15. Energetics

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15. Energetics
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Calorimetry

• Calorimetry involves the measurement of heat
  changes that occur in chemical processes or
  reactions.
• The heat change that occurs when a substance
  absorbs or releases energy is really a function
  of three quantities
• The mass
• The temperature change
• The heat capacity of the material

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Heat Capacity and Specific Heat

• The ability of a substance to absorb or retain
  heat varies widely.
• The heat capacity depends on the nature of the
  material.
• The specific heat of a material is the amount of
  heat required to raise the temperature of 1 gram
  of a substance 1 oC (or Kelvin)

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Specific Heat values for Some Common Substances
Substance CJ g-1 K-1 C J mol-1K-1
Water (liquid) 4.184 75.327
Water (steam) 2.080 37.47
Water (ice) 2.050 38.09
Copper 0.385 24.47
Aluminum 0.897 24.2
Ethanol 2.44 112
Lead 0.127 26.4
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Heat Exchange

• When two systems are put in contact with each
  other, there will be a net exchange of energy
  between them unless they are at thermal
  equilibrium, i.e. at the same temperature.

• Heat will flow from the substance at the higher
  temperature to that at a lower temperature

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Heat Changes

• The heat equation may be stated as
• DQ m C DT
• where
• DQ Change in heat
• m mass in grams
• C specific heat in J g-1 oC-1
• DT Temperature change

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Temperature Changes

• Measuring the temperature change in a calorimetry
  experiment can be difficult since the system is
  losing heat to the surroundings even as it is
  generating heat.
• By plotting a graph of time v temperature it is
  possible to extrapolate back to what the maximum
  temperature would have been had the system not
  been losing heat to the surroundings.

A time v temperature graph
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Heat Transfer Problem 1

• Calculate the heat that would be required an
  aluminum cooking pan whose mass is 400 grams,
  from 20oC to 200oC. The specific heat of
  aluminum is 0.902 J g-1 oC-1.

Solution DQ mCDT (400 g) (0.902 J g-1
oC-1)(200oC 20oC) 64,944 J
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Heat Transfer Problem 2

• What is the final temperature when 50 grams
  of water at 20oC is added to 80 grams water at
  60oC? Assume that the loss of heat to the
  surroundings is negligible. The specific heat of
  water is 4.184 J g-1 oC-1

Solution DQ (Cold) DQ (hot)
mCDT mCDT Let T final temperature
(50 g) (4.184 J g-1 oC-1)(T- 20oC)
(80 g) (4.184 J g-1
oC-1)(60oC- T) (50 g)(T- 20oC) (80
g)(60oC- T) 50T -1000 4800 80T
130T 5800 T 44.6 oC
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Phase Changes Heat

• Energy is required to change the phase of a
  substance
• The amount of heat necessary to melt a substance
  is called the Heat of fusion (?Hfus).The heat of
  fusion is expressed in terms of 1 mole or 1 gram
• It takes 6.00 kJ of energy to melt 1 mole (18
  grams) of ice into liquid water. This is
  equivalent to about 335 J per gram
• The amount of heat necessary to boil a substance
  is called the Heat of vaporization (?Hvap)
• It may be expressed in terms of 1 mole or 1 gram
• It takes 40.6 kJ of energy to boil away 1 mole
  (18 grams) of water. This is equivalent to about
  2240 J per gram.

