IB Chemistry on Gibbs Free Energy and Equilibrium constant, Kc
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The document provides a tutorial on Gibbs free energy change and equilibrium. It discusses key concepts such as dynamic equilibrium, equilibrium constant Kc, factors that affect equilibrium like temperature and pressure, and Le Chatelier's principle. It explains how temperature impacts the position of equilibrium and value of Kc for endothermic and exothermic reactions. The relationship between Gibbs free energy change (ΔG), entropy change (ΔS), enthalpy change (ΔH), and Kc is also covered. The magnitude of Kc indicates the extent of reaction and how close the system is to equilibrium. The sign of ΔG predicts spontaneity - a negative ΔG corresponds to a spontaneous process while a positive ΔG means the process
![Dynamic Equilibrium
Reversible (closed system)
Forward Rate, K1 Reverse Rate, K-1
Conc of product and reactant
at equilibrium
At Equilibrium
Forward rate = Backward rate
Conc reactants and products remain
CONSTANT/UNCHANGE
Equilibrium Constant Kc
aA(aq) + bB(aq) cC(aq) + dD(aq)
coefficient
Solid/liq not included in Kc
Conc represented by [ ]
kf
Kr
ba
dc
c
BA
DC
K
r
f
c
k
k
K
Equilibrium Constant Kc
express in
Conc vs time Rate vs time
A + B
C + D
Conc
Time
Catalyst
Factors affecting equilibrium (closed system)
TemperaturePressureConcentration
Equilibrium constant Kc ≠ Position equilibrium
forward rate constant
reverse rate constant](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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IB Chemistry on Gibbs Free Energy and Equilibrium constant, Kc
- 1. http://lawrencekok.blogspot.com Prepared by Lawrence Kok Tutorial on Gibbs Free Energy Change and Equilibrium
- 2. Dynamic Equilibrium Reversible (closed system) Forward Rate, K1 Reverse Rate, K-1 Conc of product and reactant at equilibrium At Equilibrium Forward rate = Backward rate Conc reactants and products remain CONSTANT/UNCHANGE Equilibrium Constant Kc aA(aq) + bB(aq) cC(aq) + dD(aq) coefficient Solid/liq not included in Kc Conc represented by [ ] kf Kr ba dc c BA DC K r f c k k K Equilibrium Constant Kc express in Conc vs time Rate vs time A + B C + D Conc Time Catalyst Factors affecting equilibrium (closed system) TemperaturePressureConcentration Equilibrium constant Kc ≠ Position equilibrium forward rate constant reverse rate constant
- 3. Effect of Temperature on position of equilibrium Decrease Temp ↓ • Favour exo rxn • Equi shift to right → to increase ↑ Temp •Formation Co(H2O)6 2+ (pink) Increase Temp ↑ • Favour endo rxn • Equi shift to left ← to reduce ↓ Temp •Formation of CoCl4 2- (blue) CoCl4 2- + 6H2O ↔ Co(H2O)6 2+ + 4CI – ΔH = -ve (exo) (blue) (pink) Increase Temp ↑ – Favour endo rxn – Absorb heat to reduce Temp again ↓ Decrease Temp ↓ – Favour exo rxn – Release heat to increase Temp again ↑ Increase Temperature • Rate rxn increase ↑ • Rate constant also change • Rate constant forward/reverse increase but to diff extend • Position equi shift to endo to decrease ↓ Temp • Kc, equilibrium constant change Click to view video Le Chatelier’s Principle • System in dynamic equilibrium is disturbed, position of equilibrium will shift so to cancel out the effect of change and new equilibrium can established again Effect of Temperature on equilibrium constant, Kc
