IB Chemistry on Gibbs Free Energy and Entropy
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This document provides a tutorial on Gibbs free energy and spontaneity in thermodynamics. It discusses how entropy, enthalpy, and Gibbs free energy relate to the spontaneity of chemical reactions. Standard molar entropy (S°) and standard enthalpy of formation (ΔHf°) values can be used to calculate entropy changes (ΔS) and enthalpy changes (ΔH) of reactions, and determine if they are spontaneous. Standard Gibbs free energy of formation (ΔGf°) values similarly allow calculating Gibbs free energy changes (ΔG) of reactions to predict spontaneity based on the second law of thermodynamics.
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IB Chemistry on Gibbs Free Energy and Entropy
- 1. http://lawrencekok.blogspot.com Prepared by Lawrence Kok Tutorial on Gibbs Free Energy and Spontaneity .
- 2. E = sum kinetic energy/motion of molecule, and potential energy represented by chemical bond bet atom ∆E = q + w ∆E = Change internal energy q = heat transfer w = work done by/on system Thermodynamics Study of work, heat and energy on a system ∆E universe = ∆E sys + ∆E surrounding = 0 1st Law Thermodynamics Entropy - Measure of disorder ↓ ∆S uni = ∆S sys + ∆S surr > 0 (irreversible rxn) ↓ All spontaneous rxn produce increase in entropy of universe 2nd Law Thermodynamics ∆S uni = ∆S sys + ∆S surr Isolated system - Entropy change of universe always increase Click here thermodynamics entropy Entropy Measure molecular disorder/randomness ↓ More disorder - More dispersion of matter/energy ↓ More random - Rxn toward right- Entropy Increases ↑ Direction to right- Spontaneous to right → 2nd Law Thermodynamics Embrace the chaos Over time - Entropy increase ↑ Direction to left ← Never happen ! Click here thermodynamics Energy cannot be created or destroyed > 0
- 3. ∆S = Entropy change Entropy Dispersal/Distribution Matter Energy Matter more disperse ↑ Entropy increases ↑ solid liquid gas spontaneous - entropy ↑ Over time - Entropy increase ↑ Phase change - sol liq gas→ → ↓ Entropy increase ↑ Every energy transfer - increase entropy universe Entropy universe can only go up - never go down Entropy increase - many ways energy spread out Dispersion energy as heat - increase entropy Stoichiometry- more gas/liq in product ↓ Entropy increase ↑ T Q S =∆ Heat added ↑ Phase change Stoichiometry Embrace the chaos N2O4(g) 2NO→ 2(g) 1 2 2H2O(l) 2H→ 2 (g) + O2 (g) 1 2 3 3 More gas in product - Entropy ↑ Heat added ↑ Entropy Measure molecular disorder/randomness ↓ More disorder - More dispersion of matter/energy ↓ More randon - Rxn towards right- Entropy Increases ↑ Liq more disorder than solid Gas more disorder than liq kinetic energy distributed over wide range Q = heat transfer T = Temp/K Distribution matter in space Distribution energy bet particles Direction to left ← Never happen !Direction to right- Spontaneous to right →
- 4. Statistical Entropy Entropy Measure molecular disorder/randomness ↓ More disorder - More dispersion of matter/energy ↓ More random - Entropy Increases ↑ 1st Law Thermodynamics - Doesn't help explain direction of rxn ∆S uni > 0 (+ve) More disorder - spontaneous→ ∆S uni < 0 (-ve) More order - non spontaneous→ Change sol liq gas - Higher entropy→ → Greater number particles in product - Higher entropy More complex molecule - More atoms bonded - Higher entropy Higher temp - Vibrate faster - More random - Higher entropy Why gas mixes and not unmix? Why heat flow from hot to cold? Entropy Notes on Entropy 1st Law Thermodynamics 2nd Law Thermodynamics Energy cannot be created or destroyed Transfer from one form to another ∆E universe = ∆E sys + ∆E surrounding = 0 Isolated system ↓ ∆S uni always increase ∆E = q + w Method to calculate entropy Number microstates Thermodynamic Entropy Heat + Temp involved Gas mixesSolution diffuse Heat flow hot →cold X X X ∆E = internal energy q = heat transfer w = work done ∆S = Entropy universe ∆S = Entropy system ∆S = Entropy surrounding ∆S uni = ∆S sys + ∆S surr Law Thermodynamics 1 2 ∆S = Entropy uni WkS ln=∆ ∆S = Entropy change k = boltzmann constant W = Microstate Click here statistical entropy Click here thermodynamics entropy Why solution diffuse and not undiffuse? Unit - J mol -1 K-1 surrsysuni SSS ∆+∆=∆ ∆S = Entropy sys and surr High chaos factor