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Molar Heat Data for Some Common Substances
Substance ?Qfus ?Qvap
Mercury, Hg 2.29kJ/mol 59.1kJ/mol
Ethanol, C2H5OH 5.02kJ/mol 38.6kJ/mol
Water, H2O 6.00kJ/mol 40.6kJ/mol
Ammonia, NH3 5.65kJ/mol 23.4kJ/mol
Helium, He 0.02kJ/mol 0.08kJ/mol
Acetone 5.72kJ/mol 29.1kJ/mol
Methanol, CH3OH 3.16kJ/mol 35.3kJ/mol
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Heat Transfer Problem 3
How much energy must be lost for 50.0 g of liquid
wax at 85.0C to cool to room temperature at
25.0C? (Csolid wax 2.18 J/gC, m.p. of wax
62.0 C, Cliquid wax2.31 J/gC MM 352.7
g/mol, DHfusion70,500 J/mol)
DQ
DQtota (50g)(2.31J g-1C-1)(62C-85C)
(50g/352.7gmol-1)(-70,500J mol-1) (50g)(2.18J
g-1C-1)(25C-62C)
mCliquid waxDT
n(DQfusion)
mCsolid waxDT
DQtotal (-2656.5 J) (-9994.3 J) (-4033 J)
DQtotal -16,683.8 J
DQtotal DQliquid wax DQsolidification
DQsolid wax
DQtotal mCliquid waxDT n(DQfusion)
mCsolid waxDT
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Heat Transfer Problem 4
Steam at 175C that occupies a volume of 32.75
dm3 and a pressure of 2.60 atm. How much energy
would it need to lose to end as liquid water at
20 oC?
Solution n PV/RT (2.60
atm)(32.75 dm3)
(0.0821 dm3 atm mol-1 K-1)(448 K-1)
2.315 mol DQ (2.315 mol) (37.47 J
mol-1K-1)(175oC-100oC) (2.315
mol)(40600 J mol-1) (2.315
mol)(75.327 J mol-1K-1)(100oC-20oC) DQ
6505.7J 93989 J 13950.6 J 114445.3 J
114.445 kJ
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Chemical Reactions

• In a chemical reaction
• Chemical bonds are broken
• Atoms are rearranged
• New chemical bonds are formed
• These processes always involve energy changes

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Energy Changes

• Breaking chemical bonds requires energy
• Forming new chemical bonds releases energy

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Exothermic and Endothermic Processes

• Exothermic processes release energy
• C3H8 (g) 5 O2 (g) ? 3 CO2 (g) 4H2O
  (g)
• 2043 kJ
• Endothermic processes absorb energy
• C(s) H2O (g) 113 kJ ? CO(g) H2 (g)

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Energy Changes in endothermic and exothermic
processes

• In an endothermic reaction there is more energy
  required to break bonds than is released when
  bonds are formed.
• The opposite is true in an exothermic reaction.

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Enthalpy Calculations
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Enthalpy

• Enthalpy is the heat absorbed or released during
  a chemical reaction where the only work done is
  the expansion of a gas at constant pressure

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Enthalpy

• Not all energy changes that occur as a result of
  chemical reactions are expressed as heat
• Energy Heat Work
• Work is a force applied over a distance.
• Most energy changes resulting from chemical
  reactions are expressed in a special term known
  as enthalpy

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Enthalpy

• It is nearly impossible to set up a chemical
  reaction where there is no work performed.
• The conditions for a chemical reaction are often
  set up so that work in minimized.
• Enthalpy and heat are nearly equal under these
  conditions.

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Enthalpy Changes

• The change in enthalpy is designated by the
  symbol DH.
• If DH lt 0 the process is exothermic.
• If DH gt 0 the process is endothermic.
• Sometimes the symbol for enthalpy (DH) is used
  for heat (DQ)
• In many cases where work is minimal heat is a
  close approximation for enthalpy.
• One must always remember that while they are
  closely related, heat and enthalpy are NOT
  identical

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Energy and Enthalpy Changes

• It is impractical to measure absolute amounts of
  energy or enthalpy.
• Hence we measure changes in enthalpy rather than
  total enthalpy
• Enthalpy is always measured relative to previous
  conditions.
• Enthalpy is measured relative to the system.

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Measuring Enthalpy

• The amount of heat absorbed or released during a
  chemical reaction depends on the conditions under
  which the reaction is carried out including
• the temperature
• the pressure
• the physical state of the reactants and products

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Standard Conditions

• For most thermodynamic measurements standard
  conditions are established as
• 25 oC or 298 K
• 1.0 atmosphere of pressure
• Note this is a change from the gas laws where the
  standard temperature was 0oC

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Standard State

• The pure form of a substance at standard
  conditions (25oC and 1 atmosphere) is said to be
  in the standard state.
• The most stable form of an element at standard
  conditions represents the standard state for that
  element.

