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Thermochemistry

Thermochemistry. Part 3 - Enthalpy and Thermochemical Equations. ENTHALPY. Enthalpy = a type of chemical energy (thermodynamic potential), sometimes referred to as “heat content”. Enthalpies of Reaction .

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Thermochemistry

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  1. Thermochemistry Part 3 - Enthalpy and Thermochemical Equations

  2. ENTHALPY • Enthalpy = a type of chemical energy (thermodynamic potential), sometimes referred to as “heat content”

  3. Enthalpies of Reaction • The enthalpy change (in kJ/mol) that accompanies a chemical reaction is called the enthalpy of reaction • DHrxn • Also called heat of reaction

  4. Enthalpies of Reaction • If DHrxn = negative • exothermic • heat is given off • Under conditions of constant pressure, q = ΔH < 0 (negative sign) R H P time

  5. Enthalpies of Reaction • If DHrxn = positive • endothermic • heat must be added • Under conditions of constant pressure, q = ΔH > 0 (positive sign) P R H time

  6. Thermochemical Equations • Thermochemical equationsare balanced chemical equations that show the associated enthalpy change (H) • balanced equation • enthalpy change (DHrxn)

  7. Thermochemical Equations • An example of a thermochemical equation: CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH = -890. kJ • The coefficients in the balanced equation show the # moles of reactants and products that produced the associated DH. • If the number of moles of reactant used or product formed changes, then the DH will change as well.

  8. Thermochemical Equations • For the following reaction: CH4(g) + 2O2(g) CO2(g) + 2H2O (l) DH = -890. kJ -890. kJ -890. kJ 1 mol CH42 mol O2 -890. kJ -890. kJ 1 mol CO22 mol H2O

  9. Thermochemical Equations • The thermochemical equation for burning 1 mole of CH4 (g): CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH = -890. kJ • When 1 mole of CH4 is burned, 890. kJ of heat are released. • When 2 moles of CH4 are burned, 1780. kJ of heat are released.

  10. Rules of Thermochemistry Rule #1 - The magnitude of H is directly proportional to the amount of reactant consumed and product produced.

  11. Rules of Thermochemistry Example 1: H2 + Cl22HCl H = - 185 kJ Calculate H when 1.00 g of Cl2 reacts. This thermochemical equation says that there is 185 kJ of energy released per how many moles of Cl2?(look at the balanced equation) For every 1 mole (coefficient) Cl2 = -185 kJ (released) -185 kJ = -2.61 kJ 1 mol 1 molCl2 x 70.90 g Cl2 ΔH = 1.00 g Cl2x

  12. Rules of Thermochemistry Example 2: When an ice cube weighing 24.6 g of ice melts, it absorbs 8.19 kJ of heat. Calculate H when 1.00 mol of solid water melts. 8.19 kJ = 6.00 kJ 24.6 g 18.02 gx1 mol ΔH = 1.00 molx

  13. Rules of Thermochemistry Example 3: Methanol burns to produce carbon dioxide and water: 2CH3OH + 3O2 2CO2 + 4H2O + 1454 kJ What mass of methanol is needed to produce 1820 kJ? Does “produce” mean exo or endo? + or – sign? 80.2

  14. Rules of Thermochemistry Example 4: How much heat is produced when 58.0 liters of hydrogen (at STP) are also produced? Zn + 2HCl  ZnCl2 + H2 + 1250 kJ

  15. Rules of Thermochemistry Rule #2 - H for a reaction is equal in the magnitude but opposite in sign to H for the reverse reaction. (If 6.00 kJ of heat absorbed when a mole of ice melts, then 6.00 kJ of heat is given off when 1.00 mol of liquid water freezes)

  16. Rules of Thermochemistry 2 H2O2 (l) 2 H2O (l) + O2(g) DH = -196 kJ 2 H2O (l) + O2(g) 2 H2O2 (l) DH = +196 kJ Remember “elephant’s toothpaste?”

  17. Rules of Thermochemistry Example: Given: H2 + ½O2 H2O H = -285.8 kJ Calculate H for the equation: 2H2O  2H2 + O2 …so do the same to ∆H Flip eq’n and x 2 H = 2 x –(-285.8 kJ) H = 571.6 kJ

  18. Rules of Thermochemistry Rule #3) The value of H for a reaction is the same whether it occurs in one step or in a series of steps. H for the overall equation is the sum of the H’s for the individual equations: Hess’s Law: H = H1 + H2 + …

  19. Hess’s Law: H = H1 + H2 + … Example 1:Calculate H for the reaction: C + ½O2 CO Given: 1) C + O2 CO2H = -393.5 kJ 2) 2CO + O2 2CO2H = -566.0 kJ  2flip 2) CO2  CO + O2H = +283.0 kJ C + ½O2CO H = -110.5 kJ

  20. Hess’s Law: H = H1 + H2 + … Example 2:Find the heat of reaction (enthalpy) for the following reaction NO + ½O2 NO2H = ? Given the following equation…. ½N2 + ½O2 NO H = +90.4 kJ ½N2 + O2NO2H = +33.6 flip NO  ½N2+ ½O2H = -90.4 kJ NO + ½O2 NO2 H = -56.8 kJ

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