Heat
Thermochemistry
•  energy transferred between two objects as a
result of the temperature difference between
them.
Temperature
1st Law of Thermodynamics
•  A measure of kinetic energy
•  The energy of the universe is constant.
•  i.e. the energy of the universe is conserved
ΔE = Efinal - Einitial
•  - ΔE if energy leaves system
•  + ΔE if energy enters system
•  Note the E of a system doesn’t depend on
how system got there -- i.e. it is a state
function
1
State Function
•  A function or property whose value depends
only on the present state (condition) of the
system, not on the path used to arrive at that
condition.
• E = q + w
Heat gain or loss
Work done = -PΔV
Matches our earlier
convention that Ein is + and
Eout is –
Enthalpy
Quantity of heat supplied
•  ΔH = qP = ΔE + PΔV
•  ΔH = Hfinal - Hinitial
•
= Hproducts - Hreactants
Tells how much heat is
required to change the
temp of a substance.
Temperature change
(always Tf-Ti)
Some specific heats are
Al
0.902 J/g oK
Cu
0.385 J/g oK
H2O 4.184 J/g oK
•  A 55.0 g piece of metal was heated in
boiling water to a temperature of 99.8oC and
dropped into an insulated beaker with 225
mL of water (d = 1.00 g/ml) at 21.0 oC. The
final temperature of the metal and water is
23.1oC. Calculate the specific heat of the
metal assuming that no heat was lost to the
surroundings.
2
•  Octane, C8H18, a primary constituent of
gasoline, burns in air.
•  C8H18(l) + 25/2 O2(g) ⎯→ 8 CO2(g) + 9 H2O(l)
•  Suppose that a 1.00 g sample of octane is
burned in a calorimeter that contains 1.20
kg of water. The temperature of the water
and the bomb rises from 25.00oC to
33.20oC. If the specific heat of the bomb,
Cbomb, is known to be 837 J/oC, calculate the
molar heat of reaction of C8H18.
A quantity of ice at 0oC is added to 90.0 g of
water at 80oC. After the ice melted, the
temperature of the water was 25oC. How
much ice was added?
•
•
•
•
•
specific heat of ice
specific heat of water
specific heat of steam
heat of fusion
heat of vaporization
2.06 J/goC
4.184 J/goC
2.0 J/goC
333 J/g
2226 J/g
0.91 kJ/moloC
7.54 kJ/moloC
0.92 kJ/moloC
6.01 kJ/mol
40.67 kJ/mol
•  A 33.14 g sample of copper and aluminum
was heated to 119.25oC and dropped into a
calorimeter containing 250.0 g of water at
21.00oC. The temperature rose to 23.05oC.
Assuming no heat was lost to the
surroundings, what is the percent copper in
the sample?
•  50.0 g of ice at -20.0 oC are added to 342.0
g of water at 86.0 oC. What will be the final
temperature of the sample?
•
•
•
•
•
specific heat of ice
specific heat of water
specific heat of steam
heat of fusion
heat of vaporization
2.06 J/goC
4.184 J/goC
2.0 J/goC
333 J/g
2226 J/g
0.91 kJ/moloC
7.54 kJ/moloC
0.92 kJ/moloC
6.01 kJ/mol
40.67 kJ/mol
Enthalpy
•  Enthalpy transferred out of reactants →
exothermic → ΔH = •  Enthalpy transferred into products →
endothermic → ΔH = +
3
Enthalpy
•  ΔHforward = -ΔHreverse
(For reversible reactions)
•  H2O(g) ⎯→ H2(g) + 1/2 O2(g)
ΔH = +241.8 kJ
•  H2(g) + 1/2 O2(g) ⎯→ H2O(g)
ΔH = -241.8 kJ
Enthalpy
•  The ΔH is proportional to the amount of
substance undergoing change.
•  H2O(g) ⎯→ H2(g) + 1/2 O2(g)
ΔH = +241.8 kJ
•  2 H2O(g) ⎯→ 2 H2(g) + 1 O2(g)
ΔH = +483.6 kJ
Enthalpy
•  The physical state of reactants and products
is important.