- 5. 1st Law Thermodynamics - Doesn't help explain direction of rxn ∆S uni > 0 (+ve) More disorder - spontaneous→ ∆S uni < 0 (-ve) More order - non spontaneous→ Change sol liq gas - Higher entropy→ → Greater number particles in product - Higher entropy More complex molecule - More atoms bonded - Higher entropy Higher temp - Vibrate faster - More random - Higher entropy Measure molecular disorder/randomness ↓ More disorder - More dispersion of matter/energy ↓ More random - Entropy Increases ↑ Isolated system ↓ ∆S uni always increase Entropy Why gas mixes and not unmix? Why heat flow from hot to cold? Notes on Entropy 1st Law Thermodynamics 2nd Law Thermodynamics Energy cannot be created or destroyed Transfer from one form to another ∆E universe = ∆E sys + ∆E surrounding = 0 ∆E = q + w Gas mixesSolution diffuse Heat flow hot →cold X X X ∆E = internal energy q = heat transfer w = work done ∆S = Entropy universe ∆S = Entropy system ∆S = Entropy surrounding ∆S uni = ∆S sys + ∆S surr Law Thermodynamics 3rd Law Thermodynamics Unit - J mol -1 K-1 Standard Molar Entropy, S0 Entropy perfectly crystal at 0K = 0 Std molar entropy, S0 (absolute value) ↓ S0 when substance heated from 0K to 298K Std state - 1 atm / 1M sol Temp = 298K Std Molar Entropy/S0 S0 at 298 /JK-1 mol-1 Fe (s) + 27 H2O (s) + 48 Na (s) + 52 H2O (l) + 69 CH3OH (l) + 127 H2 (g) + 130 H2O (g) + 188 CO2 (g) + 218 Solid - Order ↓ Entropy Lowest Liq - Less order ↓ Entropy Higher Gas - Disorder ↓ Entropy Highest Entropy highest Why solution diffuse and not undiffuse? High chaos factor
- 6. Entropy Why gas mix and not unmix?Why solution diffuse and not undiffuse? Why heat flow from hot to cold? Gas mixesSolution diffuse Heat flow hot →cold X X X Unit - J mol -1 K-1 Standard Molar Entropy, S0 Entropy perfectly crystal at 0K = 0 (Absolute value) ↓ S0 when substance heated from 0K to 298K Std state - 1 atm / 1M sol Temp = 298K Std Molar Entropy/S0 S0 at 298 /JK-1 mol-1 Fe (s) + 27 H2O (s) + 48 Na (s) + 52 H2O (l) + 69 CH3OH (l) + 127 H2 (g) + 130 H2O (g) + 188 CO2 (g) + 218 Solid - Order ↓ Entropy Lowest Liq - Less order ↓ Entropy Higher Gas - Disorder ↓ Entropy Highest Entropy highest Entropy Standard Molar Entropy, S0 Depend on Temp increase - Entropy increase↑ ↑ Physical/phase state Dissolving solid Molecular mass Click here thermodynamics entropy Ba(OH)2 Temp Temp/K 273 295 298 S0 for H2 + 31 + 32 + 33.2 Sol Liq Gas - Entropy increase→ → ↑ State solid liquid gas S0 for H2O + 48 + 69 + 188 entropy increase ↑ entropy increase ↑ Depend on Substance NaCI NH4NO3 S0 for solid + 72 + 151 S0 for aq + 115 + 260 More motion - entropy increase ↑ Higher mass - entropy increase ↑ Substance HF HCI HBr Molar mass 20 36 81 S0 + 173 + 186 + 198 S0 = 0 at 0K All sub > 0K, have +ve S0
- 7. Entropy perfectly crystal at 0K = 0 (Absolute value) ↓ S0 when substance heated from 0K to 298K Entropy Why gas mix and not unmix?Why solution diffuse and not undiffuse? Why heat flow from hot to cold? Gas mixesSolution diffuse Heat flow hot →cold X X X Unit - J mol -1 K-1 Standard Molar Entropy, S0 Std state - 1 atm / 1M sol Temp = 298K Std Molar Entropy/S0 S0 at 298 /JK-1 mol-1 H2O (s) + 48 Na (s) + 52 H2O (l) + 69 CH3OH (l) + 127 H2O (g) + 188 CO2 (g) + 218 Solid - Order ↓ Entropy Lowest Liq - Less order ↓ Entropy Higher Gas - Disorder ↓ Entropy Highest Entropy highest Entropy Standard Molar Entropy, S0 Depend on Temp increase - Entropy increase↑ ↑ Physical/phase state Dissolving solid Molecular mass Temp Temp/K 273 295 298 S0 for H2 + 31 + 32 + 33.2 Sol Liq Gas - Entropy increase→ → ↑ State solid liquid gas S0 for H2O + 48 + 69 + 188 entropy increase ↑ entropy increase ↑ Depend on More motion - entropy increase ↑ Click here entropy notes Click here entropy, enthalpy free energy data Click here entropy CRC data booklet Higher mass - entropy increase ↑ S0 = 0 at 0K All sub > 0K, have +ve S0 Substance NaCI NH4NO3 S0 for solid + 72 + 151 S0 for aq + 115 + 260 Substance HF HCI HBr Molar mass 20 36 81 S0 + 173 + 186 + 198
