EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC? qoverall = qice + qfusion + qwater + qboil + qsteam
q = (10.0g 2.09J/goC 15.0oC)
+ (10.0g 333J/g)
+ (10.0g 4.18J/goC 100.0oC)
+ (10.0g 2260J/g)
+ (10.0g 2.03J/goC 27.0oC)
q = (314 + 3.33×103 + 4.18×103 + 2.26×104 + 548)J
= 30.9 kJ
Heat Flow in Reactions
exothermic –reaction that gives off energy
q < 0
isolated system E=0 heat released by reaction raises the temperature of the solvent
constant T, heat is released to the surroundings
endothermic – reaction that absorbs energy
q > 0
Expansion Type Work
w = -PV system does work
P
P
Vinitial
V
V = Vfinal - Vinitial
qp = +2kJ
Do 250 J of work to compress a gas, 180 J of heat are released by the gas
What is E for the gas?
430
J70
J-7
0 J
-180
J
-250
J
0% 0% 0%0%0%
1. 430 J
2. 70 J
3. -70 J
4. -180 J
5. -250 J
0
0
130
10
Enthalpy HE = q + w
at constant V, wexpansion = 0
E = qv
at constant P, wexpansion = -PV
E = qp - PVDefinePV) = PV at constant P
Hence H = qp
EnthalpyEnthalpy heat at constant pressure or the
heat of reaction
qp = H = Hproducts - Hreactants
Exothermic ReactionH = (Hproducts - Hreactants) < 0
2 H2(g) + O2(g) 2 H2O(l) H < 0
Endothermic ReactionH = (Hproducts - Hreactants) > 0
2 H2O(l) 2 H2(g) + O2(g) H > 0
State FunctionsH and E along with P, T, V (or P, T,
V) and many others are state functions. They are the same no matter what path we take for the change.
• q and w are not state functions, they depend on which path we take between two points.
initial
final
Eq
w
q
wE=Efinal-Einitial
q and w can be anything
Path Independent Energy Changes
Which day would you like OWL quizzes due (4 AM)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
121 122 123 124 125 126 127 128 129 130
0% 0% 0%0%0%
1. Monday2. Tuesday3. Wednesday4. Thursday5. Friday
100
0
130
Stepwise Energy Changesin Reactions
Laws of Thermochemistry
1. The magnitude of is directly proportional to the amount of reaction.
H is for 1 mole of reaction as written
2 H2(g) + O2(g) 2 H2O(l) H = -571.6 kJ
H2(g) + ½ O2(g) H2O(l) H = -285.8 kJ
Can have ½ mole O2 just not ½ molecule
Laws of Thermochemistry
2. H for a reaction is equal in magnitude but opposite in sign to H for the reverse reaction.
H2(g) + ½ O2(g) H2O(l) H = -285.8 kJ
H2O(l) H2(g) + ½ O2(g) H = +285.8 kJ
Laws of Thermochemistry
3. The value of H for the reaction is the same whether it occurs directly or in a series of steps.
Hoverall = H1 + H2 + H3 + · · ·
also called Hess’ Law
Enthalpy Diagram
H2(g) + ½ O2(g) H2O(l) H = -285.8 kJ
H2O(l) H2O(g) H = +44.0 kJ
H2(g) + ½ O2(g) H2O(g) H = -241.8 kJ
Given 3 CO + 3/2 O2 3 CO2 H = -849 kJ
What is H for CO2 CO + ½ O2 ?
100
0
130
-283
kJ
+283
kJ
+849
kJ
-254
7 kJ
+254
7 kJ
0% 0% 0%0%0%
1. -283 kJ
2. +283 kJ
3. +849 kJ
4. -2547 kJ
5. +2547 kJ
Energy and Stoichiometry
• Since H is per mole of reaction we can relate heat to amount of reaction
• Given C2H6 + 7/2 O2 2 CO2 + 3 H2O H=-1559.7 kJ
• If 632.5 kJ are released to surroundings what mass of H2O is formed?