•  H2O(g) ⎯→ H2(g) + 1/2 O2(g)
ΔH = +241.8 kJ
•  H2O(l) ⎯→ H2(g) + 1/2 O2(g)
ΔH = +285.8 kJ
Hess’s Law
•  The overall enthalpy change for a reaction
is equal to the sum of the enthalpy changes
for the individual steps in the reaction.
•  Valid because enthalpy is a state function.
Enthalpy
•  Enthalpy is a state function -- it doesn’t
matter how you go from one place to
another -- enthalpy and enthalpy changes
are the same!!
•  The ΔH value is the same no matter how
you get from A→B
Determine the ΔH for the sublimation of ice to
water vapor at 0oC.
•
•
•
•
H2O(s) ⎯→ H2O(l)
ΔH = 6.02 kJ/reaction
H2O(l) ⎯→ H2O(g)
ΔH = 40.7 kJ/reaction
----------------------------------------------------H2O(s) ⎯→ H2O(g)
ΔH = 46.7 kJ/reaction
4
•  Calculate the enthalpy change for the
formation of methane, CH4, from solid
carbon (as graphite) and hydrogen gas.
•
C(s) + 2 H2(g) ⎯→ CH4(g)
•  The enthalpies for the combustion of
graphite, hydrogen gas and methane are
given.
•  C(s) + O2(g) ⎯→ CO2(g)
-393.5 kJ
•  H2(g) + ½ O2(g) ⎯→ H2O(l)
-285.8 kJ
•  CH4(g) + 2 O2(g) ⎯→ CO2(g) + 2 H2O(l)
-890.3
kJ
•  Calculate the enthalpy change for the
reaction
•
S(s) + O2(g) ⎯→ SO2(g)
•  given
•  2 SO2(g) + O2(g) ⎯→ 2 SO3(g) ΔH = -196 kJ
•  2 S(s) + 3 O2(g) ⎯→ 2 SO3(g) ΔH = -790 kJ
Standard Heat of Formation
•  The enthalpy change, ΔHfo, for the
formation of 1 mol of a substance in the
standard state from the most stable forms of
its constituent elements in their standard
states.
5
•  Benzene, C6H6, is an important
hydrocarbon. Calculate its enthalpy of
combustion; that is, find the value of ΔHo
for the following reaction.
•  C6H6(l)+15/2 O2(g) ⎯→ 6 CO2(g)+3 H2O(l)
•  Given
•
ΔHfo [C6H6(l)] = +49.0 kJ/mol
•
ΔHfo [CO2(g)] = -393.5 kJ/mol
•
ΔHfo [H2O(l)] = -285.8 kJ/mol
•  Nitroglycerin is a powerful explosive,
giving four different gases when detonated.
•
2 C3H5(NO3)3(l) ⎯→ 3 N2(g) + ½
O2(g) + 6 CO2(g) + 5 H2O(g)
•  Given the enthalpy of formation of
nitroglycerin, ΔHfo, is -364 kJ/mol, calculate
the energy liberated when 10.0 g of
nitroglycerin is detonated.
Enthalpies from Bond Energies
•  Calculate the enthalpy of formation of water
vapor from bond energies.
•
•  2 H2(g) + O2(g) ⎯→ 2 H2O(g)
•
(The experimental value is -241.8kJ/mol)
•  Oxygen difluoride, OF2, is a colorless, very
poisonous gas that reacts rapidly and
exothermically with water vapor to produce
O2 and HF. Calculate the ΔHof for OF2.
•  OF2(g) + H2O(g)  2 HF(g) + O2(g)
ΔHorxn = -318 kJ
•  The heats of formation for H2O(g) and
HF(g) are -241.8 kJ/mol and -271.1 kJ/mol
respectively.
Stoichiometry using Enthalpy
•  Consider the following reaction:
•  2 Na(s) + Cl2(g) ⎯→2 NaCl(s) ΔH = -821.8 kJ
•  Is the reaction exothermic or endothermic?