- 8. ∆Hf θ (reactant) ∆Hf θ (product) Using Std ∆Hf θ formation to find ∆H rxn ∆H when 1 mol form from its element under std condition Na(s) + ½ CI2(g) → NaCI(s) ∆Hf θ = - 411 kJ mol -1 Std Enthalpy Changes ∆Hθ Std condition Pressure 100kPa Temp 298K Conc 1M All substance at std states Std ∆Hf θ formation Mg(s) + ½ O2(g) → MgO(s) ∆Hf θ =- 602 kJ mol -1 Reactants Products O2(g) → O2 (g) ∆Hf θ = 0 kJ mol -1 ∆Hrxn θ = ∑∆Hf θ (products) - ∑∆Hf θ (reactants) ∆Hf θ (products)∆Hf θ (reactants) ∆Hrxn θ Elements Std state solid gas 2C(s) + 3H2(g)+ ½O2(g) → C2H5OH(I) ∆Hf θ =- 275 kJ mol -1 1 mole formed H2(g) + ½O2(g) → H2O(I) ∆Hf θ =- 286 kJ mol -1 Std state solid gas 1 mol liquid For element Std ∆Hf θ formation = 0 Mg(s)→ Mg(s) ∆Hf θ = 0 kJ mol -1 No product form Using Std ∆Hf θ formation to find ∆H rxn PDF version Click here chem database (std formation enthalpy) Online version Click here chem database (std formation enthalpy) C2H4 + H2 C2H6 Find ΔHθ rxn using std ∆H formation Reactants Products 2C + 3H2 Elements C2H4 + H2 C→ 2H6 ∆Hrxn θ ∆Hrxn θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) ∆Hrxn θ = Hf θ C2H6 - ∆Hf θ C2H4+ H2 = - 84.6 – ( + 52.3 + 0 ) = - 136.9 kJ mol -1 Enthalpy Formation, ∆Hf
- 9. Std ∆Gf θ formation ∆Grxn θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Grxn θ = Gf θ C2H6 - ∆Gf θ C2H4+ H2 = - 33 – ( + 68 + 0 ) = - 101 kJ mol -1 ∆Gf θ (reactant) ∆Gf θ (product) Using Std ∆Gf θ formation to find ∆G rxn o ∆Gf when 1 mol form from its element under std condition Na(s) + ½ CI2(g) → NaCI(s) ∆Gf θ = - 384 kJ mol -1 Std Free Energy Change ∆Gθ Std condition Pressure 100kPa Temp 298K Conc 1M All substance at std states Gibbs Free Energy change formation, ∆Gf Mg(s) + ½ O2(g) → MgO(s) ∆Gf θ =- 560 kJ mol -1 Reactants Products O2(g) → O2 (g) ∆Gf θ = 0 kJ mol -1 ∆Grxn θ = ∑∆Gf θ (prod) - ∑∆Gf θ (react) ∆Gf θ (product)∆Gf θ (reactant) ∆Grxn θ Elements Std state solid gas 2C(s) + 3H2(g)+ ½O2(g) → C2H5OH(I) ∆Gf θ =- 175 kJ mol -1 1 mole formed H2(g) + ½O2(g) → H2O(I) ∆Gf θ =- 237 kJ mol -1 Std state solid gas 1 mol liquid For element Std ∆Gf θ formation = 0 Mg(s)→ Mg(s) ∆Gf θ = 0 kJ mol -1 No product form Using Std ∆Gf θ formation to find ∆G rxn PDF version Click here chem database (std ∆G formation) Online version Click here chem database (std ∆G formation) C2H4 + H2 C2H6 Find ΔGθ rxn using std ∆G0 formation Reactants Products 2C + 3H2 Elements C2H4 + H2 C→ 2H6 ∆Grxn θ
- 10. ∆S sys + ve , ∆S surr - ve ↓ ∆S uni > 0 (+ve) (Rxn Spontaneous) ∆S sys - ve , ∆S surr + ve ↓ ∆S uni < 0 (-ve) (Rxn Non spontaneous) spontaneous +ve -ve = S /JK-1 ∆Ssys = + ve ∆Ssurr = + ve ∆Suni = + ve + ∆Ssys = - ve + ∆Ssurr = + ve ∆Suni = + ve = spontaneous S /JK-1 S /JK-1 ∆Ssys = + ve + ∆Ssurr = - ve = ∆Suni = + ve spontaneous C6H12O6(s) + 6O2 (g) 6CO→ 2(g) + 6H2O(l) Using ∆Hsys , ∆Suni , ∆Ssys , ∆S surr to predict spontaneity 2NO(g) + O2(g) 2NO→ 2(g) CaCO3 (s) CaO→ (s) + CO2(g) ∆H = -ve (Heat released) Difficult !! ∆S sys + ve , ∆S surr - ve ↓ ∆S uni < 0 (-ve) (Rxn Non spontaneous) ∆Ssys = + ve ∆Ssurr = - ve + = ∆Suni = - ve Non spontaneous ∆H = -ve (Heat released) ∆H = +ve (Heat absorb) CaCO3 (s) CaO→ (s) + CO2(g) ∆H = +ve (Heat absorb) ∆Ssys = + ve + ∆Ssurr = - ve ∆Suni = - ve Non spontaneous = H2(g) 2 H→ (g) ∆H = +ve (Heat absorb) H2O (l) H→ 2O(s) ∆H = -ve (Heat released) ∆Ssys = - ve + ∆Suni = - ve ∆Ssurr = + ve = ∆S sys + ve , ∆S surr - ve ↓ ∆S uni < 0 (-ve) (Rxn Non spontaneous) ∆S sys + ve , ∆S surr + ve ↓ ∆S uni > 0 (+ve) (Rxn Spontaneous) ∆S sys - ve , ∆S surr + ve ↓ ∆S uni > 0 (+ve) (Rxn Spontaneous)
- 11. ∆Hsys ∆Ssys ∆Suni Description - + > 0 (+) Spontaneous, All Temp + - < 0 (-) Non spontaneous, All Temp + + > 0 (+) Spontaneous, High ↑ Temp - - > 0 (+) Spontaneous, Low ↓ Temp Predicting Spontaneity rxn ∆G - Temp/Pressure remain constant Assume ∆S/∆H constant with temp Using ∆Hsys , ∆Suni , ∆Ssys , ∆S surr to predict spontaneity Using ∆Gsys to predict spontaneity syssyssys STHG ∆−∆=∆ Difficult !! surrsysuni SSS ∆+∆=∆ T H Ssurr ∆− =∆)()( reactfprofsys HHH ∆∑−∆∑=∆ CH4(g) + 2O2 (g) CO→ 2(g) + 2H2O(l) CH4(g) + 2 O2 (g) CO→ 2(g) + 2 H2O(l) ∆Hf 0 - 74 0 - 393 - 286 x 2 S0 + 186 +205 x 2 + 213 + 171 x 2 Reactant Product ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) 1 )tan()( 41 596555 − −=∆ −=∆ −=∆ JKS S SSS sys sys treacproductsys kJHsys 891)74(965 −=−−−=∆ 1 2990 298 )891000( − +=∆ −− =∆ ∆− =∆ JKS S T H S surr surr surr 1 2949299041 − +=+−=∆ ∆+∆=∆ JKS SSS uni surrsysuni ∆S uni > 0 spontaneous Easier Unit ∆G - kJUnit ∆S - JK-1 Unit ∆H - kJ CH4(g) + 2O2 (g) CO→ 2(g) + 2H2O(l) Only ∆S sys involved ∆S surr, ∆S uni not needed ∆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 CH4(g) + 2 O2 (g) CO→ 2(g) + 2 H2O(g) S0 + 186 +205 x 2 +213 + 188 x 2 Reactant (+596) Product (+589) kJG G STHG syssyssys 888 )007.0(298890 −=∆ −−−=∆ ∆−∆=∆ ∆Hsys = - 890 kJ kJS JKS S SSS sys sys sys reactprodsys 007.0 7 596589 1 )()( −=∆ −=∆ −=∆ −=∆ − ∆G < 0 spontaneous Entropy change ∆S greater at low temp
