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Ch 9 Energy, Enthalpy, andThermochemistry
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9.1 The nature of energy
Energy : the capacity to do work or to produce heat.
Law of conservation of energy : the energy can be
converted from one form to another but can be
neither created nor destroyed. That is, the energy
of the universe is constant.
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Energy can be classified as either potential energy
or kinetic energy.
Potential energy is due to position or composition.
I.e., water behind a dam, attractive/repulsive forces…
Kinetic energy of an object is due to the motion &depends on the mass of the object (m) & its velocity
(ν) : KE = 1/2m 2 .
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Figure 9.1: Initial positions ball A has a
higher potential energy than ball B.
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From Figure 9.1 :
1. Frictional heating
2. Work (ball A has done a work on B) : defined as
a force acting over a distance. Work is required to
raise B from its original position to a higher one.Thus, there are two ways to transfer energy :
through work & through heat.
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Regardless of the condition of the hill’s surface,
the total energy transferred will be constant. The
amounts of heat & work will differ. Energy
change is independent of the pathway; however,work & heat are both dependent on the pathway.
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A state function (property) refers to a property of
the system that depends only on its present state.
A state function (property) is that a change in this
function (property) in going from one state to
another is independent of the particular pathway taken between the two states.
Energy is a state function, but work & heat are not
state functions.
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Chemical energy :
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + energy (heat)
The universe can be divided into two parts :
the system (reactants/products of the reaction)
& surroundings (everything else).
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When a reaction results in the evolution of heat, it
is said to be exothermic.Reactions that absorb energy from the surroundings
are said to be endothermic.
The energy gained by the surroundings must be
equal to the energy lost by the system.
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In combustion of methane (CH4), the heat flow into
the surroundings results in a lowering of the
potential energy of the reaction system ( this always
holds true).
In any exothermic reaction, the potential energystored in the chemical bonds is being converted to
thermal energy (random kinetic energy) via heat.
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Figure 9.2: Combustion of methane releasesthe quantity of energy
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Figure 9.3: Energy diagram for the
reaction of nitrogen and oxygen
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The study of energy & its interconversion is
called thermodynamics. The law of
conservation of energy is often called the first
law of thermodynamics & is stated as :
the energy of the universe is constant.
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Internal energy (E) is defined as the sum of the
kinetic & potential energies of all the “particles” in
the system.
E = q + w; q : heat, w : work
The meaning of a number (magnitude of the change)& a sign (the direction of the flow & this reflects the
system’s point of view) should be remembered.
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Interconversion between system & surroundings for energy
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Figure 9.4: The piston moving a distance
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Expanded : w = -p V (0)
In dealing with “PV work” keep in mind that the
P in PΔV always refers to the external pressure -
the pressure that causes a compression or that resists an expansion.
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R = 0.08206 Latm/Kmol
= 8.3145 J/Kmol
< 0
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9.2 Enthalpy (焓)
H = E + PV
H : enthalpy, E : internal energy
P, V : properties of the systemH, E, P, & V are all state functions.
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What is exactly enthalpy (H) ?Consider a process carried out at constant pressure :
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Thus, for a process carried out at constant pressure,
H = qP.
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At constant pressure (where only PV work is allow-
ed) the change in enthalpy ( H) of the system is
equal to the energy flow as heat (qP) from the above.
H = Hproducts – Hreactants (for a chemical reaction
the enthalpy change)
H > 0 – endothermic; H < 0 – exothermic.
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9.3 Thermodynamics of ideal gases
Ideal gases : the high temp & low pressure for
real gases (PV = nRT).
(KE)avg = 3/2 RT (Ch. 5 for an ideal gas) represents
average, random, translational energy for 1 moleof gas at a given temp. Energy (“heat”) required
= 3/2 R T.
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The molar heat capacity of a substance is defined
as the energy required to raise the temp. of 1 mole
of that substance by 1 K. Thus, the molar heat
capacity of an ideal gas is 3/2 R.
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Heating an ideal gas at constant volume
CV = 3/2 R = “heat” required to change the tempof 1 mole of gas by 1 K at constant volume ( V=0,
no PV work & all the energy that flows into the gas
is used to increase the translational energies of the
gas molecules).
