Transcript Document

Calorimetry and Heats of Reaction
• Very often, the most accurate measurements
of heats of reaction are obtained using bomb
calorimeter experiments. These experiments
give us a qV or ∆U value (the change in internal
energy of a system).
• For many applications its more useful to have
a qP or ∆H value (the change in enthalpy of a
system). (Many rxns take place at constant P.)
• We must be able to “convert” a ∆U value to a
corresponding ∆H value.
7-6 Heats of Reaction: U and H
Reactants → Products
Ui
Uf
U = Uf - Ui
U = qrxn + w
In a system at constant volume (bomb calorimeter):
U = qrxn + 0 = qrxn = qv
But we live in a constant pressure world!
How does qp relate to qv?
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General Chemistry: Chapter 7
Slide 2 of 57
FIGURE 7-13
Two different paths leading to the same internal energy change in a
system
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General Chemistry: Chapter 7
Slide 3 of 57
Heats of Reaction
qV = qP + w
We know that w = - PV and U = qv, therefore:
U = qP - PV
qP = U + PV
These are all state functions, so define a new function.
Let enthalpy be
H = U + PV
Then
H = Hf – Hi = U + (PV)
If we work at constant pressure and temperature:
H = U + PV = qP
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General Chemistry: Chapter 7
Slide 4 of 57
Comparing ∆H and ∆U
• When a physical or chemical change involves
gases, the ideal gas equation, PV = nRT, is
often used to calculate the difference
between ∆H and ∆U. For a chemical change
our initial and final states will be comprised of
different substances. For a physical change
(such as a phase change) this will not be the
case.
∆H and ∆U Difference
• ∆(PV) for constant P and T:
• ∆(PV) = P(Vfinal - Vinitial) = P∆V
• We could also use ∆nGasRT. This assumes,
reasonably in most cases, that the volume
change for a system is likely to result mostly
from a change in the amount (# moles) of gas.
• P∆V = P(Vfinal - Vinitial) = (nGas,Final – nGas,Initial)RT
= ∆nGasRT
• Mention T changes?
∆H and ∆U Difference – Chemical
Change
• The next slide shows how qV and qP would
vary for a simple chemical change.
• 2 CO(g) + O2(g) → 2 CO2(g) at 298 K
• Here ∆nGasT = 2mol – (2mol+1mol) = -1mol
• Clearly, there is no PV work at constant
volume but, there is PV work at constant P.
Everything else is much less clear – and
requires some careful consideration!
Comparing Heats of Reaction
qV = U = H - PV
= -563.5 kJ/mol
w = PV = P(Vf – Vi)
= RT(nf – ni)
= -2.5 kJ
qP = H
= -566 kJ/mol
FIGURE 7-14
•Comparing heats of reaction at constant volume and constant pressure for the reaction
2 CO(g) + O2(g)
2 CO2(g)
General Chemistry: Chapter 7
Slide 8 of 57
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Tabulated Heats of Reaction
• Heats of reaction tabulations are similar to
prices encountered at the supermarket fish,
meat or produce counters. These are usually
given as $ per kg or $.kg-1. Similarly, heats of
reaction are usually tabulated as kJ/mol or
kJ.mol-1 although units such as kJ/kg or kJ/L
can be more useful. When writing
thermochemical eqtns for the combustion of
hydrocarbons a “1” usually appears before
CXHY. Why?
7-9 Fuels as
Sources of Energy
• Fossil fuels.
– Combustion is exothermic.
– Non-renewable resource.
– Environmental impact.
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General Chemistry: Chapter 7
Slide 10 of 57
Combustion Reactions –
Thermochemical Equations
• Class example: The heats (enthalpies) of
combustion of propane, C3H8(g), and butane,
C4H10(l), are -2220 kJ/mol and -2877 kJ/mol
respectively. Write complete and balanced
thermochemical equations to represent the
two combustion reactions.
Bomb Calorimetry, ∆U and ∆H
• Class example: A text example, reproduced on
the next slide, shows how a bomb calorimeter
can be used to calculate the ∆U value for the
combustion of sucrose (table sugar),
C12H22O11(s). Using this solved example,
calculate a ∆H value for the combustion of
sucrose after first writing a balanced chemical
equation for the combustion reaction.
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7 - 13
Enthalpy and Physical Changes
• Heat flows are associated with chemical
reactions and temperature changes of a solid,
a liquid and a gas. Heat flows are also
associated with phase changes. In
Newfoundland and Labrador we know, for
example, that heat must be supplied to melt
ice in the spring. Rowers (and others) must be
patient!
• Heat + H2O(s) → H2O(l) Endothermic process
Enthalpy Change (H) Accompanying a
Change in State of Matter
Molar enthalpy of vaporization:
H2O (l) → H2O(g)
H = 44.0 kJ at 298 K
Molar enthalpy of fusion:
H2O (s) → H2O(l)
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H = 6.01 kJ at 273.15 K
General Chemistry: Chapter 7
Slide 16 of 57
• George Street Phase Change?
Phase Changes and PV Work (Horton’s
and Hockey?)
• Class example: Calculate the amount of PV
work done when:
• (a) one mole of liquid water at 373 K is
transformed into steam at a pressure of 1.000
atm. Assume that the steam behaves as an
ideal gas.
• (b) one mole of ice at 273 K is melted at 1.000
atm pressure.
• What two pieces of information are missing?
PVPV
Standard States and Standard Enthalpy
Changes
•Define a particular state as a standard state.
•Standard enthalpy of reaction, H°
–The enthalpy change of a reaction in which all
reactants and products are in their standard
states.
•Standard State
–The pure element or compound at a pressure
of
1 bar and at the temperature of interest.
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General Chemistry: Chapter 7
Slide 21 of 57
FIGURE 7-15
Enthalpy Diagrams
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General Chemistry: Chapter 7
Slide 22 of 57
Enthalpy Diagrams
• The energy changes associated with chemical
reactions can be represented using either
balanced thermochemical equations or
enthalpy diagrams. In either case the simplest
examples involve the formation of a single
pure compound from its constituent
elements.
• H2(g) + ½ O2(g) → H2O(l) ∆Hof,298K = -286 kJ
Water Electrolysis
• The reaction on the previous slide is highly
exothermic and does not pollute - an ideal
way to heat our homes if there was a readily
available source of H2(g). The opposite
reaction must be endothermic (Conservation
of Energy). If we supply energy:
• H2O(l) → H2(g) + ½ O2(g) ∆Ho298K = +286 kJ
• The reaction can be used to produce both
elemental hydrogen and oxygen via
electrolysis.
Combustion of Hydrogen/Formation of Water
H2(g) + ½ O2(g)
Reactants
H
(Enthalpy)
(kJ∙mol-1)
ΔH0 = - 286 kJ
Exothermic Reaction !
(Apollo 13/Fuel Cells)
H2O(l)
Product
Electrolysis of Water/Production of Hydrogen
H2(g) + ½ O2(g)
Products
H
(Enthalpy)
(kJ.mol-1)
ΔH0 = + 286 kJ
Endothermic Reaction !
H2O(l)
Reactant