IB Chemistry on Hess's Law, Enthalpy Formation and Combustion

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Transcript of IB Chemistry on Hess's Law, Enthalpy Formation and Combustion

  1. 1. CO(g) H2 H2 H1 = -394 CO2(g) C(s) + O2(g) CO(g) + O2 C(s) + O2(g) CO2(g) A E B C D H6H5 H4H2 Hesss Law C(s) + O2(g) CO2(g) Overall H rxn is independent ofits pathway H rxn in series steps = sum of enthalpychangesfor individualsteps H1 H2 H4 H1 H3 EnergyLevel or Cycle Diagram to find H State function- propertyof system whose magnitude dependon initial and final state H3 A D A B C D H1 = H2 + H3 + H4 H A E is same regardlessof its path Final state A E A C D E H1 = H2 + H3 + H4 A E A B E H1 = H5 + H6 Initial stateInitial state Final state H3 = -283 H1 = H2 + H3 EnergyLevel Diagram 2 2 1 O H2 = H1 - H3 = -394 +283 = -111 kJ mol-1 H3 = -283 2 2 1 O EnergyCycle Diagram H1 = -394 H2 = H1 - H3 = -394 +283 = -111 kJ mol-1 Path not impt !!!! H1 = H2 + H3 C(s) + O2 CO2 (g) H1 = -394 C (s) + O2 CO(g) H2 = ??? CO(g) + O2 CO2 (g) H3 = -283+ Hesss Law Find H cannot be measured directly/experimentally C(s) + 1/2O2 CO(g) H2 ????
  2. 2. SO2(g) H2 H2 H1 = -395 SO3(g) S(s) + O2(g) SO2(g) + O2 S(s) + 3/2O2(g) SO3(g) A E B C D H6H5 H4H2 Hesss Law S(s) + 3/2O2(g) SO3(g) Overall H rxn is independent ofits pathway H rxn in series steps = sum of enthalpychangesfor individualsteps H1 H2 H4 H1 H3 EnergyLevel or Cycle Diagram to find H State function- propertyof system whose magnitude dependon initial and final state H3 A D A B C D H1 = H2 + H3 + H4 H A E is same regardlessof its path Final state A E A C D E H1 = H2 + H3 + H4 A E A B E H1 = H5 + H6 Initial stateInitial state Final state H3 = -98 H1 = H2 + H3 EnergyLevel Diagram Find H cannot be measured directly/experimentally H2 = H1 - H3 = -395 + 98 = - 297 kJ mol-1 H3 = - 98 2 2 1 O EnergyCycle Diagram H1 = -395 H2 = H1 - H3 = - 395 + 98 = - 297 kJ mol-1 Path not impt !!!! Hesss Law H1 = H2 + H3 S(s) + 3/2O2 SO3 (g) H1 = -395 S(s) + O2 SO2(g) H2 = ??? SO2(g) + O2 SO3 (g) H3 = -98+ O2 S(s) + O2 SO2(g) H2 ?????
  3. 3. N2(g) + 2O2(g)N2(g) + 2O2(g) N2O4(g) 2NO2(g) H1 N2O4(g) H1 H2 = + 33 2NO2(g) A E B C D H6H5 H4H2 Hesss Law N2(g) + 2O2(g) 2NO2(g) Overall H rxn is independent ofits pathway H rxn in series steps = sum of enthalpychangesfor individualsteps H1 H2 H4 H1 H3 EnergyLevel or Cycle Diagram to find H State function- propertyof system whose magnitude dependon initial and final state H3 A D A B C D H1 = H2 + H3 + H4 H A E is same regardlessof its path Final state A E A C D E H1 = H2 + H3 + H4 A E A B E H1 = H5 + H6 Initial stateInitial state Final state H3 = + 9 EnergyLevel Diagram Find H cannot be measured directly/experimentally H3 = + 9 EnergyCycle Diagram H2 = + 33 H1 = H2 + H3 = -33 + 9 = - 24 kJ mol-1 Path not impt !!!! Hesss Law H1 = H2 + H3 N2(g)+ 2O2 2NO2(g) H2 = +33 2NO2 N2+ 2O2 H2 = - 33 N2(g) + 2O2 N2O4(g) H3 = + 9+ 2NO2(g) N2O4(g) H1 = ? H1 = H2 + H3 H1 = H2 + H3 = -33 + 9 = - 24 kJ mol-1 inverse 2NO2(g) N2O4(g) H = -24