- 4. Le Chatelier’s Principle • System in dynamic equilibrium is disturbed, position of equilibrium will shift so to cancel out the effect of change and new equilibrium can established again Decrease Temp ↓ • Favour exo rxn • Equi shift to left ← to increase ↑ Temp •Formation N2O4 (colourless) Increase Temp ↑ • Favour endo rxn • Equi shift to right → to reduce ↓ Temp •Formation NO2 (brown) N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1 (colourless) (brown) Click to view video Effect of Temperature on position of equilibrium Effect of Temperature on equilibrium constant, Kc Increase Temp ↑ – Favour endo rxn – Absorb heat to reduce Temp again ↓ Decrease Temp ↓ – Favour exo rxn – Release heat to increase Temp again ↑ Increase Temperature • Rate rxn increase ↑ • Rate constant also change • Rate constant forward/reverse increase but to diff extend • Position equi shift to endo to decrease ↓ Temp • Kc, equilibrium constant change
- 5. N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1 Temp increase ↑ – Kc increase ↑ A ↔ B ΔH = +veReverse rate constant = k r Forward rate constant = kf Kc A B Kc r f c k k K Temp affect rate constant Temp change cK Increase Temp ↑ Position equilibrium shift to right Endo side – Absorb heat Temp decrease ↓ More product , less reactant treac product Kc tan cK Forward rate constant, kf > reverse rate, kr r f c k k K Decrease Temp ↓ Position equilibrium shift to left Exo side – Release heat Temp increase ↑ More reactant , less product treac product Kc tan Forward rate constant, kf < reverse rate, kr r f c k k K cK Conclusion : Endo rxn – Temp ↑ – Kc ↑ – Product ↑ Effect of Temperature on equilibrium constant, Kc forward rate constant reverse rate constant
- 6. Temp increase ↑ – Kc decrease ↓ A ↔ B ΔH = -ve Increase Temp ↑ Position equilibrium shift to left Endo side – Absorb heat Temp decrease ↓ More Reactant , less product treac product Kc tan cK Forward rate constant, kf < Reverse rate, kr r f c k k K Decrease Temp ↓ Position equilibrium shift to right Exo side – Release heat Temp increase ↑ More Product , less reactant treac product Kc tan Forward rate constant, kf > Reverse rate, kr r f c k k K cK Conclusion : Exo rxn – Temp ↑ – Kc ↓ – Product ↓ H2(g) + I2(g) ↔ 2HI(g) ΔH = -9.6kJmol-1 Forward rate constant = kf Reverse rate constant = k r Effect of Temperature on equilibrium constant, Kc Kc A B Kc r f c k k K forward rate constant reverse rate constant Temp affect rate constant Temp change cK
- 7. Equilibrium Constant Kc express in At equilibrium Rate of forward = Rate of reverse DCkBAk rf Forward Rate Reverse Rate Rate forward = kf [A] [B] Rate reverse = kr [A] [B] BA DC k k r f Change in Temp = r f c k k K BA DC Kc Equilibrium and Kinetics Forward Rate Reverse Rate At equilibrium Rate of forward = Rate of reverse kf = forward rate constant kr = reverse rate constant BA DC Kc r f c k k K Ratio of product /reactant conc Ratio of rate constant Change rate constant, kf/kr Change in Kc Temp increase ↑ – Kc increase ↑ Temp increase ↑ – Kc decrease ↓ A ↔ B ΔH = +ve A ↔ B ΔH = -ve Temp changes Kc with diff rxn? Endothermic rxn Exothermic rxn kf kr
