1 / 16

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

gauri
Download Presentation

Chapter 9 Chemical Quantities in Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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. 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)

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

  6. 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

  7. 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

  8. Calculations Using Heat of Reaction

  9. 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

  10. 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 kJ 3) 45 kJ

  11. 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 2Write a plan using heat of reaction and any molar mass needed. Plan: g of NO moles of NO kJ

  12. Solution (continued) STEP 3Write 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 = 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 180 kJ = 45 kJ (3) 30.01 g NO 2 mol NO

  13. 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

  14. 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

  15. 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 = 30.01 g of O2 1 mol O2 and 32.00 g O2 32.00 g O2 1 mol O2

  16. 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

More Related