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Bond Enthalpies
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Bond Enthalpies

• One approach to determining an enthalpy change
  for a chemical reaction is to compute the
  difference in bond enthalpies between reactants
  and products
• The energy to required to break a covalent bond
  in the gaseous phase is called a bond enthalpy.
• Bond enthalpy tables give the average energy to
  break a chemical bond. Actually there are slight
  variations depending on the environment in which
  the chemical bond is located

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Bond Enthalpy Table

• The average bond enthalpies for several types of
  chemical bonds are shown in the table below

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Bond Enthalpies

• Bond enthalpies can be used to calculate the
  enthalpy change for a chemical reaction.
• Energy is required to break chemical bonds.
  Therefore when a chemical bond is broken its
  enthalpy change carries a positive sign.
• Energy is released when chemical bonds form.
  When a chemical bond is formed its enthalpy
  change is expressed as a negative value
• By combining the enthalpy required and the
  enthalpy released for the breaking and forming
  chemical bonds, one can calculate the enthalpy
  change for a chemical reaction

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Bond Enthalpy Calculations

• Example 1 Calculate the enthalpy change for the
  reaction N2 3 H2 ? 2 NH3

• Bonds broken
• NN 945
• H-H 3(435) 1305
• Total 2250 kJ
• Bonds formed
• 2x3 6 N-H 6 (390) - 2340 kJ
• Net enthalpy change
• 2250 - 2340 - 90 kJ

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Hess Law and Enthalpy Calculations
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Standard Enthalpy of Formation
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• Why is the formation of a gas less exothermic
  than formation of a liquid?

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Enthalpies of Formation

• Some enthalpies of formation for common compounds

BaCO3 -1219 H2O (g) -242 HCl (g) -93
Ba(OH)2 -998 H2O (l) -286 HCl (aq) -167
BaO -554 H2O2 -188 NH3 (g) -46
CaCO3 -1207 C3H8 -104 NO 90
CaO -636 C4H10 -126 NO2 33.8
Ca(OH)2 -987 CO -110 SO2 -297
CaCl2 -796 CO2 -394 Al2O3(s) -1670
See your text Brown, LeMay and Bursten,
Chemistry the Central Science, 7th edition pages
984-987 for addition values
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Calculating Enthalpy from tables

• Enthalpies of formation represent the enthalpy
  changes when compound forms from its elements
• The enthalpy of formation for a chemical reaction
  can be expressed as the difference between the
  enthalpy state of the products and that of the
  reactants
• DHreaction S DHoproducts SDHoreactants

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Sample Problem 1

• Calcium carbonate reacts with hydrochloric acid
  according to the following equation
• CaCO3 (s) 2HCl (aq) ? CaCl2 (aq) H2O (l)
  CO2 (g)
• Calculate the enthalpy change for this reaction
• DHoreaction S DHoproducts SDHoreactants

• Solution
• DHoproducts (-796)(-286)(-394)
• -1476 kJ
• DHoreactants (-1207)(2)(-167)
• -1541 kJ
• DHoreaction -1476-(-1541) 75 kJ

DHoCaCO3 -1207
DHo HCl (aq) -167
DHoCaCl2 -796
DHo H2O (l) -286
DHo CO2 (g) -394
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Sample Problem 2

• Calculate the enthalpy change for the burning of
  11 grams of propane
• C3H8 (g) 5 O2 (g) ? 3 CO2 (g) 4 H2O (g)
• DHoreaction S DHoproducts SDHoreactants

• Solution
• DHoproducts (3)(-394)(4)(-242)
• -2150 kJ
• DHoreactants (-104)(5)(0)
• -104 kJ
• DHoreaction -2150-(-104) -2046 kJmol-1

DHo C3H8 -104
DHo O2 (g) 0
DHo H2O (g) -242
DHo CO2 (g) -394
Now 11 grams 0.25 mole of propane (11 g/44 g
mol-1) (0.25 mol )(-2046 kJ mol-1) - 511.5 kJ
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Some things to Remember