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Heating an ideal gas at constant pressure
In this case, its volume increases & PV work occurs.
Energy required = “heat”
= energy needed to change the translational energy+ energy needed to do the PV work
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The quantity of work done as the gas expanded by
ΔV is PΔV.
P V = nR T = R T
Thus for a 1K change (ΔT=1K) the work is R, &
heat required to increase the temp of 1 mole ofgas by 1 K (constant P) = 3/2 R (CV) + R = 5/2 R
= CV + R = CP
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Heating a polyatomic gas : as a polyatomic gas is
heated, the gas molecules absorb energy to increase
their rotational & vibrational motions as well as to
move through space (translate) at higher speeds
(previous discussion : the temperature of a mono-atomic ideal gas is an index of the average random
translational energy of the gas).
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Indeed, when a gas is heated, the temperature only
increases to the extent that the translational energies
of the molecules increase. Any energy that isabsorbed to increase the vibrational & rotational
energies does not contribute directly to the
translational kinetic energy. Thus, its heat capacity is
larger than 3/2 R. But, finally the CP – CV = R is also
applicable in all cases, as expected for gases that closely obey the ideal gas law.
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Table 9.1: Molar Heat Capacities of VariousGases at 298 K
R=8.3145 J/KmolCO2 v.s. N2
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Heating a gas : energy and enthalpy
E = 3/2 RT (per mole) (the average translational
energy of an ideal gas - a monoatomic ideal gas)The energy (temp dependence) of an ideal gas :
E = 3/2 R
T (per mole) Note that this expression corresponds to :
E = Cv T (per mole) or E = nC
v T (n moles)
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The CV appears because when a gas is heated at
constant volume, all the input energy (heat) goes toward increasing E (no heat is needed to do
work).
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On the other hand, when a gas is heated at constant
pressure, the volume changes & work occurs :
“heat” required = qP= nCP T = n(CV + R) T
= nCV T + nR T
- E - P V = work required
(E for an ideal gas depends only on T )
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Consider the change in enthalpy when a gas is heated
H = E + PV
H = E + (PV) = E + (nRT)= E + nR T
For a sample of ideal gas containing n moles,substituting E = nCV T
H = nCV
T + nR T = n(CV
+ R) T
= nCP T (CP = CV + R)
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Thus, we can use H = nCP T to calculate the
change in enthalpy when n moles of an ideal gas
is heated, regardless of any conditions on pressureor volume.
E T (E = 3/2 RT ); H T (PV = nRT)Proportional constant Proportional constant
CV CP
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Table 9.2:Thermodynamic properties of an ideal gas
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Figure 9.5: Summary of the two pathways
discussed in example 9.2.
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(constant P)
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H = nCP T = qP
(constant V)
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(constant V)
E = qV = nCV T
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(constant V)
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(constant P)
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E1 = E2 & H1 = H2
Fi 9 6 M i d f h k f h ( )
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Figure 9.6: Magnitude of the work for pathway (a)
and (b) is shown by the colored areas : |w| = |PΔV|
E = q + w; q, w : related to the pathway
E is a state function.
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9.4 Calorimetry
Calorimeter : a device to determine the heat
associated with a chemical reaction experimentally
( temp change & energy absorbed ).
C (heat capacity) = heat absorbed/increase in temp
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Specific heat capacity (JK -1g-1 or J°C-1g-1) : the
energy required to raise the temp of 1 g of a
substance by 1 °C.
Molar heat capacity (JK -1mol-1 or J°C-1mol-1) :
the energy required to raise the temp of 1 mol ofa substance by 1 °C.