•  Calculate the amount of heat transferred
when 8.0 g of Na(s) reacts according to this
reaction.
6
Determination of ΔH using
Hess’s Law
•  We generally expect that reactions evolving heat
should proceed spontaneously and those that
absorb heat should require energy to occur.
–  Mix barium hydroxide and ammonium chloride
Ba(OH)2•8H2O(s) + 2 NH4Cl(s)
→ BaCl2(aq) + 2 NH3(g) + 10 H2O(l)
 ΔHrxn is well known for many reactions, but it is
inconvenient to measure ΔHrxn for every reaction.
 However, we can estimate ΔHrxn for a reaction of interest
by using ΔHrxn values that are published for other more
common reactions.
° a series of
  The Standard Enthalpy of Reaction (ΔHrxn) of
reaction steps are added to lead to reaction of interest
(indirect method).
° Standard conditions (25°C and 1.00 atm pressure).
(STP for gases T= 0°C)
Hess’s Law
Calculation of ΔH by Hess’s Law
3 C(graphite) + 4 H2 (g) ⎯→ C3H8 (g) ΔH= -104
C3H8 (g) ⎯→ 3 C(graphite) + 4 H2 (g) ΔH= +104
“If a reaction is carried
out in a series of steps,
ΔH for the overall
reaction will be equal to
the sum of the enthalpy
changes for the
individual steps.”
- 1840, Germain Henri Hess
(1802–50), Swiss
ΔH= +104
3 C(graphite) + 3 O2 (g) ⎯→ 3 CO2 (g) ΔH=-1181
4 H2 (g) + 2 O2 (g) ⎯→ 4 H2O (l)
ΔH=-1143
C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)
C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)
•
Appropriate set of Equations with their ΔH values are obtained (or given), which containing
chemicals in common with equation whose ΔH is desired.
•
These Equations are all added to give you the desired equation.
•
These Equations may be reversed to give you the desired results (changing the sign of ΔH).
•
You may have to multiply the equations by a factor that makes them balanced in relation to
each other.
•
Elimination of common terms that appear on both sides of the equation .
Calculation of ΔH by Hess’s Law
Calculation of ΔH by Hess’s Law
C3H8 (g) ⎯→ 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) ⎯→ 3 CO2 (g) ΔH=-1181
4 H2 (g) + 2 O2 (g) ⎯→ 4 H2O (l)
ΔH=-1143
ΔHrxn =
+ 104 kJ
- 1181 kJ
-  1143 kJ
-  2220 kJ
Calculate heat of reaction
W + C (graphite)  WC (s)
Given data:
2 W(s) + 3 O2 (g)  2 WO3 (s)
½(2 W(s) + 3 O2 (g)  2 WO3 (s) )
W(s) + 3/2 O2 (g)  WO3 (s) )
C (graphite) + O2 (g)  CO2 (g)
C (graphite) + O2 (g)  CO2 (g)
2 WC (s) + 5 O2 (g)  2 WO3 (s) + CO2 (g)
½(2 WO3 (s) + CO2 (g)  2 WC (s) + 5 O2 (g))
WO3 (s) + CO2 (g)  WC (s) + 5/2 O2 (g)
W + C (graphite)  WC (s)
ΔH = ?
