Zumdahl’s Chapter 6
Thermochemistry
Chapter Contents
• Energy, E
– Chemical Energy
– Matter-Energy
• Chemist’s Enthalpy
– Enthalpy, H
– Calorimetry,CV=dE/dT
• Hess’s Law
– State Functions
• Standard Enthalpies of
Formation, Hf
– Elements
– Compounds
– Ions in Solution
• Energy Sources
– Efficiencies
– Environmental concern
POTENTIAL ENERGY
KINETIC ENERGY
—Fixed Total Energy
Chemical Energy
• Exothermicity
– Release of energy (usually as heat) from chemical
reactions whose products are at lower potential
energy than their reactants.
– If total energy is conserved, and it comes as
kinetic plus potential energy (E=KE+PE), then
the lowered product potential mean raised
product kinetic energies, and heat flows out.
• Chemists use enthalpy when discussing thermicity.
POTENTIAL ENERGY
KINETIC ENERGY
—Fixed Total Energy
Consumption of Energy
• Endothermicity
– If a reaction’s products are higher in potential
energy than its reactants, its kinetic energy must
be lower after the reaction.
– Heat flows in to equalize temperatures.
• Again chemists use enthalpy instead of energy.
• Both kinds of chemical energy presume the
conservation of total energy.
Conservation of Energy
• “Energy can be converted from one form to
another but can be neither created nor
destroyed.”
– Empirical observations verify, e.g.:
• Count von Rumford, boring out cannon, observed
the relationship between the mechanical energy of
the drill and the (frictional) heat of the cannon.
• True even of atomic energy if mass is E.
E = mc2 ?
• Actually E = [ m02c4 + p2c2 ]½ (Einstein)
• Valid for both matter and light.
• Light (photon):
– m0 “rest mass” is strictly zero for light.
– So E = pc (?) but light has momentum, p = h / 
• deBroglie was also correct for both light and matter.
– So E = h c /  = h … for photons
• Just as we always knew it was!
Matter’s Energy Expression
• E = { m02c4 + p2c2 }½
– but p = mv for matter.
• E = { m02c4 + m2v2c2 }½
– but m = m0 / [ 1 – (v / c)2 ]½ (Lorentz)
• which explains why you can’t go even as fast as c.
• E = { m02c4 + m02v2c2 / [ 1 – (v / c)2]½ }½
• E = m0c2 / [ 1 – (v / c)2]½ at all v … even v << c
• Indeed, given Lorentz, E  mc2 for matter at all v!
At Garden Variety Velocities
•
•
•
•
•
•
E = m0c2 / [ 1 – (v / c)2]½ when v<<c or v/c ~ 
[ 1 – 2 ]–½ ~ 1 + ½2 + ignorable terms
So E = m0c2 [ 1 + ½(v2 / c2) ] or
E = m0c2 + ½m0v2 for matter at low velocity.
E = rest mass energy + kinetic energy
What happened to potential energy?
Conservation of Mass, eh?
• Potential energy is an algebraic shorthand
for changes in mass that occur with the
juxtapositions of matter and fields. Use it.
• PE = c2m0 implies that atomic weights
vary in compounds! Are we worried?
– Formation of 1 mole of water yields 286,000 J
– m0 = 2.86105 J / (3108 m/s)2 = 3.210–12 kg
– With 1.810–2 kg / mol, we won’t miss m0 !
+w
E
Surroundings
System
–q
Pedestrian Energy
• Far from Einsteinian Esoterica, we can
observe energy changes in systems due to
two gross causes:
– Heat, q, flowing into the system raises its E.
– Work, w, done on the system raises its E.
– Together, these macroscopic components imply
• E = q + w and when we isolate a system
from surroundings, then E = 0.
Internal Energy’s Components
• HEAT
– Thermal energy flow.
– q = C T
– By Kinetic Theory, T is
proportional to kinetic
energy (& q).
– By Quantum Theory,
heat associates with
changes in energy level
populations.
q
• WORK
– Organized rather than
chaotic molecular
motion.
– Comes in many forms.
– By Quantum Theory,
work associates with
changes in the energy
levels themselves.
w
Work Work Work Work
• Electrical work drags a charge Q through an
electrical potential difference V, so +Q V
• Surface work stretches surface tension 
over larger areas, so work is +  A
• Newtonian work pushes an object a distance
x against a force F, so its work is +F x
• Pressure volume work compresses a volume
with a pressure P, so it is – P V
Pressure Volume Work
• Inescapable when doing chemistry under
the relentless atmospheric pressure if Vgas
changes during a reaction. (gas V  n)
• For constant Pext , w = –PextdV = – Pext V
– But if Pext is always the system’s Pinternal,
• w = –PdV = –nRT  V–1dV = nRT ln(V1/V2)
– which uses the Ideal Gas Law to track Pext
– and assumes constant T and is called “reversible work”
Calorimetry
• We can use E = C T to infer E from
observed T if C, heat capacity, is known.
– Conceptually, C measures the system’s number
of energy modes that can hide thermal energy.
– Since E changes with V too, we must fix V in
order to measure E by heat capacity, so
• E = CV T = qV
– and we should work in fixed V “bombs.”
Enthalpy, H, a chemist’s energy
• Since PV work is inevitable in reactions
open to the omnipresent atmosphere, we’ll
be doing a lot of PV calculations … unless
• Define H  E + PV then H = E + (PV)
• But E = qV bomb = qP – PV
At constant P
• So H = qP – PV + PV + VP = qP
• Now we get out of the bomb and onto the bench!
