break a
covalent bond when all species are in the gaseous state. These are positive
values as breaking bonds is an exothermic process that requires energy.
We can use bond enthalpy
values to estimate the
overall enthalpy changes in
reactions.
We simply take the sum of the individual bond energies for reactants and
subtract the sum of the bond enthalpies for the products (negative because
forming bonds releases energy – exothermic)
Consider this example:
CO(g) + H2O(g)  CO2(g) + H2(g)
AVERAGE BOND ENERGIES OF COMMON BONDS
The amount of energy required to break these bonds
Bond
Bond energy, kJ/mol
C–C
347
C=C
612
C≡C
838
C–O
358
C=O
746
C≡O
1077
F–F
158
Cl–Cl
243
C–H
413
H–H
436
H–O
464
O=O
498
Use these bond energy values to determine the ∆ Hrxn of methane combustion
∆ Hrxn = ∑ E (bonds broken) - ∑ E (bonds formed)
Example - Solution
CO(g) + H2O(g)  CO2(g) + H2(g)
bond enthalpy
(kJ mol-1)
C-O in carbon
monoxide
+1077
C-O in carbon
dioxide
+746
O-H
+464
H-H
+436
ΔH = 1077 + 2(464) – 2(746) – 436
ΔH = + 77 kJ mol -1
5 Energetics
Compiled by: Robert Slider (2011)
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