- 8. Magnitude of Kc Extend of reaction How far rxn shift to right or left? Not how fast cK Position of equilibrium cK Temp dependent Extend of rxn Not how fast Shift to left/ favour reactant Shift to right/ favour product cK Relationship between Equilibrium and Energetics cKRTG ln STHG Enthalpy change Entropy change Equilibrium constant Gibbs free energy change H G cK G Energetically Thermodynamically Favourable/feasible ΔGθ ln K Kc Eq mixture ΔGθ -ve < 0 Positive ( + ) Kc > 1 Product (Right) ΔGθ +ve > 0 Negative ( - ) Kc < 1 Reactant (left) ΔGθ = 0 0 Kc = 1 Equilibrium Measure work available from system Sign predict spontaneity of rxn Negative (-ve) spontaneous Positive (+ve) NOT spontaneous veG veG NOT favourable Energetically favourable Product formation NO product cKRTG ln
- 9. Magnitude of Kc Extend of reaction How far rxn shift to right or left? Not how fast cK Position of equilibrium cK Temp dependent Extend of rxn Not how fast Shift to left/ favour reactant Shift to right/ favour product cK Relationship between Equilibrium and Energetics cKRTG ln STHG Enthalpy change Entropy change Equilibrium constant Gibbs free energy change H G cK ΔGθ ln K Kc Eq mixture ΔGθ -ve < 0 Positive ( + ) Kc > 1 Product (Right) ΔGθ +ve > 0 Negative ( - ) Kc < 1 Reactant (left) ΔGθ = 0 0 Kc = 1 Equilibrium cKRTG ln STHG ∆Hsys ∆Ssys ∆Gsys Description - + ∆G = ∆H - T∆S ∆G = - ve Spontaneous, All Temp + - ∆G = ∆H - T∆S ∆G = + ve Non spontaneous, All Temp + + ∆G = ∆H - T∆S ∆G = - ve Spontaneous, High ↑ Temp - - ∆G = ∆H - T∆S ∆G = - ve Spontaneous, Low ↓ Temp Relationship bet ∆G and Kc
- 10. G Energetically Thermodynamically Favourable/feasible Sign predict spontaneity of rxn veG veG NOT favourable Energetically favourable Product formation NO product KRTG ln Predicting rxn will occur? with ΔG and Kc cK Very SMALL Kc < 1 Shift to right/ favour product Shift to left/ favour reactant Very BIG Kc > 1 veG veG KRTG ln 1cK 1cK Negative (-ve) spontaneous Positive (+ve) NOT spontaneous Relationship bet ∆G and Kc products reactants ΔGθ Kc Eq mixture ΔGθ = + 200 9 x 10-36 Reactants ΔGθ = + 10 2 x 1-2 Mixture ΔGθ = 0 Kc = 1 Equilibrium ΔGθ = - 10 5 x 101 Mixture ΔGθ = - 200 1 x 1035 Products shift to left shift to right G, Gibbs free energy A Mixture composition B 100% A 100% B ∆G decreases ↓ 30 % A 70 % B Equilibrium mixture ∆G < 0 ∆G = 0 (Equilibrium) ↓ Free energy minimum ∆G < 0 ∆G < 0 ∆G = 0 Free energy system is lowered on the way to equilibrium Rxn proceed to minimum free energy ∆G = 0 Sys seek lowest possible free energy Product have lower free energy than reactant ∆G < 0 product reactant