  4. 4. Hf (reactant) Hf (product) Using Std Hf formationto find H rxn H when 1 mol form from its element under std condition Na(s) + CI2(g) NaCI (s) Hf = - 411 kJ mol -1 Std Enthalpy Changes H Std condition Pressure 100kPa Temp 298K Conc 1M All substance at std states Hesss Law Std Hf formation Mg(s) + O2(g) MgO(s) Hf =- 602 kJ mol -1 Reactants Products O2(g) O2 (g) Hf = 0 kJ mol -1 Hrxn = Hf (products) - Hf (reactants) Hf (products)Hf (reactants) Hrxn Elements Std state solid gas 2C(s) + 3H2(g)+ O2(g) C2H5OH(I) Hf =- 275 kJ mol -1 1 mole formed H2(g) + O2(g) H2O(I) Hf =- 286 kJ mol -1 Std state solid gas 1 mol liquid For element Std Hf formation= 0 Mg(s) Mg(s) Hf = 0 kJ mol -1 No product form Using Std Hf formationto find H of a rxn Click here chem database (std formation enthalpy) Click here chem database (std formation enthalpy) C2H4 + H2 C2H6 Find H rxn using std H formation Reactants Products 2C + 3H2 Elements C2H4 + H2 C2H6 Hrxn Hrxn = Hf (pro) - Hf (react) Hrxn = Hf C2H6 - Hf C2H4+ H2 = - 84.6 ( + 52.3 + 0 ) = - 136.9 kJ mol -1
  5. 5. 2CH4(g) + 4O2(g) 4H2O + 2CO2(s) Hc = - 890 x 2 = - 1780 kJ mol -1 Std Hf formationto find H rxn H when 1 mol form from its element under std condition Na(s) + CI2(g) NaCI (s) Hf = - 411 kJ mol -1 Hesss Law Std Hf formation Mg(s) + O2(g) MgO(s) Hf =- 602 kJ mol -1 Reactants Products Hrxn = Hf (products) - Hf (reactants) Hf (products)Hf (reactants) Hrxn Elements Std state solid gas 1 mole formed Total amt energyreleased/absorbed mol reactants CH4(g) + 2O2(g) 2H2O + CO2(s) Hc = - 890 kJ mol -1 H reverse EQUAL in magnitude but opposite sign to H forward Na+ (g) + CI_ (g) NaCI(s) Hlatt = - 770 kJ mol -1 NaCI(s) Na+ (g) + CI_ (g) Hlatt = + 770 kJ mol -1 H2(g) + O2(g) H2O(I) Hf =- 286 kJ mol -1 H2(g) + O2(g) H2O(I) Hc =- 286 kJ mol -1 Compound NaF NaCI NaBr NaI Hf (kJ mol-1) -573 -414 -361 -288 More ve formation More heat releasedto surrounding More energetically stable (lower in energy) Do not decomposeeasily Subs Na2O MgO AI2O3 Hf -416 -602 -1670 Subs P4O10 SO3 CI2O7 Hf -3030 -390 +250 1 mole formed 2 mole formedx 2 Hf formationvs Hc combustion H Form - std state liquid H Comb - std state liquid More ve more stable Across Period3 H more ve Lower in energy Oxides more stable Across Period3 H more +ve Higherin energy Oxides less stable decomposeeasily Hf = Hc
  6. 6. Hrxn C2H4 + H2 C2H6 C2H6 + 3.5 O2 2CO2 + 3 H2O Hf (reactant) Hf (product) Hesss Law Std Hf formationto find H rxn Reactants Products Hrxn Hf (product)Hf (reactant) Elements Hf - 85 0 - 393 - 286 Reactants Products Hrxn = Hf (pro) - Hf (react) Hrxn = - 1644 - ( - 85 ) = - 1559 kJ mol -1 - 85 0 - 858 - 786 x 2 x 3 C2H4 + H2 C2H6 Hrxn Reactants Products 2C + 3H2 Hf + 52 o - 85 Reactants Products Hrxn = Hf (pro) - Hf (react) Hrxn = - 85 - ( + 52 ) = - 137 kJ mol -1 C2H6 + 3.5 O2 2CO2 + 3 H2O 2C + 3H2 + 3.5O2 EnergyLevel Diagram 2CO2 + 3H2O 2C + 3H2 + 3.5O2 C2H6 + 3.5 O2 Hf = -85 Hf = - 393 Hf = 0 2C + 3H2 + 3.5O2 Elements Reactants Products Hf = - 286 Hrxn= (- 393 x 2 + -286 x 3) (- 85 + 0) = - 1559 kJ mol-1 x 2 x 3 C2H4 + H2 Reactants C2H6 Hrxn = - 85 ( + 52 + 0 ) = - 137 kJ mol-1 2C + 2H2 + H2 Hf = + 52 Hf = 0 Hrxn Products 2C + 3H2 Elements Hf = - 85 Elements