ΔH = -1680.6 kJ
½(ΔH = -1680.6 kJ)
ΔH = -840.3 kJ
ΔH = -393.5 kJ
ΔH = -393.5 kJ
ΔH = -2391.6 kJ
½ (ΔH = +2391.6 kJ)
ΔH = + 1195.8 kJ)
ΔH = - 38.0
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Hess’s Law
Methods of determining ΔH
Problem: Chloroform, CHCl3, is formed by the following reaction:
Desired ΔHrxn equation: CH4 (g) + 3 Cl2 (g) → 3 HCl (g) + CHCl3 (g)
Determine the enthalpy change for this reaction (ΔH°rxn), using the
following:
2 C (graphite) + H2 (g) + 3Cl2 (g) → 2CHCl3 (g) ΔH°f = – 103.1 kJ/mol
CH4 (g) + 2 O2 (g) → 2 H2O (l) + CO2 (g)
ΔH°rxn = – 890.4 kJ/mol
2 HCl (g) → H2 (g) + Cl2 (g)
ΔH°rxn = + 184.6 kJ/mol
C (graphite) + O2 (g) → CO2(g)
ΔH°rxn = – 393.5 kJ/mol
H2 (g) + ½ O2 (g) → H2O (l)
ΔH°rxn = – 285.8 kJ/mol
answers: a) –103.1 kJ b) + 145.4 kJ
c) – 145.4 kJ
305.2 kJ
e) – 305.2 kJ f) +103.1 kJ
d) +
1.  Calorimetry (experimental)
2.  Hess’s Law: using Standard
Enthalpy of Reaction (ΔHrxn)° of a
series of reaction steps (indirect
method).
3.  Standard Enthalpy of Formation
°
(ΔHf ) used with Hess’s
Law (direct
method)
Experimental
data combined
with
theoretical
concepts
4.  Bond Energies used with Hess’s
Law
This is a hard question.
(3) Determination of ΔH using
Standard Enthalpies of Formation (ΔHf )°
Enthalpy of formation, ΔHf, is defined as the enthalpy change for the reaction in
which a compound is made from its constituent elements in their elemental forms.
C + O2  CO2
∆Hf°= -393.5 kJ/ η
° measured under standard conditions
Standard Enthalpy of formation ΔHf are
(25°C and 1.00 atm pressure).
Calculation of ΔH
CH4(g) + O2(g)  CO2(g) + H2O(g)
C + 2H2(g)  CH4(g)
C(g) + O2(g)  CO2(g)
2H2(g) + O2(g)  2H2O(g)
ΔHf = -74.8 kJ/ŋ
ΔHf = -393.5 kJ/ŋ
ΔHf = -241.8 kJ/ŋ
We can use Hess’s law in this way:
°
°
ΔH = Σ n ΔHf(products)
- Σ m ΔHf(reactants)
where n and m are the stoichiometric coefficients.
n CO2(g) + n H2O(g)
-
n CH4(g) + n O2(g)
ΔH = [1(-393.5 kJ) + 1(-241.8 kJ)] - [1(-74.8 kJ) + 1(-0 kJ)]
= - 560.5 kJ
Calculation of ΔH
C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)
ΔH = Σ n ΔHf(products) - Σ m ΔHf(reactants)
= [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]
= [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]
= (-2323.7 kJ) - (-103.85 kJ)
= -2219.9 kJ
Table of Standard Enthalpy of formation, ΔHf
(4) Determination of ΔH
using Bond Energies
ΔH
•  Most simply, the strength of a bond is measured
by determining how much energy is required to
break the bond.
•  This is the bond enthalpy.
•  The bond enthalpy for a Cl—Cl bond,
D(Cl—Cl), is measured to be 242 kJ/mol.
8
Average Bond Enthalpies (H)
•  Average bond enthalpies are positive, because bond
breaking is an endothermic process.
NOTE: These are
average bond
enthalpies, not
absolute bond
enthalpies; the C
—H bonds in
methane, CH4,
will be a bit
different than the
C—H bond in
chloroform,
CHCl3.
Hess’s Law:
ΔHrxn = Σ(bonds broken) - Σ(bonds formed)
Enthalpies of Reaction (ΔH )
•  Yet another way to
estimate ΔH for a reaction
is to compare the bond
enthalpies of bonds
broken to the bond
enthalpies of the new
bonds formed.
•  In other words,
ΔHrxn = Σ(bond enthalpies of bonds broken) Σ(bond enthalpies of bonds formed)
Bond Enthalpy and Bond Length
CH4(g) + Cl2(g) ⎯⎯→ CH3Cl(g) + HCl(g)
ΔHrxn = [D(C—H) + D(Cl—Cl) - [D(C—Cl) + D(H—Cl)
= [(413 kJ) + (242 kJ)] - [(328 kJ) + (431 kJ)]
= (655 kJ) - (759 kJ)
= -104 kJ
•  We can also measure an average bond length for
different bond types.