- 12. ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = -868 - (-51) = - 817 kJ Predicting Spontaneity rxn ∆G - Temp/Pressure remain constant Assume ∆S/∆H constant with temp Using ∆Gsys to predict spontaneity syssyssys STHG ∆−∆=∆ CH4(g) + 2O2 (g) CO→ 2(g) + 2H2O(l) Reactant (-51) Product (-868) ∆G < 0 spontaneous Easier Unit ∆G - kJ mol-1 CH4(g) + 2O2 (g) CO→ 2(g) + 2H2O(l) Only ∆Ssys involved ∆S surr, ∆S uni not needed ∆Hsys ∆Ssys ∆Gsys Description - + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at all Temp + - ∆G = ∆H - T ∆S ∆G = + ve Non spontaneous, all Temp + + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at high ↑ Temp - - ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at low ↓ Temp CH4(g) + 2 O2 (g) CO→ 2(g) + 2 H2O(g) S0 + 186 +205 x 2 +213 + 188 x 2 Reactant (+ 596) Product (+ 589) kJG G STHG syssyssys 888 )007.0(298890 −=∆ −−−=∆ ∆−∆=∆ ∆Hsys = - 890 kJ kJS JKS S SSS sys sys sys reactprodsys 007.0 7 596589 1 )()( −=∆ −=∆ −=∆ −=∆ − ∆G < 0 spontaneous Using ∆Gsys to predict spontaneity Easier CH4(g) + 2 O2 (g) CO→ 2(g) + 2 H2O(l) ∆G0 - 51 0 - 394 - 237 x 2 Method 1 Method 2 )()( reactfprofsys GGG °° ∆−∆=∆ CH4(g) + 2 O2 (g) CO2(g) + 2H2O(g) C + 2O2 + 2H2 Reactants Products ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product)Elements • Neither ∆H or ∆S can predict feasibility of spontaneous rxn • Gibbs Free Energy (∆G) – measure spontaneity and useful energy available • Gibbs Free Energy (∆G) - max amt useful work at constant Temp/Pressure • Involve ∆H sys and ∆S sys • ∆G involve only sys while ∆S uni involve sys and surr • Easier to find ∆H and ∆S for system Gibbs Free Energy change formation, ∆Gf 0 At std condition/states Temp - 298K Press - 1 atm
- 13. ∆G - Temp/Pressure remain constant Assume ∆S/∆H constant with temp Using ∆Gsys to predict spontaneity syssyssys STHG ∆−∆=∆ Easier Unit ∆G - kJCH4(g) + 2O2 (g) CO→ 2(g) + 2H2O(l) Only ∆S sys involved ∆S surr, ∆S uni not needed ∆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 CH4(g) + 2 O2 (g) CO→ 2(g) + 2 H2O(g) S0 + 186 +205 x 2 +213 + 188 x 2 Reactant (+ 596) Product (+ 589) kJG G G STHG syssyssys 888 2890 )007.0(298890 −=∆ +−=∆ −−−=∆ ∆−∆=∆ ∆Hsys = - 890 kJ kJS JKS S SSS sys sys sys reactprodsys 007.0 7 596589 1 )()( −=∆ −=∆ −=∆ −=∆ − ∆G < 0 spontaneous Gibbs Free Energy Change, ∆G ∆G sys T∆S sys Total energy change, ∆H Measure spontaneity and useful energy available Max amt useful work at constant Temp/Pressure Free Energy syssyssys STHG ∆−∆=∆ Free energy available to do work not available for work syssyssys STHG ∆−∆=∆ Free Energy Total energy change, ∆H ∆G sys T∆S sys -890kJ Free energy available to do work not available for work -888kJ +2 kJ
- 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/∆Ssys involved ∆S surr, ∆S uni not needed °°° ∆−∆=∆ syssyssys STHG Non standard conditionStandard condition or Gibbs Free Energy Change, ∆G syssyssys STHG ∆−∆=∆unisys STG ∆−=∆ veGsys −=∆ ∆Suni = +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. kJG G STHG 130 )16.0(298178 +=∆ +−+=∆ ∆−∆=∆ Predict entropy change - quatitatively Reactant Product ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) kJHsys 178)1206(1028 +=−−−=∆ ∆G uni > 0 - Decomposition at 298K - Non Spontaneous Reactant Product ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) CaCO3 (s) CaO→ (s) + CO2(g) CaCO3 (s) CaO→ (s) + CO2(g) ∆Hf 0 - 1206 - 635 - 393 S0 + 93 + 40 + 213 kJS S S SSS sys sys sys treacproductsys 16.0 160 93253 )tan()( +=∆ +=∆ −=∆ −=∆ Decomposition at 298K Decomposition at 1500K CaCO3 (s) CaO→ (s) + CO2(g) CaCO3 (s) CaO→ (s) + CO2(g) ∆Hf 0 - 1206 - 635 - 393 S0 + 93 + 40 + 213 kJHsys 178)1206(1028 +=−−−=∆ Rxn Temp dependent Spontaneous at High temp↑ 1500K 298K Decomposition limestone CaCO3 spontaneous? Gibbs Free Energy Change, ∆G kJS S S SSS sys sys sys treacproductsys 16.0 160 93253 )tan()( +=∆ +=∆ −=∆ −=∆ kJG G STHG 62 )16.0(1500178 −=∆ +−+=∆ ∆−∆=∆ ∆G uni < 0 - Decomposition at 1500K - Spontaneous ∆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
