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Heat of Reaction

Heat of Reaction. Chapter 10.3. Chemical Energy and the Universe. Warming your hands with a heat pack involves the following chemical reaction: 4Fe (s) + 3O 2(g)  2Fe 2 O 3 (s) + 1625 kJ The pack and its contents would be called the system .

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Heat of Reaction

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  1. Heat of Reaction Chapter 10.3

  2. Chemical Energy and the Universe • Warming your hands with a heat pack involves the following chemical reaction: 4Fe(s) + 3O2(g) 2Fe2O3(s) + 1625 kJ • The pack and its contents would be called the system. • Everything else is called the surroundings. • The surroundings + the system = the universe. • When energy flows from the system to the surroundings, heat is released and the reaction is exothermic. • When the flow of heat is reversed, heat flows from the surroundings into the system and the reaction is endothermic.

  3. Exothermic & Endothermic • Recall the reaction in the heat pack: 4Fe(s) + 3O2(g) 2Fe2O3(s) + 1625 kJ • The heat is on the product side—energy is released; it is exothermic. • If energy is located on the reactant side, the reaction is endothermic as in the following reaction of a cold pack: 27 kJ + NH4NO3(s) NH4+(aq) + NO3-

  4. Heat of Reaction • The heat content of a system is called the enthalpy. • Change in enthalpy (or heat) is represented by ΔHrxn ΔHrxn =H products – H reactants • ΔHrxn for exothermic reactions are always negative. • In the heat pack example: 4Fe(s) + 3O2(g) 2Fe2O3(s)ΔHrxn = - 1625 kJ • ΔHrxn for endothermic reactions are always positive.

  5. Thermochemical Equations • A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products, and the energy change, usually expressed in enthalpy (ΔH). • Heat Pack Example: 4Fe(s) + 3O2(g) 2Fe2O3(s)ΔH= - 1625 kJ • Coefficients refer to the number of moles. Therefore, -1625 kJ is the ΔH when 2 mol of Fe2O3 is formed from4Fe(s) and 3O2(g).

  6. The Rules of Thermo Equations • ΔH is directly proportional to the quantity of a substance or mass. • Example: 4Fe(s) + 3O2(g) 2Fe2O3(s)ΔH= - 1625 kJ 8Fe(s) + 6O2(g) 4Fe2O3(s)ΔH= - 3250 kJ • ΔH for a reaction is equal in magnitude but opposite in sign for the reverse reaction. • Example: 4Fe(s) + 3O2(g) 2Fe2O3(s) ΔH= - 1625 kJ 2Fe2O3(s)  4Fe(s) + 3O2(g)ΔH= + 1625 kJ

  7. Variations on ΔH • Energy released in burning is called heat of combustion (ΔHcomb) • Energy required to vaporize one mole of a liquid (liqgas) is called heat of vaporization (ΔHvap) • Energy required to melt one mole of a solid substance (sol liq) is called heat of fusion (ΔHfus) • Because melting and vaporizing are endothermic processes, the ΔH value is positive. • What sign would the ΔH have for freezing and for condensation? Are these processes endo- or exothermic?

  8. Important! • Heat can be used to change the temperature of a substance OR • Change the physical state of a substance. • Heat CANNOT do both at the same time. • For changes of state, you calculate heat with dimensional analysis using the ΔH as a conversion factor.

  9. Example 1 • How much heat is required to vaporize 343 g of liquid ethanol, CH3OH, at its boiling point? (ΔHvap= 38.6 kJ/mol) • Use dimensional analysis to solve for heat (q) • 414 kJ 343 g 1 mol 38.6 kJ =? 32.034 g 1 mol

  10. Example 2 • How much heat is evolved when 1255 g of water condenses to a liquid at 100 °C? (Hint: ΔHvap = +40.7 kJ/mol; therefore ΔHcond = -40.7 kJ/mol) • Using dimensional analysis • -2838 kJ 1255 g 1 mol -40.7 kJ = ? 18.012 g 1 mol

  11. Calculating the Heat of Reaction • To calculate the ΔH of a reaction, follow these simple steps • Balance the chemical equation • Use the numbers given in this equation: ΔHrxn =H products – Hreactants • Use your balanced equation and multiply each reactant and product by the coefficient. • Solve

  12. Example 3- Tying it together • Hydrogen and fluorine gas can be combined to create hydrofluoric acid. The heat of formation for H2 = 436 kJ/mol, F2 = 158 kJ/mol, and HF = 568 kJ/mol. What is the heat of this reaction? • H2 + F2 2HF • ΔHrxn = ( 436 + 158) – (2 x 568) = ? • ΔHrxn = -542 kJ/mol

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