Lecture 26…
ÆToday: Chemical Reaction Enthalpies
Æ Wednesday: Exam review plus Lect. 22
ÆTest Friday !
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Ksp
Acids / Bases
Nuclear Chemistry (A little)
Thermodynamics
Thermodynamics Refresher … Let’s
“wake up” those resting brain cells
Î For a chemical reaction or a process, like freezing or boiling:
ΔG = ΔH -TΔS.
Î ΔG = The change in Gibbs Free Energy, ΔH = enthalpy (heat)
change, and ΔS = entropy change.
9 If ΔG < O, reaction is spontaneous (One possibility: ΔH < 0 and ΔS is
>0)
9 If ΔG > O, reaction is non-spontaneous (ΔH > 0, ΔS < 0)
9 If ΔG = O, reaction or process is in “equilibrium” (ΔH= -TΔS)
Î If ΔS > O, More system disorder: S solid < S liquid < S gas
Î ΔH = heat flow = qp at constant pressure
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Endothermic: ΔH is positive (+)
Exothermic: ΔH is negative (-)
Î ΔH = qp = n Cp ΔT where Cp = molar heat capacity or = mass (g) x
C x ΔT where C = specific heat in kJ / g ; (n =moles)
Î qp for a phase transition = ΔH
molar heat of fusion or vaporization
xn
Enthalpy Changes During
Chemical Reactions
9 Bonds break ! (energy required)
9 Bonds form !! (energy released)
9 For a specific bond, like C-H, the energy of
bond breaking is the same as the energy
required for formation.
9 ΔHreaction = Energy required + Energy liberated (-)
9 Bond Energy = Energy required to break a mole of
bonds
• Bond Breaking: ΔH is positive (+)
• Bond Formation: ΔH is negative (-)
Changed from
Online Version
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Enthalpy of Reaction Example:
A Combustion Reaction
9Combustion: Organic compound + O2
Æ CO2 + H2O plus Heat, Light
9C3H8 (g; propane)+5O2 Æ 3CO2+4H2O
Combustion of Propane
Breaking Bonds
Exothermic
Making
Bonds
C3H8 (g; propane)+5O2 Æ
3CO2+4H2O (Balanced Eq.)
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Heat of Combustion for
Methane
• Go to Video camera
Enthalpy of Reaction (ΔH) for
Compounds: Another Approach
9 Hess’s Law is useful if the required reactions are
tabulated, otherwise use Plan B.
9 Plan B: Calculate ΔH of compound formation from
the so-called “Standard Enthalpies of Formation”
of reactants and products.
9 Standard Enthalpies of Formation (ΔH0): Enthalpy
change that takes place when one mole of a
substance in its standard state (T,P) is formed
from its elements in their standard state (=0 for
elements and diatomic gases (O2)).
Enthalpy of Reaction (ΔHrxn)
from Standard Enthalpies
9 ΔH0 reaction = Σ nΔH0formation, products - Σ mΔH0formation, reactants
n, m = number of moles in the balanced equation
9 ΔH0formation values for a compound or element in
its ionic non-standard state are found in
“tables” like Appendix A4.3 in your text.
9 Example: Calculate the Heat of Reaction
(ΔHrxn) for the combustion of propane (C3H8
(g)) from standard enthalpies of formation…
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Table 11.3: Standard Enthalpies
of Formation at 298 K
Compound
Compound
CH4 (g)
ΔHf0
(kJ/mol)
-74.8
CO (g)
ΔHf0
(kJ/mol)
-110.5
C2H6 (g)
-84.7
CO2 (g)
-393.5
C3H8 (g)
-103.8
H2O (g)
-241.8
C4H10 (g)
-125.7
H2O (l)
-285.8
O2 (g)
0.0
H2 (g)
0.0
C3H8 (g; propane)+5O2 (g) Æ
3CO2 (g) + 4H2O (g)
• Heat of Formation of Reactants:
• Propane = -103.8 kJ / mol (See Table 11.3)
• Oxygen = 5 x 0.0 = 0
• Heat of Formation of Products:
• 3 x carbon dioxide = 3 x -393.5 = -1180.5 kJ/mol
• 4 x H2O = 4 x -241.8 = -967.2
• ΔH0 reaction = ΣΔH0formation, products - ΣΔH0formation, reactants
• ΔH0 reaction = -1180.5 – 967.2 – (-103.8+0) = -2043.9
ΔH
Reactants
11_11_n.jpg
form
Unknown
=0 for O2
ΔHform Products
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A Review: Enthalpy of Formation
and Reaction
9 Standard Enthalpy of Formation: The enthalpy
change that takes place when 1 mole of a
substance in its standard state is formed from its
elements in their standard state.
9 Standard State: An element’s most stable physical
form under standard conditions, such as 25°C
(298.15K) and a pressure of 105 Pascals (~1
atm.)
9 ΔHf°, Standard Heat of Formation.
9 ΔHf° of a pure element in its most stable form is
zero.