- 16. Predict entropy change - quatitatively Reactant Product ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) ∆G uni > 0 - Decomposition at 298K - Non Spontaneous Reactant Product ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) Rxn Temp dependent Spontaneous at Low temp↓ 298K (25C) Gibbs Free Energy Change, ∆G ∆G uni < 0 - Decomposition at 1500K - Spontaneous ∆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 H2O (l) H→ 2O(s) H2O (l) H→ 2O(s) ∆Hf 0 - 286 - 292 S0 + 70 + 48 Freezing at 298K (25C) Is Freezing spontaneous? kJHsys 6)286(292 −=−−−=∆ kJS S S SSS sys sys sys treacproductsys 02.0 22 7048 )tan()( −=∆ −=∆ −=∆ −=∆ kJG G STHG 55.0 )022.0(2986 +=∆ −−−=∆ ∆−∆=∆ Freezing at 263K (-10C) H2O (l) H→ 2O(s) H2O (l) H→ 2O(s) ∆Hf 0 - 286 - 292 S0 + 70 + 48 kJHsys 6)286(292 −=−−−=∆ kJS S S SSS sys sys sys treacproductsys 02.0 22 7048 )tan()( −=∆ −=∆ −=∆ −=∆ 263K (-10C) kJG G STHG 21.0 )022.0(2636 −=∆ −−−=∆ ∆−∆=∆ At High Temp At Low Temp
- 17. C3H8(g) + 5 O2 (g) 3CO2(g) + 4H2O(l) Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X 1 syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ C3H8(g) + 5O2 (g) 3CO→ 2(g) + 4H2O(l) ∆H = -2220 kJ at 298K C3H8(g) + 5 O2 (g) 3 CO→ 2(g) + 4 H2O(l) S0 +270 +205 x 5 +213 x 3 +70 x 4 1295 919 Reactant Product kJG G STHG 2108 )376.0(2982220 −=∆ −−−=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 376.0 376 1295919 1 )tan()( −=∆ −=∆ −=∆ −=∆ − ∆H = -2220 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Is Combustion at 298K spontaneous? Using Free Energy to predict spontaneity Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = -2130 - (-23) = - 2153 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G < 0 - Combustion at 298K - Spontaneous C3H8(g) + 5 O2 (g) 3 CO→ 2(g) + 4 H2O(l) ∆G0 - 23 0 - 394 x 3 - 237 x 4 Elements 3C + 5O2 + 4H2 Reactant (-23) Product (-2130) ∆G < 0 - Combustion at 298K - Spontaneous
- 18. CH4(g) + 2O2 (g) CO→ 2(g) + 2H2O(g) ∆H = - 890 kJ at 298K CH4(g) + 2 O2 (g) CO2(g) + 2H2O(g) Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X 2 syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 596 + 589 Reactant Product kJG G STHG 888 )007.0(298890 −=∆ −−−=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 007.0 7 596589 1 )tan()( −=∆ −=∆ −=∆ −=∆ − ∆H = - 890 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Is Combustion at 298K spontaneous? Using Free Energy to predict spontaneity Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = -868 - (-51) = - 817 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G < 0 - Combustion at 298K - Spontaneous CH4(g) + 2 O2 (g) CO→ 2(g) + 2 H2O(l) ∆G0 - 51 0 - 394 - 237 x 2 Elements C + 2O2 + 2H2 Reactant (-51) Product (-868) ∆G < 0 - Combustion at 298K - Spontaneous CH4(g) + 2 O2 (g) CO→ 2(g) + 2 H2O(g) S0 + 186 +205 x 2 +213 + 188 x 2
- 19. H2O (g) H→ 2O(l) ∆H = - 44.1 kJ at 298K H2O(g) H2O(l) Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 188 + 70 Reactant Product kJG G STHG 1.9 )118.0(2981.44 −=∆ −−−=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 118.0 118 18870 1 )tan()( −=∆ −=∆ −=∆ −=∆ − ∆H = - 44.1 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = -237 - (-228) = - 9 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G < 0 - Combustion at 298K - Spontaneous H2O(g) H→ 2O(l) ∆G0 -228 - 237 Elements H2 + O2 Reactant (-228) Product (-237) ∆G < 0 - Combustion at 298K - Spontaneous Condensation steam at 298K (25C) spontaneous? H2O (g) H→ 2O(l) S0 + 188 + 70 3 Using Free Energy to predict spontaneity
- 20. H2(g) 2 H→ (g) ∆H = + 436 kJ at 298K H2(g) 2H(g) Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 130 + 230 Reactant Product kJG G STHG 406 )1.0(298436 +=∆ +−+=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 1.0 100 130230 1 )tan()( +=∆ +=∆ −=∆ −=∆ − ∆H = + 436 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = + 406 - (0) = +406 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G > 0 - Atomization at 298K - Non Spontaneous H2(g) 2H→ (g) ∆G0 0 + 203 x 2 Elements H2 Reactant (0) Product ( + 406) 4Is Atomization of H2 at 298K spontaneous? H2 (g) 2 H→ (g) S0 + 130 + 115 x 2 ∆G > 0 - Atomization at 298K - Non Spontaneous Using Free Energy to predict spontaneity