- 11. G Energetically Thermodynamically Favourable/feasible Sign predict spontaneity of rxn veG veG NOT favourable Energetically favourable Product formation NO product KRTG ln Predicting rxn will occur? with ΔG and Kc cK Very SMALL Kc < 1 Shift to right/ favour product Shift to left/ favour reactant Very BIG Kc > 1 veG veG KRTG ln 1cK 1cK Negative (-ve) spontaneous Positive (+ve) NOT spontaneous Relationship bet ∆G, Q and Kc G, Gibbs free energy A Mixture composition B 100% A 100% B ∆G decreases ↓ 30 % A 70 % B Equilibrium mixture ∆G < 0 ∆G = 0 (Equilibrium) ↓ Free energy minimum ∆G < 0 ∆G < 0 ∆G = 0 ∆G < 0 product reactant G, Gibbs free energy reactant product∆G < 0 A B ∆G decreases ↓ 100% A 100% B30 % A 70 % B ∆G = 0 Q = K ∆G < 0 Q < K ∆G > 0 ∆G < 0 Q > K ∆G > 0 A ↔ B A ↔ B Equilibrium mixture
- 12. Relationship bet ∆G and Kc G, Gibbs free energy A B 100% A 100% B ∆G decreases ↓ 30 % A 70 % B Equilibrium mixture close to product ∆G < 0 ∆G = 0 (Equilibrium) ↓ Free energy minimum ∆G < 0 ∆G < 0 ∆G = 0 ∆G < -10 Kc > 1 A ↔ B A ↔ B G, Gibbs free energy A B ∆G decreases ↓ ∆G < -100 100% A 100% B ∆G = 0 (Equilibrium) ↓ Free energy minimum Kc > 1Equilibrium mixture close to product 10 % A 90 % B ∆G < 0 ∆G < 0 ∆G = 0 ∆G very –ve → Kc > 1 → (more product/close to completion)∆G –ve → Kc > 1 → (more product > reactant) A ↔ B G, Gibbs free energy 100% A 100% B A B ∆G +ve → Kc < 1 → (more reactant > product) ∆G > +10 ∆G = 0 (Equilibrium) ↓ Free energy minimum Kc < 1 ∆G increases ↑ 70 % A 30 % B Equilibrium mixture close to reactant ∆G < 0 ∆G = 0 A ↔ B G, Gibbs free energy ∆G more +ve → Kc < 1 → (All reactant / no product at all) A ∆G = 0 (Equilibrium) ↓ Free energy minimum Kc < 1100% A 100% B Equilibrium mixture close to reactant/ No reaction. ∆G > +100 B 90 % A 10 % B ∆G increases ↑ ∆G = 0 ∆G < 0 reactant reactant reactant reactant productproduct product product
- 13. Relationship bet ∆G and Kc products reactants shift to left shift to right G, Gibbs free energy A B 100% A 100% B ∆G decreases ↓ 30 % A 70 % B Equilibrium mixture ∆G < 0 ∆G = 0 (Equilibrium) ↓ Free energy minimum ∆G < 0 ∆G < 0 ∆G = 0 Free energy system is lowered on the way to equilibrium Rxn proceed to minimum free energy ∆G = 0 System seek lowest possible free energy Product have lower free energy than reactant ∆G < -10 Kc > 1 A ↔ B A ↔ B G, Gibbs free energy A B ∆G decreases ↓ ∆G < -100 100% A 100% B ∆G = 0 (Equilibrium) ↓ Free energy minimum Kc > 1Equilibrium mixture 10 % A 90 % B ∆G < 0 ∆G < 0 ∆G = 0 ∆G very –ve → Kc > 1 → (All product/close to completion)∆G –ve → Kc > 1 → (more product > reactant) ∆G ∆G = 0 ∆G > 0 ∆G < 0 No reaction/most reactants Kc <1 Complete rxn/Most products Kc > 1 Kc = 1 (Equilibrium) Reactants = Products reactant reactant ΔGθ Kc Eq mixture ΔGθ = + 200 9 x 10-36 Reactants ΔGθ = + 10 2 x 1-2 Mixture ΔGθ = 0 Kc = 1 Equilibrium ΔGθ = - 10 5 x 101 Mixture ΔGθ = - 200 1 x 1035 Products
- 14. Gibbs Free Energy Change, ∆G ∆G - Temp/Pressure remain constant Assume ∆S/∆H constant with temp Using ∆Gsys to predict spontaneity syssyssys STHG Easier Unit ∆G - kJ mol-1 Only ∆Ssys involved ∆S surr, ∆S uni not needed Using ∆Gsys to predict spontaneity Easier Method 1 Method 2 )()( reactfprofsys GGG At std condition/states Temp - 298K Press - 1 atm Gibbs Free Energy change formation, ∆Gf 0 At High Temp ↑ Temp dependent syssyssys STHG At low Temp ↓ veG STG HST sys syssyssys STHG veG HG STH spontaneous spontaneous surrsysuni SSS T H S sys surr syssysuni STHST Deriving Gibbs Free