• The enthalpy of formation table is stated in kJ
  mol-1.
• To find the sum of enthalpies of formation for
  reactants or products, multiply the number of
  moles of each substance by the enthalpy of
  formation for that substance.
• Then find the difference Products-Reactants

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Standard Enthalpy Change of Combustion
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Hess Law Indirect Enthalpy Calculations by
Rearranging Reactions

• Hess Law provides a way to calculate enthalpy
  changes even when the reaction cannot be
  performed directly.
• If a series of reactions are added together, the
  enthalpy change for the net reaction will be the
  sum of the enthalpy change for the individual
  steps

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Techniques

• Equations may be multiplied, divided, or reversed
  and then added together to form a new equation
• If an equation is multiplied or divided the
  enthalpy of the reaction is multiplied or
  divided by the same factor
• If the direction an equation is reversed the sign
  of the enthalpy is the opposite as well
• When adding equations together the enthalpies are
  added together as well

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Hess Law Example 1

• N2 (g) O2 (g) ? 2 NO (g) DH1 181
  kJ
• 2 NO (g) O2 (g) ? 2 NO2 (g) DH2 -113
  kJ
• Find the enthalpy change for
• N2 (g) 2 O2 (g) ? 2 NO2 (g)

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Hess Law Example 1

• The required equation is really the sum of the
  two given equations
• Solution
• N2 (g) O2 (g) ? 2 NO (g) DH1 181
  kJ
• 2 NO (g) O2 (g) ? 2 NO2 (g) DH2 -113 kJ
• -------------------------------------------------
  ------------
• N2 (g) 2O2 (g) 2 NO (g) ? 2 NO (g) 2 NO2 (g)
• N2 (g) 2O2 (g) ? 2 NO2 (g)
• DH DH1 DH2 181 kJ (-113) 68 kJ

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Hess Law Example 2

• From the following reactions and enthalpy
  changes
• 2 SO2 (g) O2 (g) ? 2 SO3 (g)
  DH -196 kJ
• 2 S (s) 3 O2 (g) ? 2 SO3 (g)
  DH -790 kJ
• Find the enthalpy change for the following
  reaction
• S (s) O2 (g) ? SO2 (g)
• Solution
• 2 SO3 (g) ? 2 SO2 (g) O2 (g) DH
  196 kJ
• 2 S (s) 3 O2 (g) ? 2 SO3 (g)
  DH -790 kJ
• --------------------------------------------------
  --------------------------------------------------
  ----------
• Reversing the order of the first equation
  reverses the sign of DH

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Hess Law Example 2

• From the following reactions and enthalpy
  changes
• 2 SO2 (g) O2 (g) ? 2 SO3 (g)
  DH -196 kJ
• 2 S (s) 3 O2 (g) ? 2 SO3 (g)
  DH -790 kJ
• Find the enthalpy change for the following
  reaction
• S (s) O2 (g) ? SO2 (g)
• 2 SO3 (g) ? 2 SO2 (g) O2 (g) DH
  196 kJ
• 2 S (s) 3 O2 (g) ? 2 SO3 (g)
  DH -790 kJ
• --------------------------------------------------
  --------------------------------------------------
  ----------
• 2 SO3(g) 2 S(s) 2 3 O2 (g) ? 2 SO3 (g)2
  SO2 (g) O2 (g)
  DH -594 kJ
• 2 S(s) 2 O2 (g) ? 2 SO2 (g) DH
  -594 kJ

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Hess Law Example 2

• From the following reactions and enthalpy
  changes
• 2 SO2 (g) O2 (g) ? 2 SO3 (g)
  DH -196 kJ
• 2 S (s) 3 O2 (g) ? 2 SO3 (g)
  DH -790 kJ
• Find the enthalpy change for the following
  reaction
• S (s) O2 (g) ? SO2 (g)
• 2 SO3 (g) ? 2 SO2 (g) O2 (g)
  DH 196 kJ
• 2 S (s) 3 O2 (g) ? 2 SO3 (g)
  DH -790 kJ
• --------------------------------------------------
  --------------------------------------------------
  ----------
• 2 SO3(g) 2 S(s) 2 3 O2 (g) ? 2 SO3 (g)2
  SO2 (g) O2 (g)
  DH -594 kJ
• 2 S(s) 2 O2 (g) ? 2 SO2 (g) DH
  -594 kJ
• S(s) O2 (g) ? SO2 (g) DH
  -297 kJ