- 21. H2O (l) H→ 2O(s) ∆H = - 6 kJ at 298K H2O(l) H2O(s) Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X 2 syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 70 + 48 Reactant Product kJG G STHG 55.0 )022.0(2986 +=∆ −−−=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 022.0 22 7048 1 )tan()( −=∆ −=∆ −=∆ −=∆ − ∆H = - 6 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = -236.6 - (-237) = + 0.4kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G > 0 -Freezing at 298K - Non Spontaneous H2O(l) H→ 2O(s) ∆G0 -237 - 236.6 Elements H2 + O2 Reactant (-237) Product (-236.6) 5 H2O (l) H→ 2O(s) S0 + 70 + 48 ∆G > 0 -Freezing at 298K - Non Spontaneous Is Freezing water to ice at 298K (25C) spontaneous? Using Free Energy to predict spontaneity
- 22. H2O (l) H→ 2O(s) ∆H = - 6 kJ at 263K H2O(l) H2O(s) Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X 2 syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 70 + 48 Reactant Product kJG G STHG 21.0 )022.0(2636 −=∆ −−−=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 022.0 22 7048 1 )tan()( −=∆ −=∆ −=∆ −=∆ − ∆H = - 6 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = -237.2 - (-237) = - 0.2 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G < 0 -Freezing at 263K - Spontaneous H2O(l) H→ 2O(s) ∆G0 -237 - 237.2 Elements H2 + O2 Reactant (-237) Product (-237.2) 6 H2O (l) H→ 2O(s) S0 + 70 + 48 ∆G < 0 -Freezing at 263K - Spontaneous Is Freezing water to ice at 263K (-10C) spontaneous? Assume std condition at 263K Using Free Energy to predict spontaneity
- 23. CaCO3 (s) CaO→ (s) + CO2(g) ∆H = + 178 kJ at 298K Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 93 + 253 Reactant Product kJG G STHG 130 )16.0(298178 +=∆ +−+=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 16.0 160 93253 1 )tan()( +=∆ +=∆ −=∆ −=∆ − ∆H = + 178 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = - 999 - (- 1129) = + 130 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G > 0 - Decomposition at 298K - Non Spontaneous CaCO3(s) CaO + CO→ 2(g) ∆G0 -1129 - 604 - 395 Elements Ca + C + O2 Reactant ( -1129) Product (- 999) 7 Decomposition CaCO3 at 298K (25C) spontaneous? CaCO3 (s) CaO→ (s) + CO2(g) S0 + 93 + 40 + 213 ∆G > 0 - Decomposition at 298K - Non Spontaneous CaCO3 (s) CaO (s) + CO2(g) Using Free Energy to predict spontaneity
- 24. CaCO3 (s) CaO→ (s) + CO2(g) ∆H = + 178 kJ at 1500K Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 93 + 253 Reactant Product kJG G STHG 62 )16.0(1500178 −=∆ +−+=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 16.0 160 93253 1 )tan()( +=∆ +=∆ −=∆ −=∆ − ∆H = + 178 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = - 999 - (- 939) = - 60 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G < 0 - Decomposition at 1500K - Spontaneous CaCO3(s) CaO + CO→ 2(g) ∆G0 -939 - 604 - 395 Elements Ca + C + O2 Reactant (- 939) Product (- 999) 8 CaCO3 (s) CaO→ (s) + CO2(g) S0 + 93 + 40 + 213 CaCO3 (s) CaO (s) + CO2(g) Decomposition CaCO3 at 1500K (1227C) spontaneous? ∆G < 0 - Decomposition at 1500K - Spontaneous Assume std condition at 1500K Using Free Energy to predict spontaneity
- 25. 2NO(g) + O2(g) 2NO→ 2(g) ∆H = - 114 kJ at 298K Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 522 + 480 Reactant Product kJG G STHG 101 )042.0(298114 −=∆ −−−=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 042.0 42 522480 1 )tan()( −=∆ −=∆ −=∆ −=∆ − ∆H = - 114 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = + 104 - (174) = - 70 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G < 0 - Decomposition at 298K - Spontaneous 2 NO + O2 2NO→ 2(g) ∆G0 + 87 x 2 0 + 52 x 2 Elements N2 + O2 Reactant (+ 174) Product (+ 104) 9 2 NO(g) + O2 (g) 2NO2(g) ∆G < 0 - Decomposition at 298K - Spontaneous Is Oxidation of NO at 298K (25C) spontaneous? 2 NO(g) + O2 (g) 2NO→ 2(g) S0 + 210 x 2 + 102 + 240 x 2 Using Free Energy to predict spontaneity