Energy Change, ∆G T H SS sys sysuni ∆S sys / ∆H sys multi by -T syssyssys STHG ∆Hsys ∆Ssys ∆Gsys Description - + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at all Temp + - ∆G = ∆H - T ∆S ∆G = + ve Non spontaneous, all Temp unisys STG syssyssys STHG Only ∆H sys/∆S sys involved ∆S surr, ∆S uni not needed Non standard condition Gibbs Free Energy Change, ∆G syssyssys STHG unisys STG veGsys ∆S uni = +ve Spontaneous Spontaneous veGsys ∆H = - ve ∆S sys = +ve ∆Hsys ∆Ssys ∆Gsys Description + + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at high ↑ Temp - - ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at low ↓ Temp
- 15. Gibbs Free Energy change formation, ∆Gf 0 At High Temp ↑ Temp dependent syssyssys STHG At low Temp ↓ veG STG HST sys syssyssys STHG veG HG STH spontaneous spontaneous ∆Hsys ∆Ssys ∆Gsys Description - + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at all Temp + - ∆G = ∆H - T ∆S ∆G = + ve Non spontaneous, all Temp syssyssys STHG ∆Hsys ∆Ssys ∆Gsys Description + + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at high ↑ Temp - - ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at low ↓ Temp Relationship Equilibrium and Energetics At equilibrium ∆G = 0 S H T HST CaCO3 (s) → CaO(s) + CO2(g) CaCO3 (s) → CaO (s) + CO2(g) ∆Hf 0 - 1206 - 635 - 393 S0 + 93 + 40 + 213 At what temp will decomposition CaCO3 be spontaneous? Reactant Product ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) kJHsys 178)1206(1028 kJS JS S SSS sys sys sys treacproductsys 16.0 160 93253 )tan()( STHG S H T HST KT 1112 16.0 178 Unit ∆H – kJ Unit ∆S - JK-1 At equilibrium ∆G = 0 Click here notes from chemwiki ∆H = +ve, ∆S = +ve → Temp ↑ High → Spontaneous ∆H = -ve, ∆S = -ve → Temp ↓ Low → Spontaneous ∆Hsys ∆Ssys ∆Gsys Description + + ∆G = ∆H - T∆S ∆G = 0 Equilibrium at Temp, T - - ∆G = ∆H - T∆S ∆G = 0 Equilibrium at Temp, T ∆G = 0
- 16. kJG G STHG 130 )16.0(298178 Predict what happen at diff Temp Reactant Product ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) kJHsys 178)1206(1028 ∆G > 0 Decomposition at 298K Non Spontaneous CaCO3 (s) → CaO(s) + CO2(g) CaCO3 (s) → CaO (s) + CO2(g) ∆Hf 0 - 1206 - 635 - 393 S0 + 93 + 40 + 213 kJS JS sys sys 16.0 16093253 Decomposition at 298K Decomposition at 1500K Decomposition limestone CaCO3 spontaneous? Gibbs Free Energy Change, ∆G kJG G STHG 62 )16.0(1500178 ∆H = +ve ∆S = +ve Temp dependent ∆Hsys ∆Ssys ∆Gsys Description + + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at high ↑ Temp - - ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at low ↓ Temp At Low Temp At High Temp Unit ∆H – kJ Unit ∆S - JK-1 Equilibrium, at what Temp? ∆G = 0 ∆G < 0 Decomposition at 1500K Spontaneous STHG HST S H T KT 1112 16.0 178 Temp > 1112K rxn spontaneous Temp dependent Spontaneous at High ↑ temp
