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Hess ’ s Law Energetics

Hess ’ s Law Energetics. 1. Enthalpy. Enthalpy is the heat absorbed or released during a chemical reaction. D H endothermic. 2. Enthalpy of a reaction. - D H exothermic. Energy and Enthalpy Changes. It is impractical to measure absolute amounts of energy or enthalpy.

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Hess ’ s Law Energetics

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  1. Hess’s LawEnergetics 1

  2. Enthalpy • Enthalpy is the heat absorbed or released during a chemical reaction. • DH endothermic 2

  3. Enthalpy of a reaction • - DH exothermic

  4. Energy and Enthalpy Changes • It is impractical to measure absolute amounts of energy or enthalpy. • Enthalpy is always measured relative to previous conditions. 4

  5. Standard Enthalpy of Formation, DHfo • The enthalpy change to form 1 mole of compound from its elements with substances in their standard states. • Standard conditions: 1.0 atm pressure 1.0 M for solutions, 25.0 oC temperature Note: The standard enthalpy change for an element is zero.

  6. Standard State of substances • The pure and most stable form of a substance at standard conditions is said to be in the standard state. • For example, the standard state of oxygen is O2(g) not O3(g) because ozone is less stable that oxygen gas. 6

  7. Stability • -DHfo If the standard heat of formation is negative, the substance is stable • DHfo If the standard heat of formation is positive, the substance is not stable. • In nature, lower energy substances are more stable.

  8. Which substances are NOT stable? Substance DHfo NaCl -411 NO +90 H2O -286 CO2 -393

  9. Write the equation for the formation of: Water: H2(g) + ½ O2(g)  H2O(l) Salt: Na(s) + ½ Cl2(g)  NaCl(s) Glucose: 6 C(s) + 3O2(g) + 6 H2(g)  C6H12O6(s)

  10. Standard Enthalpy Changes • The enthalpy change that occurs when the reactants are converted to products, in their standard states is known as the standard enthalpy change. • It is designated as DHo. 10

  11. Calculations of Enthalpy with DHfo DHo = SDHfo products - S DHfo reactants In English: The enthalpy of a reaction = The sum of the enthalpies of formation of the products – The sum of the enthalpies of formation of the reactants

  12. Find the enthalpy of the reaction DHo 2C2H2+ 5O2 4CO2 + 2H2O Substance DHfo kJ/mole C2H2 227 CO2 -394 H2O -286 = (4(-394) + 2(-286)) – (2(227 + 5(0)) = -2602 kJ/reaction

  13. Calculate the energy released if 260.0 g of C2H2 were burned 2C2H2+ 5O2 4CO2 +2H2O +2602 kJ C2H2+ 5/2O22CO2 + H2O + 1301 kJ 260.0g / 26.0 g/mole = 10.0 mole 10.0 moles x 1301 kJ = 13010 kJ

  14. Hess’ Law • If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy change for the individual steps • Hess’ Law provides a way to calculate enthalpy changes even when the reaction cannot be performed directly. 14

  15. Hess’ Law: Example 1 N2 (g) + O2 (g)  2 NO (g) DH1 = +181 kJ 2 NO(g) + O2 (g)  2 NO2 (g) DH2 = -113 kJ Find the enthalpy change for N2 (g) + 2 O2 (g)  2 NO2 (g) Make the first two reactions add up to the third one and add the enthalpies 15

  16. Hess’ Law: Example 1 Solution: N2 (g) + O2 (g)  2 NO (g) DH1 = +181 kJ 2 NO(g) + O2 (g)  2 NO2 (g) DH2 = -113 kJ ------------------------------------------------------------- N2(g) +2O2 (g)  + 2 NO2 (g) DH =DH1 + DH2 = +181 kJ +(-113) = + 68 kJ 16

  17. Hess Law: Example 2 From the following reactions and enthalpy changes: 2 SO2 (g) + O2 (g)  2 SO3 (g) DH = -196 kJ 2 S (s) +3 O2 (g)  2 SO3 (g) DH = -790 kJ Find the enthalpy change for the following reaction: S (s) + O2 (g)  SO2 (g) Solution: 2 SO3 (g) 2 SO2 (g) + O2 (g)DH = +98 kJ 2 S (s) +3O2 (g)  2 SO3 (g) DH = -395 kJ -------------------------------------------------------------------------------------------------------------- Reversing first equation reverses the sign of DH. Divide this by 2 Divide the second equation by 2 17

  18. Solution SO3 (g) SO2 (g) +1/2 O2 (g) DH = +98 kJ S (s) +3/2 O2 (g)  SO3 (g) DH = -395 kJ -------------------------------------------------------------------------------------------------------------- S (s) + O2 (g)  SO2 (g) -297 kJ The reaction is exothermic Energy is released.

  19. Enthalpy Cycles • Enthalpy Cycles are used to visualize Hess’s Law.

  20. Enthalpy Cycles Look: 1 + 2 = 3

  21. 2NaHCO3 (s) → Na2CO3 (s) + CO2 (g) + H2O (l)

  22. 2NaHCO3 (s) → Na2CO3 (s) + CO2 (g) + H2O (l) ∆HΘ[reaction] = ∆HfΘ[products] −∆HfΘ[reactants] Compound ∆HfΘ (kJ mol-1) NaHCO3 (s) -951 Na2CO3 (s) -1131 CO2 (g) -394 H20 (l) -286

  23. Enthalpy Cycles and Hess’s Law ΔH    +    ΔH2   =    ΔH1 Hence:    ΔH    =     ΔH1    -     ΔH2

  24. Can you read this?

  25. C2H2 + H2 C2H6

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