- 26. N2(g) + 3H2(g) 2NH→ 3(g) ∆H = - 92 kJ at 298K Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 585 + 384 Reactant Product kJG G STHG 32 )2.0(29892 −=∆ −−−=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 2.0 201 585384 1 )tan()( −=∆ −=∆ −=∆ −=∆ − ∆H = - 92 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = - 34 - (0) = - 34 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G < 0 - NH3 production at 298K - Spontaneous N2 + 3H2 2NH→ 3(g) ∆G0 0 0 - 17 x 2 Elements N2 + H2 Reactant (0) Product (- 34) 10 N2(g) + 3H2 (g) 2NH3(g) Is Haber, NH3 production 298K (25C) spontaneous? NH3 N2(g) + 3H2 (g) 2NH→ 3(g) S0 + 192 + 131 x 3 + 192 x 2 ∆G < 0 - NH3 production at 298K - Spontaneous Using Free Energy to predict spontaneity
- 27. Fe2O3(s) + 2AI(s) 2Fe→ (s) + AI2O3(s) ∆H = - 851 kJ at 298K Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 143 + 105 Reactant Product kJG G STHG 840 )038.0(298851 −=∆ −−−=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 038.0 38 143105 1 )tan()( −=∆ −=∆ −=∆ −=∆ − ∆H = - 851 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = -1576 - (-741) = - 835 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G < 0 - AI production at 298K - Spontaneous Fe2O3 + 2AI 2Fe + AI→ 2O3 ∆G0 - 741 0 0 - 1576 Elements Fe + AI+ O2 Reactant (-741) Product (- 1576) 11Is Thermite, AI production 298K (25C) spontaneous? Fe2O3(s) + 2AI(s) 2Fe→ (s) + AI2O3(s) S0 + 87 + 28 x 2 + 27 x 2 + 51 ∆G < 0 - AI production at 298K - Spontaneous Fe2O3(s) + 2AI(s) 2Fe(s) + AI2O3(s) Using Free Energy to predict spontaneity
- 28. 4KCIO3(s) 3KCIO→ 4(s) + KCI(s) ∆H = - 144 kJ at 298K Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 572 + 535 Reactant Product kJG G STHG 133 )037.0(298144 −=∆ −−−=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 037.0 37 572535 1 )tan()( −=∆ −=∆ −=∆ −=∆ − ∆H = - 144 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = -1317 - (-1160) = - 157 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G < 0 - Decomposition at 298K - Spontaneous 4KCIO3 3 KCIO→ 4 + KCI ∆G0 - 290 x 4 - 303 x 3 - 408 Elements K + CI2 + O2 Reactant (-1160) Product (- 1317) 13 ∆G < 0 - Decomposition at 298K - Spontaneous Is decomposition KCIO3 298K (25C) spontaneous? 4KCIO3(s) 3KCIO→ 4(s) + KCI(s) S0 + 143 x 4 + 151 x 3 + 82 4KCIO3(s) 3KCIO4(s) + KCI(s) Using Free Energy to predict spontaneity
- 29. Entropy and Gibbs Free Energy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Will rxn be spontaneous ? Gas mixesSolution diffuse Heat flow hot →cold X X X syssyssys STHG ∆−∆=∆ )tan()( treacprosys SSS ∑−∑=∆ + 821 + 1698 Reactant Product kJG G STHG 3071 )877.0(2982810 −=∆ +−−=∆ ∆−∆=∆ kJS JKS S SSS sys sys sys treacproductsys 877.0 877 8211698 1 )tan()( +=∆ +=∆ −=∆ −=∆ − ∆H = - 2810 kJ Assume ∆S, ∆H at constant over Temp ∆G sys < 0 (-ve) Spontaneous→ ∆G sys > 0 (+ve) Non spontaneous→ Gibbs Free Energy, ∆G syssyssys STHG ∆−∆=∆ Unit ∆S - JK-1 Unit ∆H - kJ Unit ∆G - kJ Reactants Products ∆Gsys θ = ∑∆Gf θ (pro) - ∑∆Gf θ (react) ∆Gsys θ = -3792 - (-910) = - 2882 kJ ∆Gsys θ ∆Gf θ (reactant) ∆Gf θ (product) )()( reactfprofsys GGG °° ∆−∆=∆ ∆G < 0 Combustion sugar at 298K - Spontaneous Elements C + H2 + O2 Reactant (-910) Product (- 3792) 14 Is combustion sugar 298K (25C) spontaneous? C6H12O6(s) + 6O2 (g) 6CO→ 2(g) + 6H2O(l) ∆H = - 2810 kJ at 298K C6H12O6 (s) + 6O2(g) 6CO→ 2(g) + 6H2O(l) S0 + 209 +102 x 6 + 213 x 6 + 70 x 6 ∆G < 0 Combustion sugar at 298K - Spontaneous C6H12O6 + 6O2 6CO2 + 6H2O(l) C6H12O6 (s) + 6O2(g) 6CO→ 2(g) + 6H2O(l) ∆G0 - 910 0 - 395 x 6 - 237 x 6 Using Free Energy to predict spontaneity
- 30. Entropy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Predict entropy change - quatitatively Gas mixesSolution diffuse Heat flow hot →cold X X X Reactant Product CH4(g) + 2O2 (g) CO→ 2(g) + 2H2O(l) CH4(g) + 2 O2 (g) CO→ 2(g) + 2 H2O(l) ∆Hf 0 - 74 0 - 393 - 286 x 2 S0 + 186 +205 x 2 +213 + 171 x 2 ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) kJS JKS S SSS sys sys sys treacproductsys 041.0 41 596555 1 )tan()( −=∆ −=∆ −=∆ −=∆ − kJHsys 890)74(964 −=−−−=∆ Is Combustion at 298K spontaneous? Unit for ∆S - JK-1 Unit for ∆H - kJ C3H8(g) + 5O2 (g) 3CO→ 2(g) + 4H2O(l) C3H8(g) + 5 O2 (g) 3 CO→ 2(g) + 4 H2O(l) ∆Hf 0 - 104 0 - 393 x 3 - 286 x 4 S0 +270 +205 x 5 + 213 x 3 + 171 x 4 Reactant Product ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react)∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) kJS JKS S SSS sys sys sys treacproductsys 028.0 28 12951323 1 )tan()( +=∆ +=∆ −=∆ −=∆ − kJHsys 2219)104(2323 −=−−−=∆ 1 2 kJG G STHG 877 )041.0(298890 −=∆ −−−=∆ ∆−∆=∆ ∆G < 0 Combustion sugar at 298K - Spontaneous kJG G STHG 881 )028.0(2982219 −=∆ +−−=∆ ∆−∆=∆ ∆G < 0 Combustion sugar at 298K - Spontaneous
