11.21.2011 - Chem 260/Lecture 33 F11 Today in Chemistry 260...

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Unformatted text preview: Chem 260/Lecture 33 F11 Today in Chemistry 260 Chemistry 260 •  •  •  •  •  Lecture 33 ΔG ° = − RT ln K p Evaluating the equilibrium composition r Predicting changes in equilibrium Le Châtelier s Principle Effects of concentration, pressure, and catalysts Temperature dependence of K - the van t Hoff equation Later in Chemistry 260 November 23, 2011 Lake Michigan Chicago, IL Mt. McKinley Denali Reading: Mon, 15.1-15.4, 15.7-15.9 Wed, 11.3, 15.5-15.6, 16.1-16.4 Proton transfer equilibria Equilibria of aqueous salts Electrochemistry c Evaluating the equilibrium composition Predicting changes in equilibrium Le Châtelier s Principle Effects of concentration, pressure, and catalysts Temperature dependence of K - the van t Hoff equation d eq ⎛ PCeq ⎞ ⎛ PD ⎞ ⎜ Po ⎟ ⎜ Po ⎟ ⎠ = exp ⎡ −ΔGr ° ⎤ ⎝ ⎠⎝ KP = a b ⎢ RT ⎥ eq ⎣ ⎦ ⎛ PA ⎞ ⎛ PBeq ⎞ o⎟ ⎜ o⎟ ⎜ P⎠⎝ P⎠ ⎝ ΔGr ° = − RT ln K p Le Châtelier s Principle Today in Chemistry 260 When a system at equilibrium is disturbed, the composition adjusts so as to minimize the effect of the disturbance. Example NH4Cl ( s ) É NH3 ( g ) + HCl ( g ) ; K = PNH3 × PHCl ΔGr ° = − RT ln K p c d eq ⎛ PCeq ⎞ ⎛ PD ⎞ o⎟ ⎜ ⎜ P⎠⎝ Po ⎟ ⎠ = exp ⎡ −ΔGr ° ⎤ ⎝ KP = a b ⎢ RT ⎥ eq ⎣ ⎦ ⎛ PA ⎞ ⎛ PBeq ⎞ ⎜ Po ⎟ ⎜ Po ⎟ ⎝ ⎠⎝ ⎠   effect of adding NH3(g)?   effect of adding NH4Cl(s)?   effect of compression? Le Châtelier s Principle is only a rule of thumb can see why it works/ where it comes from by looking at the THERMODYNAMICS Movement towards equilibrium is a spontaneous process • Some changes in conditions affect the value of ΔG , K (temperature). • Others changes that do not affect the numerical value of K (pressure). The Effect of Concentration Changes Q vs. K NH4Cl ( s ) É NH3 ( g ) + HCl ( g ) The Effect of Concentration Changes Q = ( PHCl )( PNH 3 ) ( K = ( PHCl )eq PNH3 ) eq Not at equilibrium: ΔGr = ΔGr ° + RT ln Q Tells us if forward reaction is spontaneous from the current conditions of the system At equilibrium: Slope of line (change in G) with change of n (moles) ΔGr = 0 ΔGr ° = − RT ln K p As the reaction approaches equilibrium, Q  K Gibbs Energy •  •  •  •  •  •  •  •  •  •  Pure reactant ΔGr = 0 Pure product 1 Chem 260/Lecture 33 F11 The Effect of