Animated ABC Blocks

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? Laws of conservation of energy

• Internal energy (U)
• gt when receives energy (either in forms of work
  or heat)molecules store energy as either kinetic
  or potential energy
• gt kinetic energy is modes of motion such as
  translation, rotation,
• and vibration and raises when temperature raises
• gt potential energy is interaction energy
  (attraction and repulsion)
• and does not depend on temperature
• gt interaction energy can be chemical bonds or
  intermolecular
• forces such as H-bond, van der Waals interaction

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• gt kinetic and potential energy of molecules
  are called
• internal energy (U)
• gt When system receives heat or work, U is
  raised
• gt When system performs work or lose heat, U is
  declined

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• Statement of first law
• gt state of system describe by properties such as
  T, P, V
• gt change in T, P, V means change of state
• gtHeat received/lost by system causes change in T
• gt Work received/performed by system causes
  change in P, V
• gt When there are heat/work transfer to/from
  system, there will be change in state

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• gt Change in U or ?U causes change in state and
  vice versa
• gt Amounts of heat/work received or less related
  to ?U
• gt The first law of thermodynamics state that
• ?U qw
• ?U is state function (depends only on initial and
  final state not path)
• If there is no work/heat transfer to/from system
• ?U 0

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• gt Universe has no surrounding, no heat/work can
  be transferred
• to/from system
• ?Uuniverse 0
• (internal) energy of universe is constant
• energy of universe cannot be
  produced
• energy of universe cannot be
  destroyed
• gtgtThis is the law of conservation of
  energy ltlt

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gt system does work w lt 0 system receives
heat q gt 0 If system does work but not heat
absorbed (q0) ?U -w
lt 0 internal energy is reduced when system
does work gt If internal energy is used up (?U
0) no more work can be performed gt Internal
energy can be raised by adding heat/work to
system ?U gt 0
more work can be performed
more heat can be generated
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gt We can not produces more work/heat than the
input heat/work output input gt No engine
that can generate output more than input gt No
engine that can run forever without need to
refuel ?no perpetual machine of
the first kind ? gt etc.
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Internal energy work heat From 1st Law ?U
q w Ideal gas has no potential energy Its
internal energy depends only kinetic energy In
isothermal expansion , ?U 0 for ideal gas since
U depends on T only 0 q w q - w For
isothermal reversible expansion q
nRTln(Vf/Vi)
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• Example 3.4 Calculate the heat absorbed and the
  change in the internal energy for the two
  expansions of an ideal gas in Example 3.2
• Reversible isothermal expansion
• ?U 0 q w
• q -w nRTln(5/1)
• 3,988 J
• Irreversible isothermal expansion
• q -w
• pf nRT (1/Pf 1/Pi)
• 1,982 J

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• Partial differential definition of internal
  energy
• f (x,y)

total diff.
partial diff.
U (T,V)
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From dU dq dw dq PdV
heat capacity at constant volume

• For ideal gas since no
  molecular interaction
• dU CV dT
• or ?U ? CV dT CV ?T
• For adiabatic work q 0
• ?U w CV ?T
• wad CV ?T

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From dU dq PdV at constant V dU
dqV or ?U qV heat measured at constant
volume is internal energy change gt We can
measured qV using special design instrument
called Bomb Calorimeter gt With Bomb
Calorimeter, the raise in temperature of water
with known mass is measured
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Bomb Calorimeter
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• Example 3.5 0.600 g of octane is burned in a bomb
  calorimeter containing 2.5 kg of water. The bomb
  calorimeter has a heat capacity of 1.09 X 104
  J/oC. The temperature of water and the bomb
  changes from 24.984oC to 26.328oC. Calculate the
  internal energy for this process
• qbomb Cbomb DT 1.09X104
  J/oC X (26.328 24.984)oC 1.46 X104 J
• qwater mwsw DT
• 2500 g X 4.184 J/g oC (26.328 24.984)oC
  1.406X104J
• qcombustion -(qwater qbomb) -2.87X104J
• DU

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• Enthalpy (H)
• heat measured at constant pressure (atmospheric
  pressure)
• Relation to internal energy
• dq dU PdV
• H qp U PV
• ?H ?U ?PV
• ?PV (PV)f - (PV)i

H U PV
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at constant P ?PV P(Vf - Vi) P?V
?H ?U P?V For ideal gas H
U PV U nRT
H/n Hm Um RT
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-Partial differential definition of enthalpy H
U PV dH dU d(PV)
dU PdV VdP (dq PdV) PdV
VdP dH dq VdP at constant P
dH dqP ?H dqP
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enthalpy is heat and can be measured using
calorimeter H (P, T)
at constant P
Cp constant pressure heat capacity
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at constant P dH CP dT
Normally for small range of T, CP is not a
function of T ?H CP ?T
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gt Relationship between CP and CV from
H U PV For ideal gas H U
nRT H U nRT ?H - ?U nR
?T ?H/T - ?U/T nR
CP - CV nR CP,m -
CV,m R CP,m CP/n , CV,m CV/n
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Example 3.6 From example 3.4 find ?H for those
processes