- 31. Entropy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Predict entropy change - quatitatively Gas mixesSolution diffuse Heat flow hot →cold X X X Reactant Product ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) kJS JKS S SSS sys sys sys treacproductsys 118.0 118 18870 1 )tan()( −=∆ −=∆ −=∆ −=∆ − kJHsys 44)242(286 −=−−−=∆ Is Condensation/Freezing at 298K spontaneous? Reactant Product ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) kJS JKS S SSS sys sys sys treacproductsys 022.0 22 7048 1 )tan()( −=∆ −=∆ −=∆ −=∆ − kJHsys 6)286(292 −=−−−=∆ 3 4H2O (g) H→ 2O(l) H2O (l) H→ 2O(s) H2O (g) H→ 2O(l) ∆Hf 0 - 242 - 286 S0 + 188 + 70 H2O (l) H→ 2O(s) ∆Hf 0 - 286 - 292 S0 + 70 + 48 kJG G STHG 1.9 )118.0(2981.44 −=∆ −−−=∆ ∆−∆=∆ ∆G < 0 Condensation at 298K - Spontaneous kJG G STHG 55.0 )022.0(2986 +=∆ −−−=∆ ∆−∆=∆ ∆G > 0 Freezing at 298K – Non Spontaneous
- 32. Entropy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Predict entropy change - quatitatively Gas mixesSolution diffuse Heat flow hot →cold X X X Reactant Product ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) kJHsys 92)0(92 −=−−=∆ Are these rxn at 298K spontaneous? Reactant Product ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react)∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) kJHsys 168)1564(1732 −=−−−=∆ 5 6N2(g) + 3H2(g) 2NH→ 3(g) N2(g) + 3H2 (g) 2NH→ 3(g) ∆Hf 0 0 0 - 46 x 2 S0 + 192 + 131 x 3 + 192 x 2 kJS JKS S SSS sys sys sys treacproductsys 201.0 201 585384 1 )tan()( −=∆ −=∆ −=∆ −=∆ − 4KCIO3(s) 3KCIO→ 4(s) + KCI(s) 4KCIO3(s) 3KCIO→ 4(s) + KCI(s) ∆Hf 0 - 391 x 4 - 432 x 3 - 436 S0 + 143 x 4 + 151 x 3 + 82 kJS JKS S SSS sys sys sys treacproductsys 037.0 37 572535 1 )tan()( −=∆ −=∆ −=∆ −=∆ − kJG G STHG 32 )2.0(29892 −=∆ −−−=∆ ∆−∆=∆ ∆G < 0 NH3 production at 298K - Spontaneous kJG G STHG 157 )037.0(298168 −=∆ −−−=∆ ∆−∆=∆ ∆G < 0 KCIO3 production at 298K - Spontaneous
- 33. Entropy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Predict entropy change - quatitatively Gas mixesSolution diffuse Heat flow hot →cold X X X Reactant Product ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) kJHsys 178)1206(1028 +=−−−=∆ Reactant Product ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react)∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) 7 8CaCO3 (s) CaO→ (s) + CO2(g) CaCO3 (s) CaO→ (s) + CO2(g) ∆Hf 0 - 1206 - 635 - 393 S0 + 93 + 40 + 213 kJS JKS S SSS sys sys sys treacproductsys 16.0 160 93253 1 )tan()( +=∆ +=∆ −=∆ −=∆ − Decomposition at 298K Decomposition at 1500K CaCO3 (s) CaO→ (s) + CO2(g) CaCO3 (s) CaO→ (s) + CO2(g) ∆Hf 0 - 1206 - 635 - 393 S0 + 93 + 40 + 213 kJHsys 178)1206(1028 +=−−−=∆ Rxn Temp dependent Spontaneous at High temp↑ 1500K 298K (25C) Decomposition limestone CaCO3 spontaneous? kJG G STHG 130 )16.0(298178 +=∆ +−+=∆ ∆−∆=∆ ∆G > 0 Decomposition at 298K – Non Spontaneous kJS JKS S SSS sys sys sys treacproductsys 16.0 160 93253 1 )tan()( +=∆ +=∆ −=∆ −=∆ − kJG G STHG 62 )16.0(1500178 −=∆ +−+=∆ ∆−∆=∆ ∆G < 0 Decomposition at 1500 K - Spontaneous At Low Temp At High Temp
- 34. Entropy Why gas mixes and not unmix?Why conc solution diffuse and not undiffuse? Why heat flow from hot to cold? Predict entropy change - quatitatively Gas mixesSolution diffuse Heat flow hot →cold X X X Reactant Product ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) kJS JKS S SSS sys sys sys treacproductsys 022.0 22 7048 1 )tan()( −=∆ −=∆ −=∆ −=∆ − kJHsys 6)286(292 −=−−−=∆ Is Freezing spontaneous? Reactant Product ∆Ssys θ = ∑Sf θ (pro) - ∑Sf θ (react) ∆Hsys θ = ∑∆Hf θ (pro) - ∑∆Hf θ (react) kJS JKS S SSS sys sys sys treacproductsys 022.0 22 7048 1 )tan()( −=∆ −=∆ −=∆ −=∆ − kJHsys 6)286(292 −=−−−=∆ 9 10H2O (l) H→ 2O(s) H2O (l) H→ 2O(s) H2O (l) H→ 2O(s) ∆Hf 0 - 286 - 292 S0 + 70 + 48 H2O (l) H→ 2O(s) ∆Hf 0 - 286 - 292 S0 + 70 + 48 Freezing at 298K (25C) Freezing at 263K (-10C) Rxn Temp dependent Spontaneous at Low temp↓ 263K (-10C)298K (25C) kJG G STHG 55.0 )022.0(2986 +=∆ −−−=∆ ∆−∆=∆ ∆G > 0 Freezing at 298K – Non Spontaneous kJG G STHG 21.0 )022.0(2636 −=∆ −−−=∆ ∆−∆=∆ ∆G < 0 Freezing at 263K – Spontaneous At High Temp At Low Temp
- 35. 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