Useful Equations - The Clapeyron Equation

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Useful Equations - The Clapeyron Equation

• Gives the rate of change of the vapor pressure
  with temperature, dp/dT, in terms of the enthalpy
  of vaporisation, ?Hvap, volume of the liquid, Vl,
  and volume of the vapor, Vv, at temperature T and
  at a pressure equal to the vapor pressure.

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The Clapeyron Equation

• What is the change in boiling point of water at
  100oC per mm change in atmospheric pressure. The
  heat of vaporization is 539.7 cal g-1, the molar
  volume of liquid water is 18.78 mL, and the molar
  volume of steam is 30.199 liters, all at 100oC
  and 1 atm.

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The Clapeyron Equation

• Calculate the change in pressure required to
  change the freezing point of water 1oC. At 0o the
  heat of fusion of ice is 79.7 cal g-1, the
  density of water is 0.9998 g mL-1, and the
  density of ice is 0.9168. The reciprocals of the
  densities, 1.0002 and 1.0908 are the volumes in
  mL of 1 g. The volume change upon freezing (Vl
  Vs) is therfore -9.056 x 10-5 L g-1. For
  small changes of ?Hfus, T, and (Vl Vs)
  are virtually constant, so that

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The Clapeyron Equation

• A pressure of 133 atm lowers the freezing point
  of water 1oC.
• The reciprocal ?T/?V -1/133 - 0.0075 deg
  atm-1, shows that an increase in 1 atm lowers the
  freezing point 0.0075oC

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Useful Equations -The Clausius-Clapeyron Equation

• Clausius showed the Clapeyron equation could be
  simplified by assuming the vapor obeys the ideal
  gas law and by neglecting the volume of a mole of
  liquid in comparison with a mole of vapor. For
  example with water at 100oC the volume of vapor
  is 30.2 liters and the volume of liquid is 0.0188
  liter
• C is the integration constant

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The Clausius-Clapeyron Equation

• Frequently a more convenient form of the equation
  to use is
• Obtained by integrating between the limits, P2 at
  T2 and P1 at T1.

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The Clausius-Clapeyron Equation

• Using this equation, it is possible to calculate
  the heat of vaporization or the heat of
  sublimation from the vapor pressure at two
  different temperatures.
• The approximations involve the assumptions that
  the vapor is an ideal gas and the heat of
  vaporization is independent of temperature.
• Over wide temperature ranges, plots of log P
  versus 1/T are somewhat curved because ?Hvap
  varies with temperature. It is possible to
  calculate the heat of vaporization at any
  temperature from the slope of the curve by
  drawing a tangent to the curve at the required
  temperature.

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Gibbs Free Energy

• For an irreversible or spontaneous process the
  entropy must increase.
• The quantity E TS is referred to as the
  Helmholtz free energy, A. If the volume and
  temperature is constant
• The Helmholtz free energy A decrease for a
  irreversible spontaneous process.

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Gibbs Free Energy

• Physical and chemical process are usually carried
  out at constant pressure and temperature (P and
  T).
• d(E PV TS)T,P lt 0
• The quantity E PV TS is referred to as the
  Gibbs free energy and is represented by G
• G E PV TS H TS (H enthalpy)
• (dG)TP lt 0
• Thus for an irreversible process at constant T
  and P in which only pressure volume work is done,
  the Gibbs free energy decreases.

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Criteria for irreversibility and reversibility
for processes involving no work or only
pressure-volume work
These relationships may be applied to finite
changes by replacing the ds by ?s
Although the criteria show whether a change is
spontaneous it does not provide any information
on the speed (kinetics) of the process.
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Gibbs Free Energy

• At constant temperature and pressure, the Gibbs
  free energy G H TS becomes
• ?G ?H - T ?S
• Whether a process is spontaneous at constant T
  P depends on two terms, ?H and T?S.
• A change is favored if ?H is negative and ?S is
  positive. These changes correspond to a decrease
  in energy and an increase in disorder
  respectively.
• Note that at very high temperatures T ?S can
  dominate

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Criteria of Chemical Equilibrium

• Berthelot 1879 falsely concluded that reactions
  which evolve heat are spontaneous
• Spontaneous reactions can also absorb heat
• The spontaneous process leads to the minimum
  possible value of Gibbs free energy for the
  system.
• At equilibrium ?G 0
• If a system is in equilibrium it cannot undergo a
  spontaneous change under the given conditions

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Gibbs Free Energy

• If ?Go is negative, the equilibrium constant has
  a value greater than 1 and the reactants in their
  standard state will react spontaneously to give
  the products in their standard states.
• If ?Go is 0, the reactants in their standard
  state will be in equilibrium with products in
  their standard states and K 1.
• If ?Go has a positive value, the reactants in
  their standard state will not react to give
  products in their standard states ( K lt 1).
• NB It is not necessary for ?Go to be ve for a
  reaction to be useful.

