Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 9 Chemical Quantities in Reactions 9.5 Energy in Chemical Reactions Cold packs use an endothermic reaction.

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 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 = H products − H reactants

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 Endothermic Reactions In an endothermic reaction, 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 N 2 (g) + O 2 (g) kJ 2NO(g) ΔH = +181 kJ (heat added)

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 4 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) + O 2 (g) CO 2 (g) kJ ΔH = –394 kJ/mol (heat released)

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 Summary Reaction Energy Change Heat Sign of ΔH Endothermic Heat absorbed Reactant + Exothermic Heat released Product ─

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 Learning Check Identify each reaction as (Ex) exothermic or (En) endothermic. A. N 2 (g)+ 3H 2 (g) 2NH 3 (g) + 92 kJ B. CaCO 3 (s) kJCaO(s) + CO 2 (g) C. 2SO 2 (g) + O 2 (g)2SO 3 (g) + heat

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 Solution Identify each reaction as (Ex) exothermic or (En) endothermic. (Ex) A. N 2 (g)+ 3H 2 (g) 2NH 3 (g) + 92 kJ (En) B. CaCO 3 (s) kJCaO(s) + CO 2 (g) (Ex) C. 2SO 2 (g) + O 2 (g)2SO 3 (g) + heat

Basic Chemistry Copyright © 2011 Pearson Education, Inc. Calculations Using Heat of Reaction 8

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 Heat Calculations for Reactions In the reaction N 2 (g) + O 2 (g) 2NO(g) ΔH = +181 kJ 181 kJ is absorbed when 1 mol of N 2 and 1 mol of O 2 react to produce 2 mol of NO. N 2 (g) + O 2 (g) kJ 2NO(g) This can be written as conversion factors. 181 kJ 181 kJ 181 kJ 1 mol N 2 1 mol O 2 2 mol NO

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 Heat Calculations for Reactions (continued) N 2 (g) + O 2 (g) + 181kJ 2NO(g) If 15.0 g of NO is produced, how many kJ was absorbed? 1) 1400 kJ 2) 90 kJ 3) 45 kJ

Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution STEP 1 List given and needed data for the equation. Given: 15.0 g of NO produced ΔH = 180 kJ/2 mol of NO Need: kJ absorbed STEP 2 Write a plan using heat of reaction and any molar mass needed. Plan: g of NO moles of NO kJ 11

Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution (continued) STEP 3 Write the conversion factors including heat of reaction. 2 mol of NO = 180 kJ 180 kJ and 2 mol NO 2 mol NO 180 kJ 1 mol of NO = g of NO 1 mol NO and g NO g NO 1 mol NO STEP 4 Set up the problem g NO x 1 mol NO x 180 kJ = 45 kJ (3) g NO 2 mol NO 12

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 13 Learning Check How many grams of O 2 react if 1280 kJ is released in the following reaction? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) ΔH = -890 kJ 1) 92.0 g of O 2 2) 46.0 g of O 2 3) 2.87 g of O 2

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 14 Solution STEP 1 List given and needed data for the equation. Given 1280 kJ Need grams of O 2 STEP 2 Write a plan using heat of reaction and any molar mass needed. kilojoules moles of O 2 grams of O 2

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 15 Solution (continued) STEP 3 Write the conversion factors including heat of reaction. 2 mol of O 2 = 890 kJ 890 kJ and 2 mol O 2 2 mol O kJ 1 mol of O 2 = g of O 2 1 mol O 2 and32.00 g O g O 2 1 mol O 2

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 16 Solution (continued) STEP 4 Set up the problem kJ x 2 mol O 2 x g O 2 = 92.0 g of O 2 (1) 890 kJ 1 mol O 2