Figure 9 7: Coffee cup calorimeter made of two
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Figure 9.7: Coffee cup calorimeter made of two
Styrofoam cups
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For example, suppose we mix 50.0 mL of 1.0 M
HCl with 50.0 mL of 1.0 M NaOH at 25°C in a
calorimeter.H+(aq) + OH
-(aq) → H2O(l)
Energy released by reaction= energy absorbed by the solution
= specific heat capacity x mass of solution x ΔT,
where ΔT = 31.9°C → 25.0° = 6.9 °C
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Mass of solution = 100.0 mLx1.0 g/mL = 1.0 x 102 g
Energy released = (4.18J/°Cg)(1.0 x 102g)(6.9°C)
= 2.9 x 103 J
2.9 x 103 / 5.0 x 10-2 mol (mol of H+) = 5.8 x 104 J
ΔH = -58 KJ/mol
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M
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Calculation of H & E for cases in which PV
work occurs
(1) ΔE = q (P) + W = ΔH + 0
(at constant pressure, where ΔV=0, W = 0 & thus
ΔE = ΔH = q P / solution reaction)
(2) 2SO2(g) + O2(g) → 2SO3(g) ( gas reaction)
ΔH = q P (at constant pressure)
W = -PΔV (ΔV
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Figure 9.8: Schematic to show the change
in volume for the reaction
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H = qP= -198 kJ
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Calorimetry experiments can also be performed
at constant volume
ΔE = q V + W (=0) = q V (constant volume)
To measure the energy of combustion of octane
(C8H18). A 0.5269 g of octane is placed in a bomb
calorimeter (heat capacity = 11.3 kJ/°C)
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Figure 9 9: Schematic of a bomb calorimeter
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Figure 9.9: Schematic of a bomb calorimeter
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9.5 Hess’s law
In going from a particular set of reactants to products, the change in enthalpy is the same
whether the reaction takes place in one step
or in a series of steps (Hess’s law).
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Figure 9.10: Principle of Hess's law
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Characteristics of enthalpy changes
Two characteristics of H for a reaction :
1. If a reaction is reversed, the sign of H is also
reversed.
2. H is directly proportional to the quantities(coefficients) of reactants & products in a reaction.
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(a) - (b) + 3(c) + 3(d)
過
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Explanations :
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Explanations :
1. Large heat capacity of feet because of water.
2. Although the surface of the coals has a very high
temp, the red-hot layer is very thin. Temp reflects
the intensity of the random kinetic energy in a
given sample of matter. The amount of energy available for heat flow, depends on the quantity
of matter at a given temp.
3. Vaporization of the moisture consumes some ofthe energy from the hot coals.
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9.6 Standard enthalpies of formationSummary : definitions of standard states
• For a gas the standard state is a pressure ofexactly 1 atm.
• For a solution the standard state is a conc of exactly 1 M (1 atm).
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• For a liquid & solid ( condensed state), the
standard state is the pure liquid or solid.
• For an element the standard state is the form
in which the element exists (is more stable)
under conditions of 1 atm & 25°C.
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The importance of tabulated ΔH°f values is thatenthalpies for many reactions can be calculated.
Fig. 9.11: Schematic diagram of the energy changes
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CH4(g) + 2O2(g) → CO2(g) + 2H2O(l )
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Summary : key concepts for doing enthalpy calculations
• When a reaction is reversed, the magnitude ofH
remains the same, but the sign changes.
• When the balanced eq for a reaction is multiplied
by an integer, the value of H must be multiplied by the same integer.
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• The change in enthalpy for a given reaction can
be calculated from the enthalpies of formation of
the reactants & products :
• Elements in their standard states are not included in the ΔHreaction calculation. That is, ΔH°f for an
element in its standard state is zero.
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Figure 9.12: Pathway for the combustion
f i
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of ammonia
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Th th l f b ti f t i
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The enthalpy of combustion per gram of octane isabout twice that per gram of methanol. On this
basis, gasoline appears to be superior to methanol
for use in a racing car, where weight considerations
are usually very important. Why, then, is methanol
used in racing cars? The answer is that methanol burns much more smoothly than gasoline in high-
performance engines, and this advantage more
than compensates for its weight disadvantage.
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(煤
)
(噴射機燃料)
(潤滑 )(瀝青)
(柴
)
9 8 New energy resources
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9.8 New energy resources
• Coal conversion
• Hydrogen as a fuel
• Other energy alternatives - ethanol & methanol
Figure 9.15: Coal gasification
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Possible solution:
H2(g) + M(s) → MH2(s)
MH2(s) → M(s) + H2(g)