Sep. 17, 2010!
assignments: reaction enthalpy, T dependence
of enthalpy change
Enthalpy change in reactions
Chemical reactions involve state changes
(matter changes from reactants to products).
readings: Ch. 4.4-4.6
ΔH = Hprod - Hreact
Half-exam #1 (30 min) will be Friday 9/24 in class !
State 1:
reactants
State 2:
products
nA A + nB B → nC C + nD D
Today: enthalpy of reactions, enthalpy of formation!
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Example: adding reactions
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Example: adding reactions
What is ΔH for the reaction
C(graphite) + 2 H2(g) + ½ O2(g) → CH3OH(ℓ)?
What is ΔH for the reaction
C(graphite) + 2 H2(g) + ½ O2(g) → CH3OH(ℓ)?
Data:
C(graphite) + O2(g) → CO2(g),
ΔH = -393.5 kJ/mol
H2(g) + ½ O2(g) → H2O(ℓ),
ΔH = -285.8 kJ/mol
CH3OH(ℓ) + 3/2 O2(g) → CO2(g) + 2 H2O(ℓ), ΔH = -726.4 kJ/mol
Data:
C(graphite) + O2(g) → CO2(g),
ΔH = -393.5 kJ/mol
H2(g) + ½ O2(g) → H2O(ℓ),
ΔH = -285.8 kJ/mol
CH3OH(ℓ) + 3/2 O2(g) → CO2(g) + 2 H2O(ℓ), ΔH = -726.4 kJ/mol
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The standard state
Enthalpy of formation for compounds
Standard state conditions for reactions are:
• T = 25 °C
• P = 1 atm
• Phase is the most stable at this T and P
• 1 M concentration if a solute
nA A + nB B → nC C
If the species A and B above are
elements, then C is a compound
containing atoms A and B.
nA A + nB B → nC C + nD D
The enthalpy change in this sort
of reaction is the enthalpy of
formation of compound C, ΔH0f
ΔH0 indicates a reaction under
standard state conditions
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Enthalpy of formation for elements
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Enthalpies of formation are tabulated
The enthalpy of formation for any
element under standard state
conditions is zero, ΔH0f = 0.
This works because we are interested in enthalpy
changes for reactions. The absolute value of
enthalpy for each reactant and product is not as
useful.
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Species
ΔH0f (kJ mol-1)
S0 (J K-1 mol-1)
ΔG0f (kJ mol-1)
bicarbonate- (aq)
-691.99
91.2
-586.77
butyric acid (s)
-533.9
226.4
-377.69
carbon (s,graphite)
0
5.74
0
carbon dioxide (g)
-393.51
213.74
-394.36
carbon dioxide (aq)
-413.80
117.6
-385.98
creatine (s)
-537.18
189.5
-246.93
L-cysteine (s)
-533.9
169.9
-343.97
L-cystine (s)
-1051.9
280.58
-693.33
ethane (g)
-84.68
229.60
-32.82
ethanol (ℓ)
-276.98
160.67
-174.14
formaldehyde (g)
-115.90
218.78
-109.91
water (ℓ)
-285.83
69.95
-237.13
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Example: using enthalpies of formation
Bond energy and reaction enthalpy
What is ΔH0 for the reaction!
CO2(aq) + H2O(ℓ) → HCO-(aq) + H+(aq)?!
The enthalpy change in a chemical reaction
comes from the difference in total bond energy of
reactants minus that of products.
ΔH = Σ BE(reactants) – Σ BE(products)
ΔHrxn = energy input to break bonds +
energy released as bonds form
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Example: bond energy → reaction enthalpy
Example: bond energy → reaction enthalpy
Photosynthetic formation of glucose from H2O and CO2.
Photosynthetic formation of glucose from H2O and CO2.
6 H2O(ℓ) + 6 CO2(g) → C6H12O6 (s) + 6 O2(g)
6 H2O(ℓ) + 6 CO2(g) → C6H12O6 (s) + 6 O2(g)
ΔHo = 2801 kJ/mol
ΔHo = 2801 kJ/mol
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Temperature dependence of ΔH
Temperature dependence of ΔH
aA+bB →cC+dD
Data tables give ΔH0f, which are
for standard temperature, 25 °C.
What if your reaction is at a
different temperature?
aA+bB →cC+dD
T > 25 °C
aA+bB →cC+dD
T = 25 °C
The way to solve this problem is to
use the fact that H is a state function,
i.e. ΔH = 0 around a complete cycle.
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Example: ΔH for oxidation of ethanol at 37 °C
Example: ΔH for oxidation of ethanol at 37 °C
CH3CH2OH(ℓ) + 3 O2(g) → 2 CO2(g) + 3 H2O(ℓ)
ΔH0 = -1367.5 kJ/mol
CH3CH2OH(ℓ) + 3 O2(g) → 2 CO2(g) + 3 H2O(ℓ)
ΔH0 = -1367.5 kJ/mol
CP,water = 75.3 J K-1 mol-1
CP,EtOH = 112 J K-1 mol-1
CP,oxygen = 29 J K-1 mol-1
CP,carbon dioxide = 37 J K-1 mol-1
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Recap
Chapters 2, 3 & 4 are about the First Law of
thermodynamics: the total energy in the
Universe is constant (or, energy is conserved).
• Internal energy E
• Enthalpy H
• Ideal gas as a useful system
• Heat (heat capacity and latent heat)
• Work (force and displacement)
Next up: the Second Law, entropy and Gibbs
free energy
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