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Born Haber Cycle
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Born-Haber Cycle

• When an ionic compound is formed from its
  elements, the overall reaction can be broken down
  into separate steps.
• Most of the steps are endothermic but the overall
  is exothermic due to the high lattice enthalpy.

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Born Haber Cycle for formation of NaCl
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Some Definitions

• The enthalpy of atomization is the enthalpy
  change that occurs when one mole of gaseous atoms
  is formed from the element in the standard state
  under standard conditions
• Example ½ Cl2 (g) ? Cl (g) DHoat
  121 kJ mol-1
• The electron affinity is the enthalpy change that
  occurs when an electron is added to an isolated
  atom in the gaseous state
• O (g) e- ? O- (g) DHo -142 kJ
  mol-1
• O- (g) e- ? O2- (g) DHo 844 kJ
  mol-1
• The lattice enthalpy is the enthalpy change that
  occurs from the conversion of an ionic compound
  in the gaseous state into its gaseous ions
• LiCl (g) ? Li (g) Cl- (g)
  DHo 846 kJ mol-1

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Magnesium chloride

• http//chemistry.tutorvista.com/inorganic-chemistr
  y/reaction-pathways-born-haber-cycle.html

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Lattice Energy

• Ionic compounds are usually solids. The release
  of energy on forming the solid, called the
  lattice energy is the driving force for the
  formation of ionic compounds.
• Because of high lattice energies, ionic solids
  tend to be hard and have high melting points.
  Ionic compounds are insulators in the solid
  state, because electrons are localized on the
  ions, but conduct when molten or in solution, due
  to flow of ions (not electrons).
• Lattice energies can be calculated using Hesss
  law, via a Born-Haber Cycle.

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• Step 1 Convert elements to atoms in the gas
  stateatomization and bond energy
• e.g. for Li, Li (s) ? Li (g) DH1
  DHatomization
• for F, 1/2 F2 (g) ? F (g) DH2 1/2 (Bond
  Energy)
• Step 2 Electron transfer to form (isolated)
  ions Ionization Energy and Electron Affinity
• Li (g) ? Li (g) e DH3 IE1
• F (g) e ? F (g) DH4 EA1
• Step 3 Ions come together to form solid lattice
  enthalpy
• Li (g) F (g) ? LiF (s) DH5 Lattice
  Energy
• Overall Li (s) 1/2 F2 (g) ? LiF (s) DH
  DHf S(DH15)
• Lattice Energy DHf (DH1 DH2 DH3 DH4)

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Born Haber Cycle Diagram Finding the Enthalpy
of Formation of NaCl

• The stepwise energy changes for the formation of
  NaCl

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Born Haber Cycle for NaCl

• The formation of NaCl can be considered as a
  five step process
• Na (s) 1/2 Cl2 (g) ?NaCl (s)
• The vaporization of sodium metal to form the
  gaseous element.
• The dissociation of chlorine gas to gaseous
  chlorine atoms is equal to one half of the bond
  energy for a Cl-Cl covalent bond
• The ionization of gaseous sodium atoms to Na(g)
  ?Na
• The ionization of chlorine atoms. (This quantity
  is the negative electron affinity for the element
  chlorine.)
• The lattice energy on the formation of sodium
  chloride from the gaseous ions