• Isothermal reversible expansion
• ?H ?U ?PV
• since for constant T PfVf PiVi
• ?PV PfVf - PiVi 0
• ?U 0 for ideal gas
• ?H 0 0 0
• Isothermal irreversible expansion
• ?H ?U ?PV 0
• ?H does not depends on path
• ?H is the state function

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• Direction of DH
• Since DH is q
• DH lt 0 system releases heat
• exothermic
• DH gt 0 system absorbs heat
• endothermic

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• ? Thermochemistry
• enthalpy is the state function
• depends only on initial and final
  state not path
• Physical change (phase transition)
• initial final

?H1 Hg - Hl
H2O (l)
H2O (g)
It can also happen this way
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It can also happen this way
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gt Chemical change
?Hc
We can write reaction in stepwise manner
?H1
(1)
?H2
(2)
?H3
(3)
adding (1), (2), (3), we obtain
the same as combustion of C3H6 According to the
1st law
?Hc ?H1 ?H2 ?H3
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- Hesss law
The standard enthalpy of a reaction is the sum
of the standard enthalpies of the reactions in to
which the overall reaction may be divided
For example
H2O (g)
?HO 44.01 kJ
25 Oc, 1 atm H2O (l)
From Hesss Law
is the same as the combination of
?HO 44.01 6.97 50.98 kJ
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In the case of combustion of C3H6 (g)
At 298 K and 1 atm
?H01 -124 kJ
?H02 -2220 kJ
?H03 286 kJ
?H0c -2058 kJ
?H0 ? standard enthalpy change enthalpy change at
1 atm
gtgt We can calculate unknown ?H from known
values gtgt It is unnecessary for a reaction to
follow the path used in the calculation of ?H
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Example 3.7 calculate ?UO for combustion of C3H6
(g)
?H -2058 kJ
?HO ?UO ?(PV) ?UO ?H - ?(PV)
?(PV) (PV)f (PV)i
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-Bond enthalpies (BE) heat used for dissociation
of chemical bond in gas molecule
Bond enthalpy of H-Cl bond (BE(H-Cl)) 431
kJ Bond enthalpy always has positive values
(endothermic) Concept of BE can be used to
calculate ?H
Different molecules with the same bond type have
different BE Mean bond enthalpy which is average
value of BE of the same bond typefrom different
molecules is normally used
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Example 3.8 Estimate the standard enthalpy change
for the reaction
?HO1
?HO2
?HO3
?HO4
?HO5
C (s,graphite) 2H2 (g) 1/2O2 (g)
CH3OH (l)
?HO
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?HO ?HO1 ?HO2 ?HO3 ?HO4 ?HO5
?HO1 atomization of graphite 716.68 kJ
heat of sublimation
?HO2 2BE (H-H) 2 x 436 872 kJ
?HO3 1/2BE (OO) 1/2 x 497 248.5 kJ
-?HO4 3BE (C-H) BE(C-O) BE (O-H)
3 x 412 360 463 2059 kJ
?HO5 -?Hvap -38.00 kJ
?HO -259.82 kJ (-239.00kJ expt.)
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Enthalpy of formation (?Hf)
is heat or enthalpy associated with the formation
of 1 mole of a compound form stable form
(reference state)
Elements
1 mol of compound ?H ?Hf
at 298 K
?HO-285.83 kJ
(1) H2 (g) 1/2O2 (g)
H2O (l)
?HOf (H2O,l) -285.83 kJ/mol
?HOf (H2O,g) -241.82 kJ/mol
(2) H2 (g) 1/2O2 (g)
H2O (l)
(2)-(1)
?HO ?HOf (H2O,g) - ?HOf (H2O,l)
-241.82 (-285.83) 44.01 kJ
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Enthalpy of formation of elements in stable form
0
DHof (H2,g) 0 DHof (O2,g) 0 DHof
(C,graphite) 0
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Example 3.9 Find standard enthalpy of reaction at
298 K for
(6mol)(-393.51 kJ/mol) (3mol)(-285.83 kJ/mol)
(1mol)(49kJ/mol) -3268 kJ
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Temperature dependent
from
---------- ()
aA bB cC
dD
?H (cHm,C dHm,D) - (aHm,A
bHm,B)
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Using ()
Kirchhoff s Law
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Example 3.9 The standard enthalpy of gaseous
water and liquid water at 250C are -241.82 and
-285.83 kJ/mol, respectively. Find the standard
heat of vaporization of water.Given Cpm,l
75.29 and Cpm,g 33.58 J/Kmol
H2O(l)
H2O(g) ?HOvap ?
?HOvap ?HO(1000C) ?HO(250C)

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