Transcript CH07B1

7-5 The First Law of Thermodynamics
Internal Energy, U.
 Total energy (potential and kinetic) in a system.
•Translational kinetic energy.
•Molecular rotation.
•Bond vibration.
•Intermolecular attractions.
•Chemical bonds.
•Electrons.
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First Law of Thermodynamics
 A system contains only internal energy.
 A system does not contain heat or work.
 These only occur during a change in the system.
U = q + w
 Law of Conservation of Energy
 The energy of an isolated system is constant
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First Law of Thermodynamics
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Functions of State
Any property that has a unique value for a
specified state of a system is said to be a
State Function.
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Water at 293.15 K and 1.00 atm is in a specified state.
d = 0.99820 g/mL
This density is a unique function of the state.
It does not matter how the state was established.
Functions of State
U is a function of state.
 Not easily measured.
U has a unique value between two states.
 Is easily measured.
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Path Dependent Functions
 Changes in heat and work are not functions of
state.
 Remember example 7-5, w = -1.24  102 J in a one step
expansion of gas:
 Consider 2.40 atm to 1.80 atm and finally to 1.20 atm.
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Path Dependent Functions
w = (-1.80 atm)(1.36-1.02)L – (1.30 atm)(2.04-1.36)L
= -0.61 L atm – 0.82 L atm = -1.43 L atm
= -1.44  102 J
Compared -1.24  102 J for the two stage process
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7-6 Heats of Reaction: U and H
Reactants → Products
Ui
Uf
U = Uf - Ui
 = qrxn + w
In a system at constant volume:
U = qrxn + 0 = qrxn = qv
But we live in a constant pressure world!
How does qp relate to qv?
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Heats of Reaction
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Heats of Reaction
qV = qP + w
We know that w = - PV and U = qP, therefore:
U = qP - PV
qP = U + PV
These are all state functions, so define a new function.
Let
H = U + PV
Then
H = Hf – Hi = U + PV
If we work at constant pressure and temperature:
H = U + PV = qP
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Comparing Heats of Reaction
qP = -566 kJ/mol
= H
PV = P(Vf – Vi)
= RT(nf – ni)
= -2.5 kJ
U = H - PV
= -563.5 kJ/mol
= qV
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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)
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H = 6.01 kJ at 273.15 K
EXAMPLE 7-8
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 10.0°C to vapor at 25.0°C.
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.
Set up the equation and calculate:
qP = mcH2OT + nHvap
= (50.0 g)(4.184 J/g °C)(25.0-10.0)°C +
= 3.14 kJ + 122 kJ = 125 kJ
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50.0 g
44.0 kJ/mol
18.0 g/mol
Standard States and Standard Enthalpy
Changes
Define a particular state as a standard state.
Standard enthalpy of reaction, H°
 The enthalpy change of a reaction in which all
reactants and products are in their standard states.
Standard State
 The pure element or compound at a pressure of
1 bar and at the temperature of interest.
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Enthalpy Diagrams
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7-7 Indirect Determination of H:
Hess’s Law
 H is an extensive property.
 Enthalpy change is directly proportional to the amount of
substance in a system.
N2(g) + O2(g) → 2 NO(g)
½N2(g) + ½O2(g) → NO(g)
H = +180.50 kJ
H = +90.25 kJ
 H changes sign when a process is reversed
NO(g) → ½N2(g) + ½O2(g)
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H = -90.25 kJ
Hess’s Law
 Hess’s law of constant heat summation
 If a process occurs in stages or steps (even hypothetically),
the enthalpy change for the overall process is the sum of
the enthalpy changes for the individual steps.
½N2(g) + ½O2(g) → NO(g)
H = +90.25 kJ
NO(g) + ½O2(g) → NO2(g)
H = -57.07 kJ
½N2(g) + O2(g) → NO2(g)
H = +33.18 kJ
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