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Limitations of Thermodynamics

• The Gibbs free energy can only predict if a
  chemical reaction is possible.
• It cannot predict how fast a reaction will occur.
• The Gibbs free energy for 1 mole of carbon and 1
  mole of oxygen at one atmosphere, 25oC, is
  greater than the Gibbs free energy for carbon
  dioxide under the same conditions. However the
  reaction is so slow it virtually does not occur.
• C O2 ? CO2 (possible)
• The reverse reaction of CO2 forming C and O2 will
  not occur spontaneously.
• CO2 ? C O2 (not possible)
• Note at equilibrium all species would be present.

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Chemical Equilibria

• (X) indicates the concentration of the reactants
  and products, but to be strictly correct it is
  the activity of reactants and products that
  should be used.

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Chemical Equilibria

• Since the chemical potentials of the reactants
  and products in their standard states are equal
  to their standard molar Gibbs fee energies, then
• This equation is very important because it links
  the standard Gibbs free energy for a reaction
  with the value of the equilibrium constant.

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Gibbs Free Energy

• The standard Gibbs free energy change, ?Go, for a
  reaction is the change in G when the indicated
  number of moles of reactants, each at unit
  activity, is converted into the indicated number
  of moles of products, each at unit activity.
• The standard Gibbs free energy can be calculated
  from calorimetric and thermodynamic data.
• Thus the equilibrium constant can be obtained
  without experimental information

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Gibbs Free Energy

• Methods of determining ?Go
• From the equilibrium constant ?Go -RTlnK
• ?Ho obtained calorimetrically and ?So obtained
  from third law entropies using ?Go ?Ho - T?So
• Using statistical mechanics and spectroscopic data

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Gibbs Free Energy of Formation

• The Gibbs free energy of formation of a substance
  is the Gibbs free energy change for the reaction
  in which the substance in its standard state at
  25oC is formed from its elements in their
  standard states at 25oC.
• As was the case for ?Ho, ?Go the Gibbs free
  energy for a reaction can be determined by adding
  and subtracting the reactions for which ?Go is
  known.

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Gibbs Free Energy of Formation at 25oC
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Calc. of ?G0 from

• Calculate ?Go and Kp at 25oC for
• CO(g) H2O(g) CO2(g) H2(g)
• ?Go (-94.2598 0) (-32.8079 54.6357)
• -6.8162 kcal
• ?Go -(1.987)(298.1)(2.303)log Kp

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Calculation of Gibbs Free Energy and Equilibrium
Constant from Enthalpy of Formation and Entropy
Values

• Calculate Kp for the following reaction at 25oC
• C (graphite) 2H2 (g) CH4 (g)
• ?Ho -17,889 cal
• ?So 44.5 2(31.211) 1.3069
• -19.28 cal deg-1
• ?Go ?Ho - T ?So -17,889 (298.1)(-19.28)
• -12,140 cal

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Enthalpy of Formation
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Third Law Entropies _at_ 25oC
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Calculation of Gibbs Free Energy and Equilibrium
Constant

• ?Go -RTlnK
• -12,140 -(1.987)(298.1)(2.303)logKP
• logKP 8.91
• KP 8.1 x 108
• Although the equilibrium constant for this
  reaction is large at ambient temperature it is
  not possible to carry out this reaction as no
  catalyst is known.
• At high temperature the reaction is unfavourable

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Influence of Temperature on ?G

• The values for CP for reactants and products are
  known over a wide range of temperature.
• With this information it is possible to calculate
  ?Go at any temperature in the range that CP is
  known

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Influence of Temperature on ?G

• The enthalpy of reaction at temperature T
• The Gibbs free energy change for may be
  calculated from this equation if the heat
  capacity for each reactant and product is known
  as a function of temperature from 25oC to the
  desired temperature, ie. the value of the
  constants a, b and c are known

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Influence of Temperature on ?G

• Calculate ?Go and KP for the reaction
• C(graphite) H2O(g) CO(g) H2(g)
• ?Go298K -32.8079 (54.6357)
• 21.8278 kcal
• ?Ho298K -26.4157 (-57.7979)
• 31.3822 kcal
• CP graphite 3.81 1.56 x 10-3T and the values
  of the gases are obtained from the Table

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Gibbs Free Energy of Formation at 25oC
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Enthalpy of Formation
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Influence of Temperature on ?G
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Influence of Temperature on ?G

• This general equation can now be used to
  calculate the Gibbs free energy change for the
  reaction at any temperature in the range for
  which the heat capacity equations are valid
  (300-1,500oK)
• At 1000oK, ?Go1000 -1,330
• NB that the reaction occurs at 1000oK and not at
  298oK