Practical E vs. H Relation
• Chemical energies, E, are best measured in
bomb calorimeters, but enthalpies, H, are
most conveniently used. So relate them:
• H = E + (PV) = qV bomb + RT(ngas)
– Which makes the (quite defensible) assumption
that all gases are ideal (enough).
• Since their non-idealities can be determined, we can
spruce this up at will.
Temperature Dependence of H
• Just as E = CV T (for constant CV) …
• H = CP T (for constant CP)
• So we can extrapolate H at T other than at
25°C (from standard tables) if we know CP.
– But neither C is truly independent of T, so
– H =  CP(T) dT and it’s so common …
– We find tables of CP ~ a + bT + c / T 2
• Molar C (J/mol K) vs. “Specific Heat” C (J/g K)
Hess’s Law
• “State functions” are thermodynamic
variables (like energy or enthalpy) that have
the same value when you return the system
to the same state (same P,V,T,n).
• Hess: “Enthalpy changes between reactants
and products are not dependent upon how
the reaction is brought about.”
• Otherwise return to reactants wouldn’t undo H!
Hess’s Joyful Consequences
• C2H6 + 3.5 O2  2 CO2 + 3 H2O H1
– Then Hess’s Law guarantees that
• 2 CO2 + 3 H2O  C2H6 + 3.5 O2 has –H1
– Even though the latter’s not a feasible reaction.
– Since H is extensive (scales with # of moles),
• 4 CO2 + 6 H2O  2 C2H6 + 7 O2 is –2H1
– Which permits us to do algebra with reactions.
Chemical Algebra
• Suppose calorimetry gave us the following:
– C2H6 + 3.5 O2  2 CO2 + 3 H2O
– C2H4 + 3 O2  2 CO2 + 2 H2O
– H2 + ½ O2  H2O
H1
H2
H3
• If we reverse the first reaction and add,
– C2H4 + H2  C2H6 results and has a
– Where H = H2 + H3 – H1
H
• We find ethene’s hydrogenation enthalpy without
having to hydrogenate ethene! Just torch it.
Chemical Reference Points
• We just used the fully oxidized forms of the
compounds to do thermochemistry on a
reaction that didn’t even involve oxygen!
– And we did it because calorimetry is easy.
– But Hess lets us use any consistent reference
point, and standard elemental states are best.
• We turn calorimetric results into Standard
Enthalpies of Formation from Elements.
Std. Enthalpy of Formation, Hf
• Elements will be as their most stable
allotrope at 1 bar and 25°C, symbolized ⊖.
– These are standard conditions for thermo.
• Hf of all stable allotropes is ZERO.
– Each element is its own reference point.
• This works because reactions don’t destroy atoms.
• H2 + ½O2  H2O(l)
Hf = –286 kJ/mol
• The Standard Enthalpy of Formation of liquid water.
Compounds in Solution
• While standard enthalpies apply to pure
solids and gases at 1 bar (25°C), solution
concentrations will scale their enthalpies.
• So choose 1 M as a standard concentration
(activity, to be precise) for a solute.
– Ions present a special case (because of counterions):
• So only [H+(aq)] = 1M is defined to have 0 enthalpy
of formation, and all other ions measure against it.
Reaction Enthalpy from Hf
• Hrxn =  nproduct Hf –  nreactant Hf
– where n are stoichiometric coefficients from the
reaction and the sums are over all products and
all reactants. (Not bothering with the 0 elements.)
– Reactants are subtracted because they are
being destroyed, not formed, by the reaction!
• If some T other than 25°C is needed, use CP
obtained just like Hrxn above, a difference of
stoichiometrically-scaled molar heat capacities.
Two Examples
• 2 Ag(s) + S(s)  Ag2S(s)
– Hrxn = Hf(Ag2S) – 2 Hf(Ag) – Hf(S)
– Hrxn = (–32 kJ/mol) – 2(0) – 0 = – 32 kJ/mol
• but
• 2 Ag+(aq) + S2–(aq)  Ag2S(s)
– Hrxn = Hf(Ag2S) – 2 Hf(Ag+) – Hf(S2–)
– Hrxn = –32 – 2(+105) – (+33) = –275 kJ/mol
• Why so different? (Think “lattice energies.”)
Plundering the Legacy
• H2SO4 is our highest volume chemical
commodity unless you consider petroleum.
• While we do use petroleum in syntheses
(feedstock of plastics), mostly we torch it.
– For the Joules.
– Fossil ferns are the cheapest source of mobile
energy; the costs don’t take into consideration
the socioeconomic disaster of Global Warming.
Solar Energy Conversions
• We don’t photosynthesize.
– We can’t yet match the efficiency of things that
do. They reverse carbohydrate combustion.
– But when we can split water into its elements
photolytically, we’ll have solved The Energy
Problem. H2O is as mobile as gasoline.
– And resultant H2 will be our mobile fuel.
– It won’t even upset Earth’s solar E budget!
Energy Efficiencies
• What’s a tank of water worth as gasoline?
– For automobile fuel, we consider weight.
– Each kg of water holds 55.5 moles of H2.
– Our solar cars extract it while sitting in parking lots.
• That’s worth 55.5(286 kJ) ~ 16 MJ
– But a kg of isooctane has 8.77moles of C8H18.
• That’s worth 8.77(5461 kJ) ~ 48 MJ
• A recycling water car needs 3 the fuel wt.
• But only 1/3 the fuel weight when the O2 is ejected!