Concentration Changes Q = ( PHCl )( PNH 3 ) ( K = ( PHCl )eq PNH3 ) eq ΔGr = ΔGr ° + RT ln Q ⎛Q⎞ ΔGr = RT ( ln Q − ln K p ) = RT ln ⎜ ⎜ Kp ⎟ ⎟ ⎝ ⎠ Q as Q → K p , → 1, Kp ⎛Q ln ⎜ ⎜ Kp ⎝ ⎞ ⎟ → 0, ΔGr → 0 ⎟ ⎠ Gibbs Energy ΔGr = − RT ln K p + RT ln Q ΔGr = ? Pure reactant ..is the differential/variance of Gibbs free energy with the change of the number of moles Not at equilibrium: ⎛Q⎞ ΔGr = RT ln ⎜ ⎜ Kp ⎟ ⎟ ⎝ ⎠ Not at equilibrium: ΔGr ΔGr = ΔGr ° + RT ln Q Gibbs Energy Q vs. K NH4Cl ( s ) É NH3 ( g ) + HCl ( g ) The Effect of Concentration Changes As the reaction approaches equilibrium, Q  K ΔGr = ? ΔGr = 0 Q<K Pure reactant ΔGr = 0 ΔGr < 0 Pure product Effect of Concentration Changes H2 ( g ) + I2 ( g ) É 2HI ( g ) Pure product ΔGr > 0 Effect of Concentration Changes H2 ( g ) + I2 ( g ) É 2HI ( g ) Q>K K=Q Q= ( PHI )2 ( PH2 )( PI2 ) Q= Pure reactant ΔGr = ΔGr ° + RT ln Q Gibbs Energy Gibbs Energy ΔGr = ΔGr ° + RT ln Q ( PHI )2 ( PH2 )( PI2 ) Pure product * If add reactants, drive the reaction to the right. * If remove products, drive the reaction to the right. * If add products, drive the reaction to the left. * If remove reactants, drive the reaction to the left. As the reaction approaches equilibrium, Q  K Pure reactant Pure product The Effect of Pressure According to Le Châtelier s Principle, When a system at equilibrium is compressed, the composition of a gas-phase equilibrium adjusts so as to reduce the number of molecules in the gas phase. N2 ( g ) + 3H2 ( g ) É 2NH3 ( g ) The Effect of Pressure ΔG is the standard Gibbs energy of reaction and is therefore defined for the standard pressure Po = 1 atm !G ! = " RT ln K Hence, K is independent of the pressure at which the reaction is carried out. If the NH3 synthesis reaction is compressed isothermally the equilibrium constant will remain unchanged. 2 Chem 260/Lecture 33 F11 The Effect of Pressure eq However, pressure CAN have a large effect on composition ⎛ PNH N2 ( g ) + 3H2 ( g ) É 2NH3 ( g ) Kp = 1 Imagine that the volume is doubled Px = Pxeq 2 2 1 ⎛ ⎞ NH 3 3 N2 H2 ( eq = ) eq ⎜ PNH3 ⎟ PNH3 2 ⋅ 23 ⎝2 ⎠ = 2⋅ 3 eq eq 2 ⎛ 1 eq ⎞⎛ 1 eq ⎞ PN2 PH 2 ⎜ PN2 ⎟⎜ PH 2 ⎟ ⎝2 ⎠⎝ 2 ⎠ ⎞ Po ⎟ ⎠ 2 eq eq ⎛ PN2 ⎞⎛ PH 2 ⎞ ⎜ P o ⎟⎜ Po ⎟ ⎝ ⎠⎝ ⎠ 3 2 The Presence of a Catalyst = 4K p ( )( ) ⎛Q⎞ ⎛ 4K p ΔGr = RT ln ⎜ = RT ln ⎜ ⎜ Kp ⎟ ⎟ ⎜ Kp ⎝ ⎠ ⎝ = RT ln ( 4 ) = 1.387 RT 3 3 