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Hess’s Law

Hess’s Law. Hess’s Law. Some enthalpy changes can not be measured directly in the lab. Hess’s Law is used to calculate these enthalpy changes Use the information provided to calculate ∆H f CH 4. ∆H f CH 4. C (gr) + 2H 2(g). CH 4(g). 2 x H c H 2 2 x -286 = -572. H c C -394.

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Hess’s Law

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  1. Hess’s Law

  2. Hess’s Law • Some enthalpy changes can not be measured directly in the lab. • Hess’s Law is used to calculate these enthalpy changes • Use the information provided to calculate ∆Hf CH4

  3. ∆Hf CH4 C(gr) + 2H2(g) CH4(g) 2 x Hc H2 2 x -286 = -572 Hc C -394 + 2O2(g) Hc CH4 -890 CO2(g) + 2H2O(g)

  4. ∆Hf CH4 = -76kJmol-1

  5. ∆Hf CH4 C(gr) + 2H2(g) CH4(g) 2 x Hc H2 2 x -286 = -572 Hc CH4 -890 Hc C -394 CO2(g) + 2H2O(g) ∆Hf CH4 = Hc C + 2 x Hc H2- Hc CH4 ∆Hf CH4 = -394 + (2 x -286) - (-890) = -76kJmol-1

  6. Hess’s Law • Hess’s Law states • In a reaction the total enthalpy change is independent of the route

  7. Hess’s Law Diagrams • Write unknown equation across the top • Look at info provided & complete triangle • Balance equations (ignore O2) • Think about definitions of ∆Hf & ∆Hc • Make sure arrows point in the correct directions • Do you need to multiply any values? • Write figures on arrows if this helps

  8. Hess’s Law Diagrams • Draw overall direction arrows • Write out in full ∆H= expression without no. • In same direction as arrow + • Against direction of arrow - • Put numbers into ∆H= expression, keep ( ) • Be very careful with signs • Calculate value – check over !!!

  9. Hess’s Law Diagrams • Draw a Hess’s Law diagram to calculate the Hr of the coffee can reaction • CaO(s) + H2O(l) →Ca(OH)2 (s) Hf CaO(S) =-635 Hf H2O(l) = -286 Hf Ca(OH)2(S) = -986

  10. ∆Hr CaO(s) + H2 O(l) Ca(OH)2(s) Hf H2O -286 Hf CaO -635 Hf Ca(OH)2 -986 Ca(s) + O2(g) + H2(g) ∆Hr = -Hf CaO(s)- Hf H2O(l)+ HfCa(OH)2 (s) ∆Hf CH4 = -(-635) – (-286) + (-986) = -65kJmol-1

  11. Hess’s Law • Hess’s Law states • In a reaction the total enthalpy change is independent of the route • Draw a Hess’s Law diagram to calculate the Hr of the decomposition of NaHCO3(s) • 2NaHCO3(s) →Na2CO3(s) + CO2(g) + H2O(l)

  12. ∆Hr 2NaHCO3(s) Na2CO3(s) + H2 O(l) + CO2(g) Hf Na2CO3 -1131 Hf CO2 -394 2 x Hf NaHCO3 2 x -948 Hf H2O -286 2Na(s) + 3O2(g) + H2(g) + 2C(gr) ∆Hr = - 2 x Hf NaHCO3(s)+ Hf Na2CO3+ Hf H2O(l)+ HfCO2 (g) ∆Hr = -(2 x -948) + (-1131) + (-286) + (-394) = +85kJmol-1

  13. Hess’s Law • Draw a Hess’s Law diagram to calculate the following : • Hr of the formation of propane C3H8(g) • 3C(gr) + 4H2(g) →C3H8(g) • Hr of photosynthesis • 6CO2(g) + 6H2O →C6H12O6(aq) + 6O2(l)

  14. ∆Hf C3H8 3C(gr) + 4H2(g) C3H8(g) + 3O2(g) 4 x Hc H2 4 x -286 = -1144 3 x Hc C 3 x -394 = -1182 Hc C3H8 -2220 3CO2(g) + 4H2O(g) ∆Hf C3H8 = 3x Hc C + 4 x Hc H2- Hc C3H8 ∆Hf C3H8 = (3 x -394) + (4 x -286) - (-2220) = -106kJmol-1

  15. ∆Hr 6CO2(g) + 6H2O(l) C6H12O6 (s) + 6O2 (g) Hf C6H12O6 -1273 6 x Hf CO2 6 x -394 6 x Hf H2O 6 x -286 6C(gr) + 9O2(g) + 6H2(g) ∆Hr = - 6 x Hf CO2(g)– 6 x Hf H2O(l)+ Hf C6H12O6 ∆Hr = -(6 x -394) - (6 x -286) + (-1273) = +2807kJmol-1

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