•  As the number of bonds between two atoms
increases, the bond length decreases.
2003 B Q3 9
2005 B 2002 Entropy
•  The amount of randomness, or molecular
disorder, in a system.
•  S = more positive to indicate greater
disorder.
10
Predict which has greater entropy
•  O2(g) at 5 atm of O2 at 0.5 atm
•  Br2(l) or Br2(g)
•  1 mol N2 (g) in 22.4 L or 1 mol N2(g) in
2.24 L
•  CO2(g) or CO2(aq)
•  + ΔS is entropy favored
•  - ΔS is entropy disfavored
•  - ΔH is enthalpy favored
•  + ΔH is enthalpy disfavored
Energy Units
•  1 calorie = 4.184 J
•  1 food calorie = 1 Cal = 1 kcal = 1000 cal
•  Given the reaction below for the combustion of
glucose to form carbon dioxide and water,
calculate the Calories/g for carbohydrates.
•  C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
ΔHrxn = -2801.6 kJ
Predict entropy changes for
•  freezing of one mole of water
•  evaporation of 1 mol of Br2
•  precipitation of BaSO4 upon mixing of
aqueous solutions of Ba(NO3)2 and H2SO4
•  2 C(s) + O2(g) ⎯→ 2 CO(g)
•  2 K(s) + Br2(l) ⎯→ 2 KBr(s)
•  2 MnO2(s) ⎯→ 2 MnO(s) + O2(g)
•  O(g) + O2(g) ⎯→ O3(g)
Gibbs Free Energy, ΔG
•  Determines whether a reaction is
spontaneous and at what temperature it
becomes spontaneous.
•  Spontaneous -- A process that proceeds on
its own with out any continuous external
influence.
•  M & M candies consist of 70%
carbohydrates, 21% fat, and 4.6% protein as
well as other ingredients that do not have
caloric value. What quantity of energy is
generated if 47.9 g of M&Ms (1 small
package) were burned in a bomb
calorimeter? How long will a I need to
walk to use up the value of the M&Ms if 1
hour of walking uses up 400 Cal?
•  4 Cal/g carbs
•  4 Cal/g protein
•  9 Cal/g fat
11
ΔG = ΔH - TΔS
•  If ΔH = + and ΔS = - never spontaneous
ΔG = +
•  If ΔH = - and ΔS = + always spontaneous
ΔG = •  If ΔH = + and ΔS = + or if ΔH = - and ΔS = temperature determines spontaneity
•  At T where ΔG = - reaction is spontaneous
•  At T where ΔG = + reaction is nonspontaneous
•  NH3(g) + 2 O2(g) → HNO3(aq) + H2O(l)
ΔH = -413 kJ ΔS = -386J/K
•  0 = -413kJ - T(-0.386J/K)
•  T = 1069K = 796oC
•  Reaction is spontaneous below 796oC
•  Ca(s) + Cl2(g) → CaCl2(s)
ΔH = -59.8 kJ
ΔS = -273J/K
•  ΔG = ΔH - TΔS spontaneous at low T nonspontaneous at high T, entropy takes precedence
•  Reaction becomes spontaneous at temperature
where ΔG becomes zero -- or when ΔG = zero
reaction is spontaneous in neither direction -equilibrium!
•  0 = -59.8 kJ - T(-0.273kJ/K)
•  T = 219K or -53oC
•  Reaction is spontaneous below -53oC
•  C6H12O6(s) → 2 C2H5OH(l) + 2 CO2(g)
ΔH = -70 kJ ΔS = +780J/K
•  0 = -70kJ - T(+0.78J/K) →→ T = -90K
•  spontaneous at all temperatures -- would
need an impossible temperature to become
non-spontaneous!!
•  C6H12(l) + 6 O2(g) → 3 CO2(g) + 6 H2O(g)
ΔH = ΔS = +
•  6 CO2(g)+6 H2O(g) →C6H12O6(s)+9 O2(g)
ΔH = +
ΔS = -
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