• 632.5 kJ released means H = -632.5 kJ for this much H2O
OHg92.21OHmol1
OHg016.18
reactionmol1
OHmol3
kJ7.1559
reactionmol1kJ5.632 2
2
22
Bomb Calorimeter measure qv
qrxn + qcal = 0qrxn = -qcal
qrxn = - ccalTErxn = qrxn/moles rxnErxn ≈ Hrxn
H = E + (PV)H = E + RTngas
@298K RT = 2.5 kJ/mol
“Coffee Cup” Calorimeter qp
Photo by George Lisensky
Measuring H
• When 25.0 mL of 1.0 M H2SO4 are added to 50.0 mL of 1.0 M KOH, both initially at 24.6C the temperature rises to 33.9C. What is H for H2SO4 + 2 KOH K2SO4 + 2 H2O ? (Assume d = 1.00 g/mL, c = 4.18 J/g.C)
• qsoln = mcT
• m = (25.0 + 50.0)mL×1.00g/mL = 75.0 g
Measuring H cont.
• q=mcT
• qsoln = 75.0 g × 4.18 J/g.C × (33.9-24.6)C
• qsoln = 2916 J
• qrxn + qsoln = 0
• qrxn = -2916 J
Hrxn = qrxn/moles rxn
Measuring H cont
• How many moles rxn?
• 1 mol rxn / 1 mol H2SO4
• 1 mol rxn / 2 mol KOH
4242 SOHmol025.0
mL1000
L1
L1
SOHmol00.1mL0.25
KOHmol050.0mL1000
L1
L1
KOHmol00.1mL0.50
Stoichiometric mixture so 0.025 mol rxn
Measuring H cont
Hrxn = qrxn/moles rxn
Hrxn = -2916 J / 0.025 mol rxn
Hrxn = -116622 J / mol rxn
Hrxn = -117 kJ
H is per mole of reaction as written
If excess Al is added to 50 mL of 0.250 M H2SO4 how many moles of
the following reaction occur?2 Al + 3 H2SO4 Al2(SO4)3 + 3 H2
10
0
0
130
0% 0% 0%0%0%
1. 0.0125 mol
2. 0.0375 mol
3. 0.025 mol
4. 0.00625 mol
5. 0.00417 mol
Hess’s Law
• Can find H for an unknown, or hard to measure, reaction by summing measured H values of known reactions.
EXAMPLE
H for formation of CO cannot readily be measured since a mixture of CO and CO2 is always formed.
C (s) + ½ O2 (g) CO (g) H = ?
C (s) + O2 (g) CO2 (g) H1 = -393.5 kJ
CO (g) + ½ O2 (g) CO2 (g) H2 = -283.0 kJ
C (s) + ½ O2 (g) CO (g) H = H1 - H2
H = H1 - H2 = -393.5 – (-283.0) = -110.5 kJ
Standard Enthalpy of Formation
the enthalpy associated with the formation of 1 mol of a substance from its constituent elements under standard state conditions at the specified temperature
For an element this is a null reaction
O2 (g) O2 (g) H = 0
Hf = 0 for all elements in their standard states
For which one of these reactions is ΔHºrxn
= ΔHºf?
0% 0% 0%0%0%
1. N2(g) + 3 H2(g) 2 NH3(g)
2. C(graphite) + 2 H2(g) CH4(g)
3. C(diamond) + O2(g) CO2(g)
4. CO(g) + ½ O2(g) CO2(g)
5. H2(g) + Cl2(g) 2 HCl(g)
100
0
130
Calculation of Ho
Ho = mols Hfoproducts – mols Hf
oreactants
We can always convert products and reactants to the elements.
Hess’s law says H is the same whether we go directly from reactants to products or go via elements
Example What is the value of Hrxn for the reaction:2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
from Appendix J TextC6H6(l) Hf
o = + 49.0 kJ/mol
O2(g) Hfo = 0
CO2(g) Hfo = - 393.5
H2O(g) Hfo = - 241.8
Hrxn mols Hfoproduct
– mols Hforeactants
ExampleWhat is the value of Hrx for the reaction:2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
from Appendix J TextC6H6(l) Hf
o = + 49.0 kJ/mol; O2(g) Hfo = 0
CO2(g) Hfo = - 393.5; H2O(g) Hf
o = - 241.8 Hrxn molsHf
oproduct - mols Hf
oreactants
Hrxn - 393.5) + 6(- 241.8)product
- 2(+ 49.0 ) + 15(0)reactants kJ/mol
= - 6.2708 103 kJ
Fossil Fuels
coal
petroleum
natural gas
Energy Resources in the U.S.
Caloric Value of Some Foods
Top Related