9 ΔH0 reaction = Σ ΔH0formation, products - Σ ΔH0formation, reactants
Enthalpy (ΔH) of Reactions:
Hess’s Law
9 The enthalpies of many chemical reactions have
been determined experimentally under “standard”
conditions (T,P, 1 mol) and these are complied in
Tables (not your text, unfortunately).
9 For Example: CH4 (g) + H2O (g) → CO (g) + 3 H2
ΔH = 206 kJ
9 Hess’s Law: The enthalpy change for a reaction
that is the sum of two or more other reactions is
equal to the sum of the enthalpy changes of the
constituent reactions….Useful for estimating the
heats of combustion, heat of formation, etc. if
ΔH values are not compiled.
Hess’s Law Beyond Here and
Calorimetry is NOT Required for
FCH152-2006…
• See Problem “Hess’s Law Made
Simple”……..
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Use Hess’s Law for this reaction
to determine the ΔH …
• CH4 + 2H20 → CO2 + 4H2 (ΔHrxn = ?)
• Reaction 1: CH4 + H20 → CO + 3H2 (ΔH1)
• Reaction 2: CO + H20 → CO2 + H2 (ΔH2)
• Add Reaction 1 and 2: Eliminate CO from
each side:
CH4 + 2H20 + CO → CO + CO2 + 4H2
CH4 + 2H20 → CO2 + 4H2
ΔHtotal, rxn = ΔH1+ ΔH2
Hess’s Law Problem
• Overall Reaction: Step 1 plus Step 2
• CH4 + H20 +CO +H20 →
CO2 + H2 + CO + 3H2 = CO2 + CO + 4H2
• Cancel Out CO’s on Each Side of
Equation….
• Overall Equation: CH4 + 2H20 →
CO2 + 4H2
• Add ΔH for each step = Overall ΔH
11_12.jpg
Step 2:
Data Found in
Reference “Tables”
Reaction of Unknown
Enthalpy Change ↓
CO + H20 → CO2 + H2
Step 1:
Data Found in
Reference “Tables”
Application of Hess’s Law
Overall Reaction:
Sum of Two Others
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Hess’s Law Made Really Easy ….
• You want to determine the ΔHreaction for a unknown reaction….
For example:
A + B → C + D ; ΔHreaction is unknown
• Find some reactions in a Chemical Reference Book for which
ΔHreaction is known:
– Reaction 1: C + E → A + F ; (ΔH1 = 30 kJ)
– Reaction 2: A + F → C + E ; (ΔH2 = -30 kJ; Flip !)
– Reaction 3: B + E → F + D ; (ΔH3 = -45 kJ)
• Add Reactions 2 and 3: Eliminate E,F from each side:
A+ B + E + F → C + D + E + F
A+ B → C + D (Target Reaction)
So, ΔHtotal = ΔH2+ ΔH3 = -15 kJ
Another example of using Hess’s
Law to determine the ΔH …
• 2 C (s) + O2 (g) → 2 CO (g) ; ΔH = ?
• Reaction 1: C (s) + O2 (g) → CO2 (-393.5 kJ)
– Problem #1: Need to double the number of carbons
– Two times Reaction 1 = 2 x -393.5 = - 787 kJ
• Reaction 2: 2 CO (g) + O2 (g) → 2 CO2 (g) (-566
kJ)
– Problem #2: No CO on Right Side !
– Do the Flip !!
– 2 CO2 (g) → 2 CO (g) + O2 (g) (566 kJ)
Another example of using Hess’s
Law to determine the ΔH …
• 2 C (s) + O2 (g) → 2 CO (g) ; ΔH = ?
• Reaction 1 (Doubled): 2C (s) + 2O2 (g) → 2
CO2 (-787 kJ)
• Reaction 2 (Flipped): 2 CO2 (g) → 2 CO (g)
+ O2 (g) (566 kJ)
• Overall Reaction (Add 1 and 2):
– 2 CO2 (g) + 2 C + 2O2 (g) → 2 CO (g) + O2 (g) + 2 CO2 =
-787 + 566 kJ = -221 kJ = ΔH )
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11_12_n.jpg
Known
Unknown
Known
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Calorimetry:
Measuring Heat Flow
Example: Heat of
Combustion
(1) Measure
temperature
increase in water
(2) Calculate heat
required to generate
this increase
Click on Image
“Heat Capacity” of the Calorimeter
“Calibrate” with Benzoic Acid (1g):
•ΔHcombustion
= qp = 26.38 kJ
C = q/ΔT
•Measure
ΔT
in calorimeter.
•In this case ΔT = 7.3 °C.
C = 26.38kJ / 7.3°C
= 3.64 kJ/°C
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Calorimeter Exercise
How much heat would be generated by 1
gram of coal if it were combusted in the
calorimeter and a temperature rise of
6.76°C was observed?
qp = C
.
ΔT
qp = 3.64 kJ/°C
.
6.76°C
qp = 24.6 kJ
Calorimetry Exercise….
• ….
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