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Born-Haber Cycle for NaCl

• The stepwise energy changes for the formation of
  NaCl
• The vaporization of sodium metal to form the
  gaseous element.
• Na (s) ? Na (g)
  ?Hsublimation 109 kJ mol-1
• The dissociation of chlorine gas to gaseous
  chlorine atoms is equal to one half of the bond
  energy for a Cl-Cl covalent bond
• 1/2 Cl2 (g) ? Cl (g)
  ?Hdiss 122 kJ mol-1
• The ionization of gaseous sodium atoms to
• Na (g) ? Na (g) e-
  ?Hionization 496 kJ mol-1
• The ionization of chlorine atoms. (This quantity
  is the negative electron affinity for the element
  chlorine.)
• Cl (g) e- ? Cl- (g)
  ?Helect.affinity - 368 kJ mol-1
• The lattice energy on the formation of sodium
  chloride from the gaseous ions
• Na (g) Cl- (g) ? NaCl (s)
  ?Hlattice - 770 kJ mol-1

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• The lattice enthalpy is a measure of the strength
  of the attraction between ions.Depends on
• The charge of the ions
• The size of the ions
• The lattice enthalpy of magnesium oxide is much
  larger than sodium chloride because the ion is
  smaller

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Periodic Trends in Lattice Energy
Coulombs Law
charge A X charge B
electrostatic force a
distance2
http//hyperphysics.phy-astr.gsu.edu/hbase/electri
c/elefor.html

(since energy force X distance)

• So, lattice energy increases, as ionic radius
  decreases (distance between charges is smaller).
• Lattice energy also increases as charge increases.

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Figure 9.7
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Entropy
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Entropy

• Entropy is defined as a state of disorder or
  randomness.
• In general the universe tends to move toward
  release of energy and greater entropy.

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Entropy

• The statistical interpretation of
  thermodynamics was pioneered by James Clerk
  Maxwell (18311879) and brought to fruition by
  the Austrian physicist Ludwig Boltzmann
  (18441906).

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Entropy

• Spontaneous chemical processes often result in a
  final state is more Disordered or Random than the
  original.
• The Spontaneity of a chemical process is related
  to a change in randomness.
• Entropy is a thermodynamic property related to
  the degree of randomness or disorder in a
  system.

Reaction of potassium metal with water. The
products are more randomly distributed than the
reactants
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Entropy and Thermodynamics

• According to the second law or thermodynamics the
  entropy of the universe is always increasing.
• This is true because there are many more
  possibilities for disorder than for order.

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Entropy is Disorder

• Disorder in a system can take many forms.
  Each of the following represent an increase in
  disorder and therefore in entropy
• Mixing different types of particles. i.e.
  dissolving salt in water.
• A change is state where the distance between
  particles increases. Evaporation of water.
• Increased movement of particles. Increase in
  temperature.
• Increasing numbers of particles. Ex.
• 2 KClO3 ? 2 KCl 3O2

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Entropy States

• The greatest increase in entropy is usually
  found when there is an increase of particles in
  the gaseous state.
• The symbol for the change in disorder or entropy
  is given by the symbol, DS.
• The more disordered a system becomes the more
  positive the value for DS will be.
• Systems that become more ordered have negative DS
  values.

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Entropy, S

• The entropy of a substance depends on its state
  
• S (gases) gt S (liquids) gt S (solids)

So (J/K-1mol-1) H2O (liquid) 69.95 H2O
(gas) 188.8
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Entropy and States of Matter
S(Br2 liquid) lt S(Br2 gas)
S(H2O solid) lt S(H2O liquid)
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Entropy, Phase Temperature
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Entropy and Temperature

• The Entropy of a substance increases with
  temperature.

Molecular motions of heptane at different temps.
Molecular motions of heptane, C7H16
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Standard Entropy Values

• The standard entropy, DSo, of a substance is the
  entropy change per mole that occurs when heating
  a substance from 0 K to the standard temperature
  of 298 K.
• Unlike enthalpy, absolute entropy changes can be
  measured.
• Like enthalpy, entropy is a state function. The
  change in entropy is the difference between the
  products and the reactants
• DSo S So (products) - S So (reactants)

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Standard Entropy Values
Some standard enthalpy values

• The amount of entropy in a pure substance depends
  on the temperature, pressure, and the number of
  molecules in the substance.
• Values for the entropy of many substances at
  have been measured and tabulated.
• The standard entropy is also measured at 298 K.
  

