Chapter 9 Energy, Enthalpy and Thermochemistry - PowerPoint PPT Presentation

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Chapter 9 Energy, Enthalpy and Thermochemistry

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Chapter 9 Energy, Enthalpy and Thermochemistry The study of energy and its interconversions is called thermodynamics. Kinetic Energy: energy due to the motion of the ... – PowerPoint PPT presentation

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Title: Chapter 9 Energy, Enthalpy and Thermochemistry


1
Chapter 9Energy, Enthalpy and Thermochemistry
  • The study of energy and its interconversions is
    called thermodynamics.
  • Kinetic Energy energy due to the motion of the
    object (1/2 mv2)
  • Potential Energy energy due to position or
    composition
  • Heat the transfer of energy between two objects
    due to a temperature difference
  • Work a force acting over a distance

2
Figure 9.1 (a) Initial position of balls
3
Figure 9.1 (b) Final position of balls
4
Energy of Matters
5
State Function
  • A property of the system depends only on its
    present state. A state function does not depend
    in any way on the systems past.
  • Energy is a state function, but work and heat are
    not state function

6
Combustion of methane
7
Nitrogen/oxygen
8
First Law of Thermodynamics
  • The energy of the universe is constant

In closed system ?E ?Uqw (?????) q the
heat added to the system during the process w
the work done on the system during the
process qgt0 heat flows into the system from the
surroundings qlt0 an outflow of heat from the
system to the surroundings wgt0 work is done on
the system by the surroundings wlt0 the system
does work on the surroundings
9
P-V Work
10
P-V Work
11
Enthalpy (?)
  • The heat qp absorbed in a constant-pressure
  • process equals the systems enthalpy change.

12
  • For a chemical reaction
  • ?H?Hproducts-?Hreactants
  • If ?Hreactantslt?Hproducts (endothermic)
  • If ?Hreactantsgt?Hproducts (exothermic)
  • Consider a constant-volume process
  • dw-PdV0 (???????)
  • ?Uqwqv
  • ?Uqv

13
Heat Capacity (???)
14
Thermodynamics of Ideal Gases
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Heat Capacity of Heating an Ideal Monatomic Gas
  • Under constant volume, the energy flowing into
    the gas is used to increase the translational
    energy of the gas molecules.

17
Heat Capacity of Ideal Monatomic Gases
Gas Cv (J/K mol) Cp (J/K mol)
He 12.47 20.8
Ne 12.47 20.8
Ar 12.47 20.8
18
Heat Capacity of Diatomic Gases
Gas Cv (J/K mol) Cp (J/K mol)
H2 20.54 28.86
N2 20.71 29.03
19
Heat Capacity of Polyatomic Gases
Gas Cv (J/K mol) Cp (J/K mol)
N2O 30.38 38.70
CO2 28.95 37.27
C2H6 44.60 52.92
20
Heat Capacity of Heating a Polyatomic Gas
  • Polyatomic gases have observed values for Cv that
    are significantly greater than 3/2 R.
  • This larger value for Cv results because
    polyatomic molecules absorb energy to excite
    rotational and vibrational motions in addition to
    translational motions.

21
Cv and Cp of molecules
22
Cv and Cp of Monatomic Gas
23
Cv and Cp of H2O at 373K
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2 mol of monatomic ideal gas Calculate q, w, ?U
and ?H for both pathway
26
  • TA122K, TC366K, TB183K, TD61K
  • Cv3/2R, Cp5/2R
  • Path 1(A?C)
  • w1-P?V-2atm(30-10)L101.3J/Latm-4.05103J
  • q1qpnCp?T25/2(R)(366-122)1.01104J?H1
  • ?U1nCv?T 23/2(R)(366-122)6.08103J
  • Path 2(C?B)
  • q2qvnCv?T23/2(R)(183-366)-4.56103J?U2
  • ?H2nCp?T 25/2(R)(183-366)-7.6103J
  • ?V0 w2 -P?V 0

27
  • Path 3(A?D)
  • q3qvnCv?T23/2(R)(61-122)-1.52103J?U3
  • ?H3nCp?T 25/2(R)(61-122)-2.53103J
  • ?V0 w2 -P?V 0
  • Path 4(D?B)
  • w1-P?V-1atm(30-10)L101.3J/Latm-2.03103J
  • q4qpnCp?T25/2(R)(183-61)5.08104J?H4
  • ?U4nCv?T 23/2(R)(183-61)3.05103J

28
Summary
  • Path 1
  • qpath1q1q25.5 103J
  • wpath1w1w2 -4.05103J
  • ?Hpath1 ?H1 ?H2 2.55103J
  • ?Upath1 ?U1 ?U2 1.52103J
  • Path 2
  • qpath2q3q43.56103J
  • wpath2w3w4 -2.03103J
  • ?Hpath2 ?H3 ?H4 2.55103J
  • ?Upath2 ?U3 ?U4 1.52103J

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30
Calorimetry
  • Specific heat capacity with unit JK-1g-1
  • Molar heat capacity with unit JK-1mol-1

31
Coffee Cup Calorimeter
  • A constant-pressure calorimetry is used in
    determining the change in enthalpy equals the
    heat.
  • ?HqpnCp?T

32
Bomb Calorimeter
  • ?V0
  • ?Uqv0qv
  • ?UqvnCv ? T

33
2SO2(g)O2(g)?2SO3(g) ?H-198 KJ 2 mol. 1
mol. 2 mol. Calculate ?H and ?U
34
  • P is constant, ?Hqp-198 KJ (energy flow out of
    system)
  • ?U qp w
  • w-P?V and ?V?n(RT/P)
  • T and P are constant, ?nnfinal-ninitial-1 mol
  • So w-P?V-P?n (RT/P)
  • - ?nRT-(-1)(8.314)(298)2.48 kJ
  • ?U qp w-198 kJ2.48 kJ-196 kJ

35
Hesss Law
  • If a reaction is carried out in a series of
    steps, ?H for the reaction will be equal to the
    sum of the enthalpy changes for the individual
    steps
  • The overall enthalpy change for the process is
    independent of the number of steps or the
    particular nature of the path by which the
    reaction is carried out.

36
  • Consider the combustion reaction of methane to
  • form CO2 and liquid H2O
  • CH4(g) 2O2(g)? CO2(g) 2H2O(l)
  • ?H1 -890KJ/mol
  • This reaction can be thought of as occurring in
  • two steps
  • CH4(g) 2O2(g)? CO2(g) 2H2O(g)
  • ?H2 -802 kJ/mol
  • 2H2O(g)?2H2O(l)
  • ?H3 -88KJ/mol
  • ?H1 ?H2 ?H3

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38
Standard Enthalpies of Formation
  • The change in enthalpy that accompanies the
    formation of 1 mole of a compound from its
    elements with all substances in their standard
    states.
  • The superscript zero indicates that the
    corresponding process has been carried out under
    standard conditions.

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42
Present Sources of Energy
  • Petroleum and Natural Gas Coal

New Energy Sources
Coal Conversion Hydrogen as a fuel
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CO2 capture
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