Gibbs Energy (P ) (P )(P ) 2 Q= ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Q<K Q>K K=Q Q will decrease until Q = K \ NH3(g) will be converted to N2(g) + H2(g) ΔGr < 0 ΔGr > 0 The Presence of a Catalyst A catalyst is a substance that accelerates a reaction without itself appearing in the chemical reaction. A catalyst speeds up a reaction, by lowering the energy barrier. (foreshadowing notice: kinetics) N2 ( g ) + 3H2 ( g ) The Effect of Temperature Catalysts alter reaction pathways. However, ΔG of the reaction does not change, since ΔG is a state function Since ΔG is unaltered by the catalyst, K is also unaltered 2NH3 ( g ) !G = " RT ln K ! The presence of a catalyst does not change the equilibrium constant of a reaction. The Effect of Temperature The van t Hoff equation According to Le Châtelier s Principle, when temperature is increased The equilibrium composition of an endothermic reaction will shift toward products. If endothermic, ΔH > 0, reactants + heat If exothermic, ΔH < 0, reactants ΔGr ° ΔH r ° − T ΔSr ° −ΔH r ° ΔSr ° =− = + RT RT RT R At a certain temperature, T products products + heat Why do reactions at equilibrium respond this way? The origin of this effect is the dependence of ΔG (and therefore K) on T. ΔGr = ΔGr ° + RT ln Q Now let s consider the effect of temperature on K itself. ΔGr ° = ΔH r ° − T ΔSr ° The equilibrium composition of an exothermic reaction will shift toward reactants. ⎡ −ΔGro ⎤ K P = exp ⎢ ⎥ ⎣ RT ⎦ ln K = − At another temperature, T ln K ′ = − ΔGr ° −ΔH r ° ΔSr ° = + RT ′ RT ′ R The difference between the two is the van t Hoff equation: ⎛ −ΔH r ° ΔSr ° ⎞ ⎛ −ΔH r ° ΔSr ° ⎞ −ΔH r ° ΔH r ° ln K − ln K ′ = ⎜ + + + ⎟−⎜ ⎟= R ⎠ ⎝ RT ′ R⎠ RT RT ′ ⎝ RT ⎛ K ⎞ −ΔH r ° ⎛ 1 1 ⎞ ln ⎜ ⎟ = ⎜−⎟ R ⎝ T T′ ⎠ ⎝ K′ ⎠ 3 Chem 260/Lecture 33 F11 The van t Hoff equation ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞ K ⇒ = exp ⎜ ⎜ − ⎟⎟ K′ ⎝ R ⎝ T T′ ⎠⎠ ⎛ K ⎞ −ΔH r ° ⎛ 1 1 ⎞ ln ⎜ ⎟ = ⎜−⎟ R ⎝ T T′ ⎠ ⎝ K′ ⎠ The van t Hoff equation: (ln (k) Temp. dependence) ⎛ K ⎞ −ΔH r ° ⎛ 1 1 ⎞ ln ⎜ ⎟ = ⎜−⎟ R ⎝ T T′ ⎠ ⎝ K′ ⎠ ⎛1 1 ⎞ if T ′ > T , ⎜ − ⎟ > 0 ⎝ T T′ ⎠ ln K = − ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞ K = exp ⎜ ⎜ − ⎟⎟ K′ ⎝ R ⎝ T T′ ⎠⎠ When ΔH > 0 (endothermic) K is larger more product favored - + K′ > K ΔGr ° RT ΔGr ° = ΔH r ° − T ΔSr ° −ΔH r ° ΔSr ° −ΔH r ° ⎛ 1 ⎞ ΔSr ° ln K = + = ⎜ ⎟+ RT R R ⎝T ⎠ R <1 ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞ K = exp ⎜ ⎜ − ⎟⎟ K′ ⎝ R ⎝ T T′ ⎠⎠ K is smaller more reactant favored + + intercept = ln(K) When ΔH < 0 (exothermic) K′ < K () () ; K = PX o () ⎛P ln ⎜ X ⎝ Po Clausius-Clapeyron Equation Use to determine vapor pressure at some temperature o ⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞ ln ⎜ X ⎟ = ⎜−⎟ R T Tb ⎠ ⎝ 1atm ⎠ ln ( PX ) = −ΔH R ⎛1 1 ⎞ ⎜−⎟ ⎝ T Tb ⎠ Mountain Top Cooking ⎛ P ⎞ −ΔH ln ⎜ X ⎟ = R ⎝ Po ⎠ o vap ⎛1 1 ⎞ ⎜−⎟ ⎝ T Tb ⎠ Barometric Pressure Ph = Po e − mgh () −mgh kbTa ⎞ o ⎟ = −ΔH vap ⎟ R ⎟ ⎠ = ) 3 −1 1⎞ ⎛ P ⎞ − 40.66 × 10 J mol ⎛ 1 ln ⎜ ⎜− ⎟ ⎟= 8.314 J K −1 mol −1 ⎝ Th 373.15 ⎠ ⎝ 1atm ⎠ ⎛1 1⎞ ⎛P ⎞ ln ⎜ X ⎟ = −4890 K ⎜ − ⎟ ⎝ 1atm ⎠ ⎝ Th 373.15 ⎠ Double the Pressure ⎛1 1⎞ ⎛ 2atm ⎞ ln ⎜ ⎟ ⎟ = −4890 K ⎜ − ⎝ 1atm ⎠ ⎝ Th 373.15 ⎠ 0.693 = −4890 K + 13.1 Th Th = 394.1K = 121o C () Mountain Top Cooking () H 2O l ! H 2O g () H 2O l ! H 2O g o −ΔH vap ⎛ 1 1 ⎞ −mgh = ⎜−⎟ kbTa R ⎝ Th Tb ⎠ ⎛ 1.661 × 10 −27 kg ⎞ − ⎜ 29amu * ⎟ 9.8m / s * h − ( 40.657 × 103 J mol −1 ) ⎛1 1amu 1⎞ ⎝ ⎠ = ⎜− ⎟ −23 2 −2 8.314472 J K −1 mol −1 ⎝ Th 373.15 ⎠ (1.381 × 10 m kg s K −1 )Ta Mt. McKinley Denali h = 6194 m kbT ⎛1 h 1 ⎞ 1.43 ×105 K = 1.43 ×105 K ⎜ − − 383.2 ⎟= Ta Th ⎝ Th 373.15 ⎠ Mt. McKinley Denali h = 6194 m ⎛1 1⎞ ⎜−⎟ ⎝ Th Tb ⎠ o −ΔH vap ⎛ 1 1 ⎞ ⎜−⎟ R ⎝ Th Tb ⎠ o ⎞ −ΔH vap ⎛ 1 1 ⎞ ⎟= ⎜−⎟ R ⎝ T Tb ⎠ ⎠ Use to determine boiling point at some external pressure o ⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞ ln ⎜ h ⎟ = ⎜−⎟ R ⎝ Th Tb ⎠ ⎝ Po ⎠ ⎛ − mgh kbTa Pe ln ⎜ o ⎜ Po ⎜ ⎝ () ( Using Normal Boiling Point at 1atm o vap March, 1952 Pressure Cooking H 2O l ! H 2O g ⎛ K ⎞ −ΔH ⎛ 1 1 ⎞ ln ⎜ ⎟ = ⎜−⎟ R ⎝ T T′ ⎠ ⎝ K′ ⎠ o r ⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞ ln ⎜ X ⎟ = ⎜−⎟ ′ R ⎝ T T′ ⎠ ⎝ PX ⎠ −ΔH ro R 1/T Vapor Pressure: Special case is the Clausius Clapeyron equation X l !X g ΔSro R slope = >1 Jacobus Henricus van 't Hoff (1852 - 1911) First Nobel Prize in Chemistry On top of Mt. McKinley at external temp of 0o C 6194m 1.43 ×10 K = − 383.2 273.15K Th 5 On Lake Michigan at external temp of 10o C 177m 1.43 ×105 K = − 383.2 283.15K Th Th = 352.3K Lake Michigan Chicago, IL h = 177 m Th = 372.6 K = 79.2o C = 99.4o C Lake Michigan Chicago, IL h = 177 m 4 ...
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