11.28.2011 - 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 ln ⎜ ⎜ Kp ⎝ ⎞ ⎟ → 0, ΔGr → 0 ⎟ ⎠ Gibbs Energy ΔGr = − RT ln K p + RT ln Q ⎛Q⎞ ΔGr = RT ( ln Q − ln K p ) = RT ln ⎜ ⎜ Kp ⎟ ⎟ ⎝ ⎠ Q as Q → K p , → 1, Kp Not at equilibrium: ⎛Q⎞ ΔGr = RT ln ⎜ ⎜ Kp ⎟ ⎟ ⎝ ⎠ Not at equilibrium: ..is the differential/variance of Gibbs free energy with the change of the number of moles ΔGr = ΔGr ° + RT ln Q Gibbs Energy Q vs. K NH4Cl ( s ) É NH3 ( g ) + HCl ( g ) ΔGr The Effect of Concentration Changes As the reaction approaches equilibrium, Q  K ΔGr = ? ΔGr = ? ΔGr = 0 Q<K Pure reactant ΔGr = 0 Pure reactant 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 ConcepTest * 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 ConcepTest At T=1200 C, the reaction: At T=1200 C, the reaction: P4(g) ⇆ 2 P2(g) P4(g) ⇆ 2 P2(g) has an equilibrium constant K = 0.612 has an equilibrium constant K = 0.612 If the initial partial pressure of P4 is 5 atm and that of P2 is 2 atm, which way will the reaction proceed? A)  left B)  right C)  neither Q= 2*2/5= 0.8 If the initial partial pressure of P4 is 5 atm and that of P2 is 2 atm, which way will the reaction proceed? A)  left B)  right C)  neither 2 Chem 260/Lecture 33 F11 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. 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 2 eq ⎜ PNH3 ⎟ PNH3 2 ⋅ 23 ⎝2 ⎠ = 2⋅ 3 eq eq 2 1 eq ⎞⎛ 1 eq ⎞ ⎛ PN2 PH 2 ⎜ PN2 ⎟⎜ PH 2 ⎟ ⎝2 ⎠⎝ 2 ⎠ 3 3 ⎞ Po ⎟ ⎠ 2 eq eq ⎛ PN2 ⎞⎛ PH 2 ⎞ ⎜ P o ⎟⎜ Po ⎟ ⎝ ⎠⎝ ⎠ 3 The Presence of a Catalyst = 4K p ⎛Q⎞ ⎛ 4K p ⎞ ΔGr = RT ln ⎜ = RT ln ⎜ ⎜ Kp ⎟ ⎟ ⎜ Kp ⎟ ⎟ ⎝ ⎠ ⎝ ⎠ = RT ln ( 4 ) = 1.387 RT Gibbs Energy (P ) (P )(P ) 2 Q= ⎜ ⎝ Q will decrease until Q = K \ NH3(g) will be converted to N2(g) + H2(g) Q<K Q>K K=Q 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) 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 !G = " RT ln K ! N2 ( g ) + 3H2 ( g ) The Effect of Temperature 2NH3 ( g ) The presence of a catalyst does not change the equilibrium constant of a reaction. 3 Chem 260/Lecture 33 F11 The Effect of Temperature The van t Hoff equation According to Le Châtelier s Principle, when temperature is increased ΔGr ° = ΔH r ° − T ΔSr ° The equilibrium composition of an exothermic reaction will shift toward reactants. The equilibrium composition of an endothermic reaction will shift toward products. reactants + heat If exothermic, ΔH < 0, reactants products products + heat The origin of this effect is the dependence of ΔG (and therefore K) on T. ⎡ −ΔGro ⎤ K P = exp ⎢ ⎥ ⎣ RT ⎦ The van t Hoff equation ⎛ K ⎞ −ΔH r ° ⎛ 1 1 ⎞ ln ⎜ ⎟ = ⎜−⎟ R ⎝ T T′ ⎠ ⎝ K′ ⎠ ⇒ ln K = − At another temperature, T ln K ′ = − ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞ K = exp ⎜ ⎜ − ⎟⎟ K′ ⎝ R ⎝ T T′ ⎠⎠ ⎛ −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′ ⎠ 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′ ⎠ When ΔH > 0 (endothermic) K is larger more product favored ln K = − ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞ K = exp ⎜ ⎜ − ⎟⎟ K′ ⎝ R ⎝ T T′ ⎠⎠ - + K′ > K ln K = ΔGr ° RT K is smaller more reactant favored + + intercept = ln(K) ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞ K = exp ⎜ ⎜ − ⎟⎟ K′ ⎝ R ⎝ T T′ ⎠⎠ K′ < K ln K ; K = PX ⎛ K ⎞ −ΔH ⎛ 1 1 ⎞ ln ⎜ ⎟ = ⎜−⎟ R ⎝ T T′ ⎠ ⎝ K′ ⎠ o r ⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞ ln ⎜ X ⎟ = ⎜−⎟ ′ R ⎝ T T′ ⎠ ⎝ PX ⎠ o Clausius-Clapeyron Equation Using Normal Boiling Point at 1atm o ⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞ ln ⎜ X ⎟ = ⎜−⎟ R ⎝ T Tb ⎠ ⎝ 1atm ⎠ ln ( PX ) = ConcepTest o −ΔH vap ⎛ 1 1 ⎞ ⎜−⎟ R ⎝ T Tb ⎠ Use to determine vapor pressure at some temperature Use to determine boiling point at some external pressure B A C ln K () −ΔH ro R 1/T Vapor Pressure: Special case is the Clausius Clapeyron equation () Jacobus Henricus van 't Hoff (1852 - 1911) First Nobel Prize in Chemistry ΔSro R slope = >1 X l !X g ΔGr ° = ΔH r ° − T ΔSr ° −ΔH r ° ΔSr ° −ΔH r ° ⎛ 1 ⎞ ΔSr ° + = ⎜ ⎟+ RT R R ⎝T ⎠ R <1 When ΔH < 0 (exothermic) ΔGr ° −ΔH r ° ΔSr ° = + RT ′ RT ′ R The difference between the two is the van t Hoff equation: Why do reactions at equilibrium respond this way? ΔGr = ΔGr ° + RT ln Q ΔGr ° ΔH r ° − T ΔSr ° −ΔH r ° ΔSr ° =− = + RT RT RT R At a certain temperature, T ln K If endothermic, ΔH > 0, Now let s consider the effect of temperature on K itself. 1/T 1 /T 1/T Which statement(s) is(are) true: I.  A is an endothermic reaction II.  DSr=0 for reaction B III.  Reaction C is product favored at high temperature A.  I only B.  III only C.  I and III only D.  I, II, and III E.  None of these are true 4 Chem 260/Lecture 33 B March, 1952 Pressure Cooking ln K C ln K A ln K F11 () () H 2O l ! H 2O g 1/T 1 /T 1/T Which statement(s) is(are) true: I.  A is an endothermic reaction II.  DSr=0 for reaction B III.  Reaction C is product favored at high temperature (endothermic) A.  I only B.  III only C.  I and III only D.  I, II, and III E.  None of these are true ln(K) intercept = ΔS R o ⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞ ln ⎜ X ⎟ = ⎜−⎟ R ⎝ T Tb ⎠ ⎝ Po ⎠ ( slope = Double the Pressure ) 3 −1 1⎞ ⎛ P ⎞ − 40.66 × 10 J mol ⎛ 1 ln ⎜ ⎜− ⎟ ⎟= 8.314 J K −1 mol −1 ⎝ Th 373.15 ⎠ ⎝ 1atm ⎠ o r −ΔH R o r ⎛1 1⎞ ⎛ 2atm ⎞ ln ⎜ ⎟ ⎟ = −4890 K ⎜ − ⎝ 1atm ⎠ ⎝ Th 373.15 ⎠ ⎛1 1⎞ ⎛P ⎞ ln ⎜ X ⎟ = −4890 K ⎜ − ⎟ ⎝ 1atm ⎠ ⎝ Th 373.15 ⎠ 0.693 = Th = 394.1K = 121o C 1/T Mountain Top Cooking ⎛P ln ⎜ X ⎝ Po ⎞ −ΔH ⎟= R ⎠ o vap ⎛1 1 ⎞ ⎜−⎟ ⎝ T Tb ⎠ Barometric Pressure Ph = Po e − mgh () ⎛ Pe ln ⎜ o ⎜ Po ⎜ ⎝ −mgh kbTa kbTa ⎞ o ⎟ = −ΔH vap ⎟ R ⎟ ⎠ = = −ΔH R (1.381 × 10 kbT −23 ) m2 kg s −2 K −1 Ta = ( ) − 40.657 × 103 J mol −1 ⎛ 1 1⎞ ⎜− ⎟ 8.314472 J K −1 mol −1 ⎝ Th 373.15 ⎠ ⎛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 ⎠ kbTa () H 2O l ! H 2O g o vap ⎛1 1⎞ ⎜−⎟ ⎝ Th Tb ⎠ ⎛ 1.661 × 10 −27 kg ⎞ − ⎜ 29amu * ⎟ 9.8m / s * h 1amu ⎝ ⎠ −mgh Mt. McKinley Denali h = 6194 m o ⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞ ln ⎜ h ⎟ = ⎜−⎟ Po ⎠ R ⎝ Th Tb ⎠ ⎝ − mgh () Mountain Top Cooking () H 2O l ! H 2O g −4890 K + 13.1 Th On top of Mt. McKinley at external temp of 0o C 6194m 1.43 ×105 K = − 383.2 273.15K Th 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 5 ...
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