  7. 7. Hf (reactant) Hf (product) Hf (reactant) Hf (product) Using Std Hf formationto find H rxn Hesss Law Reactants Products Hrxn = Hf (products) - Hf (reactants) Hf (products)Hf (reactants) Hrxn Elements 2H2S + SO2 3S + 2H2O Find H rxn using std H formation Reactants Products 3S + O2 + 2H2 Elements Hrxn Hrxn = Hf (pro) - Hf (react) Hrxn = - 572 - ( - 338 ) = - 234 kJ mol -1 2H2S + SO2 3S + 2H2O Hf - 20.6 - 297 0 - 286 - 41.2 - 297 0 - 572 x 2 x 2 Reactants Products Using Std Hf formationto find H rxn Reactants Products Hrxn Hf (product)Hf (reactant) 4C + 12H2 + 9N2 + 10O2 Elements Hf + 53 - 20 - 393 - 286 0 Reactants Products Hrxn = Hf (pro) - Hf (react) Hrxn = - 5004 - ( +112 ) = - 5116 kJ mol -1 4CH3NHNH2 + 5N2O4 4CO2 + 12H2O + 9N2 4CH3NHNH2 + 5N2O4 4CO2 + 12H2O + 9N2 + 212 - 100 - 1572 - 3432 0 x 4 x 5 x 4 x 12 NH4NO3 N2O + 2H2O Hrxn Reactants Products N2 + 2H2 + 3/2O2 NH4NO3 N2O + 2H2O Hf - 366 + 82 - 286 x 2 Reactants Products Hrxn = Hf (pro) - Hf (react) Hrxn = - 488 - ( - 366 ) = - 122 kJ mol -1 Elements
  8. 8. Hc (reactant) Hc (product) Hc combustion to find H formationof hydrocarbon H when 1 mol completelyburnt in oxygen under std condition Std Enthalpy Changes H Std condition Pressure 100kPa Temp 298K Conc 1M All substance at std states Hesss Law Std Hc combustion C(s) + O2(g) CO2(g) Hc = - 395 kJ mol -1 Reactants Products Hrxn = Hc (reactant) - Hc (product) Hc (product)Hc (reactant) Hrxn Combusted products 1 mole combusted H2(g) + O2(g) H2O(I) Hc = - 286 kJ mol -1 Std Hc combustionto find H of a rxn Click here chem database (std combustion enthalpy) Click here chem database (std combustion enthalpy) Find H formation using std H comb Reactants Products 2CO2 + 3H2O Hrxn Hrxn = Hc (react) - Hc (pro) Hrxn = Hc ( C + H2 )- Hc C2H6 = 2 x (-395) + 3 x (-286) (-1560) = - 88 kJ mol -1 H2(g) + O2(g) H2O(I) Hc = - 286 kJ mol -1 C(s) + O2(g) CO2(g) Hc = - 395 kJ mol -1 Combustion H2 = formationof H2O Hc = Hf Combustion C = formationof CO2 Hc = Hf Using combustion data 2C(s) + 3H2(g) C2H6(g) Hf = - 84 kJ mol -1 2C(s) + 3H2(g) + O2(g) C2H5OH(I) Hf = - 275 kJ mol -1 2C + 3H2 C2H6 + 3.5O2 + 3.5O2 x 2 x 3 How enthalpy formationhydrocarbonobtained ?
  9. 9. -790 -858 - 1371 2 C + 3H2 + 3.5O2 C2H5OH Hc (reactant) Hc (product) Hrxn = Hc (react) - Hc (pro) = 2 x (-395) + 3 x (-286) (-1560) = - 88 kJ mol -1 Reactants Products 2 C + 3H2 + 3.5O2 C2H5OH 2C + 3H2 C2H6 2C + 3H2 C2H6 Hc comb to find Hf formationof hydrocarbon Hrxn Hesss Law Reactants Products Hrxn Hc (product)Hc (reactant) Hc -395 -286 - 1560 Reactants Products - 790 -858 - 1560 x 2 Hrxn Reactants Products EnergyLevel Diagram Hc = -395 Hc = - 1560 Hc = -286 Combustedproducts Reactants Products Hrxn = (- 395 x 2 + -286 x 3) (-1560 ) = - 88 kJ mol-1 Reactants Hrxn Products Hc = - 1371 2CO2 + 3H2O x 3 2C + 3H2 C2H6 2CO2 + 3H2O 2CO2 + 3H2O x 2 x 3 2 C + 3H2 + 3.5O2 2CO2 + 3H2O Hc -395 -286 - 1371 Hrxn = Hc (react) - Hc (pro) = 2 x (-395) + 3 x (-286) (-1371) = - 275 kJ mol -1 x 2 x 3 C2H5OH Hc = -395 x 2 Hc = -286 x 3 2CO2 + 3H2O Combustedproducts 2CO2 + 3H2O Hrxn = (- 395 x 2 + -286 x 3) (-1371 ) = - 275 kJ mol-1 + 3.5O2 + 3.5O2 + 3.5O2 + 3.5O2
  10. 10. 3C2H2 C6H6 Hc (reactant) Hc (product) Hrxn = Hc (react) - Hc (pro) = 6 x (-395) + 3 x (-286) (-3271) = + 49 kJ mol -1 Reactants Products 3 C2H2 C6H6 6C + 3H2 C6H6 6C + 3H2 C6H6 Hc comb to find Hf formationof hydrocarbon Hrxn Hesss Law Reactants Products Hrxn Hc (product)Hc (reactant) Hc -395 -286 - 3271 Reactants Products -2370 -858 - 3271 x 6 Hrxn Reactants Products EnergyLevel Diagram Hc = -395 Hc = - 3271 Hc = -286 Combustedproducts Reactants Products Hrxn = (-