- 17. Predict what happen at diff Temp Reactant Product ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) ∆G > 0 Freezing at 298K Non Spontaneous Freezing at 298K (25C) Freezing at 263K (-10C) Is freezing water spontaneous? Gibbs Free Energy Change, ∆G ∆H = -ve ∆S = -ve Temp dependent At High Temp At Low Temp Unit ∆H – kJ Unit ∆S - JK-1 Equilibrium, at what Temp? ∆G = 0 ∆G < 0 Freezing at 263K (-10C) Spontaneous STHG HST S H T )0(273 022.0 010.6 CKT H2O (l) → H2O(s) Is Freezing spontaneous? H2O (l) → H2O(s) ∆Hf 0 - 286 - 292 S0 + 70 + 48 kJHsys 010.6)286(292 kJS JS sys sys 022.0 227048 kJG G STHG 55.0 )022.0(298010.6 ∆Hsys ∆Ssys ∆Gsys Description + + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at high ↑ Temp - - ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at low ↓ Temp Water start to freeze Temp < 0 , Spontaneous kJG G STHG 22.0 )022.0(263010.6
- 18. Relationship between Equilibrium and Energetics KRTG ln HG cK G Energetically Thermodynamically Favourable/feasible Sign predict spontaneity of rxn Negative (-ve) spontaneous Positive (+ve) NOT spontaneous veG veG NOT favourable Energetically favourable Product formation NO product KRTG ln STHG H2(g) + O2(g) → H2O2(l) Energetically feasible ΔG / ΔH = -ve Predicting if rxn will occur? veG veH f Energetic favourable (-ve) Product H2O2 more stable ΔG and ΔH = -negative Reaction wont happen!!!!!! Kinetically unfavourable/stable due to HIGH activation energy H2(g) + O2(g) H2O2(l) Energy barrier Will rxn occur? depends Kinetically feasible low activation energy+ To occur ΔG < 0 (-ve) Activation energy LOW Energetically favourable Kinetically favourable ΔGθ ln K Kc Eq mixture ΔGθ -ve < 0 Positive ( + ) Kc > 1 Product (Right) ΔGθ +ve > 0 Negative ( - ) Kc < 1 Reactant (left) ΔGθ = 0 0 Kc = 1 Equilibrium
- 19. Energetic favourable (-ve) Graphite more stable Relationship between Equilibrium and Energetics KRTG ln HG cK G Energetically Thermodynamically Favourable/feasible Sign predict spontaneity of rxn Negative (-ve) spontaneous Positive (+ve) NOT spontaneous veG veG NOT favourable Energetically favourable Product formation NO product KRTG ln STHG Energetically feasible ΔG / ΔH = -ve Predicting if rxn will occur? veG veH f ΔG and ΔH = -negative Reaction wont happen!!!!!! Kinetically unfavourable/stable due to HIGH activation energy Energy barrier Will rxn occur? depends Kinetically feasible low activation energy+ To occur ΔG < 0 (-ve) Activation energy LOW Energetically favourable Kinetically favourable ΔGθ ln K Kc Eq mixture ΔGθ -ve < 0 Positive ( + ) Kc > 1 Product (Right) ΔGθ +ve > 0 Negative ( - ) Kc < 1 Reactant (left) ΔGθ = 0 0 Kc = 1 Equilibrium Diamond(s) → Graphite(s) Diamond forever Diamond(s) Graphite(s)
- 20. Click here to view free energy Predicting Spontaneity of Rxn Thermodynamic, ΔG Equilibrium, Kc 1cK 1cK KRTG ln G veG cK 1cK Energetically favourable 0G Predicting rxn will occur? N2(g) + 3H2(g) ↔ 2NH3(g) H2O(l) ↔ H+ (aq)+ OH- (aq) Shift toward reactants Energetically unfavourable Non spontaneous Mixture reactant/product Equilibrium veG Spontaneous Shift toward product 79G 33G 6 10G 14 101 cK 5 105cK Fe(s) + 3O2(g) ↔ 2Fe2O3(s) 261 101cK Shift toward reactants Energetically unfavourable Shift toward product Energetically favourable Energetically favourable Kinetically unfavourable/(stable) Rate too slow due to HIGH activation energy Rusting Process Energy barrier Shift toward product Click here for notes
