• Concepts Activity Fugacity Equilibrium constant Solid vapor equilibrium Ellingham diagrams Sievert’s law Control of oxygen partial pressure
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1 Activity/Fugacity
• For ideal gases and mixtures of condensed phases, the “activity” of a species is equal to its mole fraction. That means that its influence on the properties of the system is its mole fraction (X) multiplied by the total system property. For a gas mixture, the partial pressure of A is: = PXP AA • In real systems, the influence of one component on the properties of the system may be different than its mole fraction would suggest. Activity is given by the ratio of the fugacity of a substance to its fugacity in a standard state (usually pure substance at 1 atm, same T) = fA aA o fA • You may also see this expressed in terms of an activity coefficient an mole fraction, but the idea is the same. The proportional effect of the amount of substance A on the whole system property in question is not simply given by its mole fraction =γ aX AAA WS2002 2
2 Activity/Fugacity
• For real gases, the fugacity is the effective pressure corrected for non-ideality
• For solids and liquids, fugacity is equal to the vapor pressure of in equilibrium with the condensed phase since, at equilibrium, the partial molar free energy of the solid and gas must be equal • For gases, the ratio of fugacity to partial pressure must approach one (ideal behavior) as total pressure approaches zero.
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3 Activity
=− + • For any substance dGAA S dT V A dP At constant temperature = dG AA V dP • The difference between the Gibbs’ free energy of a material under any conditions can be determined by integrating that equation.
• First V is replaced by RT/P since the fucacity = RT dGA dP of any phase is related to the vapor pressure P
• Integrating gives the difference in free GA f A energy between the state of interest and ∫∫dG= RT dln f o f o the standard state GA A ∆ =−=o fA = GGA GA RTln RT ln aA Xf f o ==fA A A, pure A • For a real solution aA o o fA fA
• For an ideal solution aA = XA WS2002 4
4 Chemical Equilibrium
• A chemical reaction can be written bB +⇔ cC dD + eE • The free energy change can be δW=∆ G = dGDD + eG − bG BC − cG expressed in terms of the partial rev molar free energy for each component =+o • For each component, the partial molar G B GB RTln( aB ) free energy is given in terms of activity
• Substituting for each term
aad e ∆==∆+GGRT0lno D E ∆ b c • In the case of G = O (equilibrium) aaB C the ratio of acitivities is the equilibrium d e ∆=−o aaD E =− WS2002constant Ka GRTln RT ln K aab c 5 a B C
5 Gaseous Equilibrium Example
1 OO⇔ • Consider the equilibrium 2 2 • If heat is added, equilibrium will shift the reaction in the direction that absorbs heat (this reaction is endothermic) • Equilibrium constant
• Free energy of reaction ∆=∆−∆GGo o 1 Go fO,,2 fO2 at 1000 K ∆=−o GJ1000 187, 000 • Free energy is related to equilibrium constant by • If the total pressure is 1 atm, the equilibrium composition at 1000 K is -10 XO = 1.5 x 10 atm XO2 = ~1 atm
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6 Solid-Vapor Equilibrium
⇔ • Consider the reaction 4 Cu(s) + O2(g) 2 Cu2O(s)
• We can write and expression of Ka • The activity of pure, condensed phases is unite The activity of an ideal gas is its partial pressure
• ∆G° vs. T is known • The partial pressure of oxygen for the equilibrium can be calculated at any temperature. At 1000 K
• Below the equilibrium PO2, the metal is stable. Above, the oxide is stable
• Plot at all temperatures
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7 Variation in ∆G° with T
∆=∆−∆o o o • For a chemical reaction at any temperature T GHTST T T T • For each component of the reaction oo=+ HHT 298 ∫ CdTp 298 T C SSoo=+∫ P dT T 298 T 298 • For the reaction T ∆ ∆=∆+∆oo where CP is the difference in heat HHT 298 ∫ CdTp capacity between products and reactants 298 T ∆C ∆=∆+SSoo∫ P dT T 298 T 298
∆oo =∆ +∆ − −∆o −∆ − • So, combining GHT 298 CTPT(298 ) TSTCTp (ln ln 298 ) ∆=G A + BT + CTln T ∆≈∆−∆GHTSoo o T WS2002 8
8 Ellingham Diagrams
• Specific for reactions of the form
(2x)M + O2 = 2MxO • Plot previous page on ∆Go vs. T axes, slope =-∆S°, intercept = ∆ H°
∆=−GRTKo ln • Lines of equal pO2 can be plotted since: a 2 2 ()aMO ()1 1 = x = = Ka 2x 2x ()()() ()()PO aaM O 1 PO 2 22 1 ∆=−GRTo ln = RT ln() P • So: ()P O2 O2
• Known for many metals Below line, metal stable Above line
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9 Sievert’s Law
• When a diatomic gas dissolves in a metal, it dissociates ⇒ H2(g) 2H (dissolved in copper) a2 • We know that K = H a a H2 • At low concentrations of H if it behaves ideally, the activity of monatomic hydrogen in the metal is equal to its concentration 2 []H K = a P H2 2 []HKP= aH2 []HKP= 12// 12 aH2
• Sievert’s law states that the amount of a diatomic gas dissolved in metal is proportional to the square root of its pressure WS2002 10
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