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Chapter 9 Chemical Quantities in Reactions

Cold packs use an endothermic reaction. Chapter 9 Chemical Quantities in Reactions. 9.5 Energy in Chemical Reactions. Heat of Reaction. The heat of reaction , is the amount of heat absorbed or released during a reaction at constant pressure

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Chapter 9 Chemical Quantities in Reactions

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  1. Cold packs use an endothermic reaction. Chapter 9 Chemical Quantities in Reactions 9.5 Energy in Chemical Reactions

  2. Heat of Reaction The heat of reaction, • is the amount of heat absorbed or released during a reaction at constant pressure • is the difference in the energy of the reactants and the products • is shown as the symbol ΔH ΔH = Hproducts − Hreactants

  3. Endothermic Reactions In an endothermicreaction, • heat is absorbed • the sign of ΔH is + • the energy of the products is greater than the energy of the reactants • heat is a reactant N2(g) + O2(g) + 181 kJ 2NO(g) ΔH = +181 kJ (heat added)

  4. Endothermic Reactions • http://www.youtube.com/watch?v=ubvo5AKzltY • Mix equal quantities of NaHCO3 (sodium hydrogen carbonate) with some citric acid (C6H8O7). Add water and the reaction will start. The reaction is endothermic, so the mixture and container cools down. Na3C6H5O7 (sodium citrate), carbon dioxide and water are the products, which have a total entropy greater than that of the reactants and surroundings. This allows this reaction to be spontaneous, even though it is endothermic.

  5. Endothermic Reactions Ba(OH)2(s)+2NH4SCN(aq) Ba(SCN)2(s)+2NH3(g)+H2O(l) ΔH = +

  6. Exothermic Reactions In an exothermic reaction, • heat is released • the sign of ΔH is - • the energy of the products is less than the energy of the reactants • heat is a product C(s) + O2(g) CO2(g) + 394 kJ ΔH = –394 kJ/mol (heat released)

  7. Exothermic Reaction • 2Al (s)+ 3CuSO4 (aq) Al2(SO4)3 (aq) +3Cu (s) ΔH = ─ • http://www.youtube.com/watch?v=AgBvRMFXHis&feature=related

  8. Summary Reaction Energy Change Heat Sign of ΔH Endothermic Heat absorbed Reactant + Exothermic Heat released Product ─

  9. Learning Check Identify each reaction as (Ex) exothermic or (En) endothermic. A. N2(g)+ 3H2(g) 2NH3(g) + 92 kJ B. CaCO3(s) + 556 kJ CaO(s) + CO2(g) C. 2SO2(g) + O2(g) 2SO3(g) + heat

  10. Solution Identify each reaction as (Ex) exothermic or (En) endothermic. (Ex) A. N2(g)+ 3H2(g) 2NH3(g) + 92 kJ (En) B. CaCO3(s) + 556 kJ CaO(s) + CO2(g) (Ex) C. 2SO2(g) + O2(g) 2SO3(g) + heat

  11. Calculations Using Heat of Reaction

  12. Heat Calculations for Reactions In the reaction N2(g) + O2(g)2NO(g) ΔH = +181 kJ 181 kJ is absorbed when 1 mol of N2 and 1 mol of O2 react to produce 2 mol of NO. N2(g) + O2(g) + 181 kJ 2NO(g) This can be written as conversion factors. 181 kJ181 kJ 181 kJ 1 mol N2 1 mol O2 2 mol NO

  13. Heat Calculations for Reactions (continued) N2(g) + O2(g) + 181kJ 2NO(g) If 15.0 g of NO is produced, how many kJ was absorbed? 1) 1400 kJ 2) 90.4 kJ 3) 45.2 kJ

  14. Solution STEP 1 List given and needed data for the equation. Given: 15.0 g of NO produced ΔH = 181 kJ/2 mol of NO Need: kJ absorbed STEP 2Write a plan using heat of reaction and any molar mass needed. Plan: g of NO moles of NO kJ

  15. Solution (continued) STEP 3Write the conversion factors including heat of reaction. 2 mol of NO = 181 kJ 181 kJ and 2 mol NO 2 mol NO 181 kJ 1 mol of NO = 30.01 g of NO 1 mol NO and 30.01 g NO 30.01 g NO 1 mol NO STEP 4Set up the problem. 15.0 g NO x 1 mol NO x 181 kJ = 45.2 kJ (3) 30.01 g NO 2 mol NO

  16. Learning Check How many grams of O2 react if 1280 kJ is released in the following reaction? CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -890 kJ 1) 92.0 g of O2 2) 46.0 g of O2 3) 2.87 g of O2

  17. Solution STEP 1 List given and needed data for the equation. Given 1280 kJ Need grams of O2 STEP 2Write a plan using heat of reaction and any molar mass needed. kilojoules moles of O2 grams of O2

  18. Solution (continued) STEP 3Write the conversion factors including heat of reaction. 2 mol of O2 = 890 kJ 890 kJ and 2 mol O2 2 mol O2 890 kJ 1 mol of O2 = 32.00 g of O2 1 mol O2 and 32.00 g O2 32.00 g O2 1 mol O2

  19. Solution (continued) STEP 4Set up the problem. 1280 kJ x 2 mol O2 x 32.00 g O2 = 92.0 g of O2 (1) 890 kJ 1 mol O2

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