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Factors That Determine Entropy States

• The greater the disorder or randomness in a
  system the larger the entropy.
• Some generalizations
• The entropy of a substance always increases as it
  changes from solid to liquid to gas and vice
  versa.
• When a pure solid or liquid dissolves in a
  solvent, the entropy of the substance increases.
• When gas molecules escape from a solvent, the
  entropy increases.
• Entropy generally decreases with increasing
  molecular complexity

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Gibbs Free Energy
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Spontaneity

• A chemical reaction is spontaneous if it results
  in the system moving form a less stable to a more
  stable state.
• Decreases in enthalpy and increases in entropy
  move a system to greater stability.
• The combination of the enthalpy factor and the
  entropy factor can be expressed as the Gibbs Free
  Energy.

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Gibbs Free Energy

• The standard free energy change is defined by
  this equation
• DGo DHo T DSo
• Where
• DHo the enthalpy change
• DSo the entropy change
• T Kelvin temperature
• A chemical reaction is
  spontaneous if it results
  in a negative free energy change.

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Gibbs Free Energy

• Possible Combinations for free energy change
• DGo DHo T DSo

DG DH DS DH-TDS
Always Spontaneous lt 0 (-) lt 0 (-) gt 0 () Always (-)
Never Spontaneous gt 0 () gt 0 () lt 0 (-) Always ()
Spontaneous at High Temperature lt 0 (-) gt 0 () gt 0 () gt 0 () (-) if T large () if T small
Spontaneous at Low Temperature gt 0 () lt 0 (-) lt 0 (-) lt 0 (-) () if T large (-) if T small
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Free Energy Problem 1

• A certain chemical reaction is exothermic with a
  standard enthalpy of - 400 kJ mol-1.
• The entropy change for this reaction is 44 J
  mol-1 K-1. Calculate the free energy change for
  this reaction at 25 oC.
• Is the reaction spontaneous?

87
88
Free Energy Problem 1

• A certain chemical reaction is exothermic with a
  standard enthalpy of - 400 kJ mol-1. The
  entropy change for this reaction is 44 J mol-1
  K-1. Calculate the free energy change for this
  reaction at 25 oC. Is the reaction spontaneous?
• Solution
• Convert the entropy value to kJ. 44 J mol-1 K-1
  0.044 kJ mol-1 K-1
•  DG - 400 kJ mol-1 (298 K)(0.044 kJ
  mol-1 K-1)
•  DG - 400 kJ mol-1 13.1 kJ mol-1
• DG - 413.1 kJ mol-1 . Since DG is
  negative the
• 
  reaction is spontaneous.
• Note. Because DH lt0 and DS gt0, this
  reaction is spontaneous at all temperatures.

88
89
Free Energy Problem 2

• A certain chemical reaction is endothermic with a
  standard enthalpy of 300 kJ mol-1. The entropy
  change for this reaction is 25 J mol-1 K-1.
  Calculate the free energy change for this
  reaction at 25 oC. Is the reaction spontaneous?

89
90
Free Energy Problem 2

• A certain chemical reaction is endothermic with a
  standard enthalpy of 300 kJ mol-1. The entropy
  change for this reaction is 25 J mol-1 K-1.
  Calculate the free energy change for this
  reaction at 25 oC. Is the reaction spontaneous?
• Solution
• Convert the entropy value to kJ. 25 J mol-1 K-1
  0.025 kJ mol-1 K-1
•  DG 300 kJ mol-1 (298 K)(0.025 kJ
  mol-1 K-1)
•  DG 300 kJ mol-1 7.45 kJ mol-1
• DG 292.55 kJ mol-1 . Since DG is
  positive the
• 
  reaction is non-spontaneous.
• Note. Because DH gt0 and DS gt0, this
  reaction is non-spontaneous at low temperatures.
  It the temperature were substantially increased
  it would become spontaneous.

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91
http//ww2.chemistry.gatech.edu/wilkinson/Class_n
otes/spring_2004_1311_page/slides/Solubility20of
20ionic20compounds20and20intermolecular20force
s20220up.pdf

• Why NaCl dissolves in water?