- 21. IB Questions Esterification produce ethyl ethanoate. ΔG = -4.38kJmol-1 Cal Kc CH3COOH(l) + C2H5OH(l) ↔ CH3COOC2H5(l) + H2O(l) Kc = 5.9 cKRTG ln RT G Kc ln 29831.8 4380 ln cK Kc at 1338K is 0.0118. Cal Kc at 1473K A + B ↔ C + D kJH 3.177 Qualitative (Le Chatelier Principle) Quantitatively Formula 1473 1 1338 1 31.8 177300 0118.0 ln 2K K2 = 0.051 Temp increase ↑ – Kc increase ↑ Endothermic rxn A + B ↔ C + D Kc at 1000K and 1200K is 2.44 and 3.74. Cal ΔH. ?H 211 2 11 ln TTR H K K 1200 1 1000 1 31.844.2 74.3 ln H ΔH = 21.3kJmol-1 2 ?cK ?cK Temp decrease ↓ again Temp increase ↑ Shift to right → - absorb heat 211 2 11 ln TTR H K K NO oxidized to NO2. Kc = 1.7 x 1012. Cal ∆G at 298K 1 3 4 2NO + O2 ↔ NO2 ?G cKRTG ln 11 12 7.6969772 )107.1ln(298314.8 kJmolJmolG G
- 22. Van’t Hoff Equation cKRTG ln Relationship bet Temp and Kc STHG STHKRT ln R S RT H Kc ln c RT H Kc ln Heat absorbed, ΔH +ve Temp increase ↑ – Kc increase ↑ Heat released, ΔH –ve Temp increase ↑ – Kc decrease ↓ Gibbs free energy change Equilibrium constant Enthalpy change Entropy change N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1 Temp increase ↑ – Kc increase ↑ H2(g) + I2(g) ↔ 2HI(g) ΔH = -9.6kJmol-1 Temp increase ↑ – Kc decrease ↓ c TR H K 1 ln Endothermic rxn c RT H K ln Temp increase ↑ – Kc increase ↑ Exothermic rxn c RT H K ln c TR H K 1 ln Temp increase ↑ – Kc decrease ↓ Temp/K 250 400 650 1000 Kc 800 160 50 24 ΔH= +ve ΔH= -ve Temp/K 350 400 507 550 Kc 3.89 47.9 1700 6030 Gibbs free energy change
- 23. Van’t Hoff Equation cKRTG ln Relationship bet Temp and Kc R S RT H K ln c RT H K ln Gibbs free energy change Equilibrium constant Enthalpy change Entropy change N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1 Temp increase ↑ – Kc increase ↑ Endothermic rxn c TR H K 1 ln Plot Kc against Temp Temp/K 350 400 507 550 Kc 3.89 47.9 1700 6030 ln Kc 1.36 3.87 7.44 8.7 1/T(x 10-3) 2.86 2.50 1.97 1.82 RT H c eK Plot ln Kc against 1/T N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1 Endothermic rxn Using Kc and Temp to find ΔH Conclusion R H Gradient 31.8 007.0 H JH 58170 Temp increase ↑ Kc increase ↑Endo rxn =+ΔH Relationship bet Temp, Kc and ΔH ΔH=+ve -0.007 STHG Gibbs free energy change STHKRT ln
- 24. Van’t Hoff Equation KRTG ln Relationship bet Temp and Kc R S RT H K ln c RT H K ln Gibbs free energy change Equilibrium constant Enthalpy change Entropy change Exothermic rxn c TR H K 1 ln Plot Kc against Temp ln Kc 6.9 3.45 -3.3 -9.5 1/T(x 10-3) 2.9 2.6 2 1.4 RT H c eK Plot ln Kc against 1/T Exothermic rxn Using Kc and Temp to find ΔH Conclusion R H Gradient 31.8 11316 H JH 94000 Temp increase ↑ Kc decrease ↓Exo rxn =-ΔH Relationship bet Temp, Kc and ΔH H2(g) + I2(g) ↔ 2HI(g) ΔH = -9.6kJmol-1 H2(g) + I2(g) ↔ 2HI(g) ΔH = -9.6kJmol-1 Temp/K 345 385 500 700 Kc 1000 31.6 0.035 0.00007 Temp increase ↑ – Kc decrease ↓ ΔH=-ve +11316 STHKRT ln STHG Gibbs free energy change
- 25. Acknowledgements Thanks to source of pictures and video used in this presentation Thanks to Creative Commons for excellent contribution on licenses http://creativecommons.org/licenses/ Prepared by Lawrence Kok Check out more video tutorials from my site and hope you enjoy this tutorial http://lawrencekok.blogspot.com