Heats of Reaction: U and H

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Heats of Reaction ?U and ?H
Fixed V vs. Fixed P
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Heats of Reaction ?U and ?H
SO.. qV qP w We know that w - P?V and ?U
qV, therefore ?U qP - P?V qP ?U P?V U
and P and V are state functions, so define a new
function. Let H U PV If we work at constant
pressure and temperature ?H ?U P?V qP
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Comparing Heats of Reaction
2CO O2? 2CO2
Constant V ?U qV -563.5 kJ/mol Evaluate
the PV work P?V P(Vf Vi) RT(nf ni)
-2.5 kJ (small!) Therefore qP -566 kJ/mol
?H
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Exo and Endothermic
DH Hfinal Hinitial Hproducts - Hreactants

• Heat exchange accompanies chemical reactions
• Exothermic Heat flows out of the system (to the
  surroundings)
• when DH is negative
• when Hproducts lt Hreactants
• Endothermic Heat flows into the system (from
  the surroundings)
• when DH is positive
• when Hproducts gt Hreactants

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Exo and Endothermic
qsystem gt 0
qsystem lt 0
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Enthalpy Diagrams Another way to see it
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An Example from a Previous Exam
Consider the reaction S(s) 3/2O2(g) ?
SO3(g) DHf(SO3) -395.7 kJ/mol Is this an
exothermic or endothermic event? exothermic How
much heat is evolved when 10.0 g of sulfur is
burned in 11.0 atm of oxygen in a 1.00 L flask
initially at 0.00C? n S 10.0 g / 32.06 g/mol
0.312 mol n O2 PV/RT 11.00(1.00) /
0.08206(273.15) 0.491 mol If all of the S
reacts we would need 0.312 mol S x 3/2(mol O2/mol
S) 0.468 mol O2 We have more than this! Sulfur
is the limiting reagent DH -395.7 kJ/mol
(0.312 mol) - 123.5 kJ What is the final
pressure of the system at 0.00C? The final
system will consist of SO3 gas and the excess O2
gas There should be 0.312 moles SO3 and
0.491-0.468 0.023 mol O2 P O2
(0.023)(0.08206)(273.15) / 1.00 0.516 atm P
O2 P SO3 0.312(0.08206)(273.15) / 1.00 6.99
atm P SO3 P total P O2 P SO3 0.51 6.99
7.5 atm P total
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Heat Capacity, Enthalpy and Changes of State
Changes of state involve energy Ice 333 J/g ?
Water (heat of fusion)
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Heating / Cooling Curve of Water
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Changes of State of Matter
Molar enthalpy of vaporization
H2O (l) ? H2O(g) ?H 44.0 kJ at 298 K
Molar enthalpy of fusion
H2O (s) ? H2O(l) ?H 6.01 kJ at 273.15 K
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Example
Enthalpy Changes Accompanying Changes in States
of Matter. Calculate ?H for the process in which
50.0 g of water is converted from liquid at
10.0C to vapor at 25.0C.
Break the problem into two steps Raise the
temperature of the liquid first then completely
vaporize it. The total enthalpy change is the
sum of the changes in each step.
3.14 kJ 122 kJ 125 kJ
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Standard Enthalpy Values An anchor point

• Measure the enthalpy change under standard
  conditions DH values are labeled DHo
• Define a standard state (at the temperature of
  interest often 298.15K)
• Compound
• For a gas, pressure is exactly 1 atmosphere.
• For a solution, concentration is exactly 1 molar.
• Pure substance (liquid or solid), it is the pure
  liquid or solid.
• Element
• The form N2(g), K(s) in which it exists at 1
  atm.

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Standard Enthalpy Values
NIST (Natl Institute for Standards and
Technology) gives values of DHof standard
molar enthalpy of formation This is the enthalpy
change when 1 mol of compound is formed from
elements under standard conditions.
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Hesss LawDetermining DH

• Reactants Products
• One of the real strengths in using enthalpies is
  that we can determine the enthalpies of many
  different processes using a much smaller subset
  of enthalpy values.
• The change in enthalpy is the same whether the
  reaction takes place in one step or a series of
  steps.(A State Function)

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Calculations via Hesss Law

• 1. If a reaction is reversed, DH is also
  reversed.
• N2(g) O2(g) 2NO(g) DH 180 kJ
• 2NO(g) N2(g) O2(g) DH -180 kJ
• 2. If the coefficients of a reaction are
  multiplied by an integer, DH is multiplied by
  that same integer (an extensive property).
• 6NO(g) 3N2(g) 3O2(g) DH -540 kJ
• -180(3)kJ

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DH A State Function
S solid
O
2
DH1

3/2 O
-320.5 kJ
2

DH
-395.7 kJ
SO
gas
2
1/2 O
2
SO
gas
3
DH2
-75.2 kJ
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Using Enthalpy
Consider the decomposition of water H2O(g) 242
kJ/mole ? H2(g) 1/2 O2(g) This is an
Endothermic reaction heat is taken up by the
system DH 242 kJ/mole
Reactants Products The change in enthalpy is
the same whether the reaction takes place in one
step or a series of steps.
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Hesss Law
Making H2 from H2O involves two steps (1 mole
reaction) H2O(liq) 44 kJ ? H2O(g) H2O(g)
242 kJ ? H2(g) 1/2 O2(g) H2O(liq) 286 kJ ?
H2(g) 1/2 O2(g) Example of HESSS LAW The net
DH is the sum of the DHs of the individual steps
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Another Example From Test 1 2004
Determine the enthalpy for the reaction N2H4(l)
2H2O2(l) ? N2(g) 4H2O(l) From these
data N2H4(l) O2(g) ? N2(g) 2H2O(l) DH
-622.2kJ H2(g) 1/2O2(g) ? H2O(l) DH
-285.8kJ H2(g) O2(g) ? H2O2(l) DH
-187.8kJ N2H4(g) O2(g) ? N2(g) 2H2O(l) DH
-667.0kJ 2H2O2(l) H2(g) ? 2H2O(l) DH
-383.8kJ N2H4(g) H2(g) ? 2NH3(g) DH
-187.6kJ Solution N2H4(l) O2(g) ? N2(g)
2H2O(l) DH -622.2kJ 2(H2(g) 1/2O2(g) ?
H2O(l)) DH 2(-285.8kJ) 2(H2O2(l) ? H2(g)
O2(g)) DH 2(187.8kJ) N2H4(l) 2O2(g) 2H2(g)
2H2O2(l) ? N2(g) 4H2O(l) 2H2(g) 2O2(g)
N2H4(l) 2H2O2(l) ? N2(g) 4H2O(l) DH
-622.2 2(-285.8kJ) 2(187.8kJ) -818.2kJ