1 / 18

Lesson Objectives Classify reactions as endothermic or exothermic ( 11C )

Lesson Objectives Classify reactions as endothermic or exothermic ( 11C ) Complete calculations using thermochemical equations ( 11C ) Complete calculations involving specific heat, temperature change, mass, and heat ( 11D ). Thermochemical Equations.

shika
Download Presentation

Lesson Objectives Classify reactions as endothermic or exothermic ( 11C )

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. Lesson Objectives • Classify reactions as endothermic or exothermic (11C) • Complete calculations using thermochemicalequations (11C) • Complete calculations involving specific heat, temperature change, mass, and heat (11D) • Thermochemical Equations

  2. Endothermic and Exothermic Reactions • Energy can be absorbed or released • Endothermic– energy is absorbed by a reaction • Exothermic– energy is released from a reaction

  3. Energy is released when bonds form • Energy is absorbed to break bonds • Releasing and Absorbing Energy Ex) 2H2+ O2 2H2O Energy absorbed Energy released Reactants Products

  4. Exothermic ⇒ less E is absorbed to break bonds than is released when bonds form Bond Energy reactant < Bond Energy product • Endothermic ⇒ more E is absorbed to break bonds than is released when bonds form Bond Energy reactant > Bond Energy product • Classifying Reactions

  5. Enthalpy change – (∆H); amount of energy released or absorbed as heat by a system when the pressure is constant • ∆Hrxn= Hproducts–Hreactants • Exothermic⇒ ∆H = negative values • Endothermic⇒ ∆H = positive values • Enthalpy Exothermic Endothermic Enthalpy Enthalpy Reaction Progress Reaction Progress Hreactants> Hproducts Hreactants< Hproducts

  6. Thermochemical equation–chemical equation that includes the enthalpy change • Coefficients represent the number of moles • ∆H is directly proportional to the number of moles • Include physical states • Thermochemical Equations Ex) 4Fe(s) + 3O2(g)2Fe2O3(s) + 1625 kJ or 4Fe(s) + 3O2(g)2Fe2O3(s) ∆H= –1625 kJ

  7. Standard enthalpy values for specific chemicals can be found in reference tables • Phase of the chemical will affect the enthalpy value • Elements in their natural state will have ∆H0f of zero • Enthalpy Values

  8. Enthalpy of reaction equals the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants • Calculating ∆Hrxn from ∆H0f Ex) CH4(g) + 2Cl2(g)  CCl4(l ) + 2H2(g) (–74.6 + (2×0)) ∆Hrxn= (–139 + (2×0)) – ∆Hrxn= ∆Hf0 (products) - ∆Hf0 (reactants) ∆Hrxn= –64.4 kJ/mol

  9. Problem Use Table C-13 in your book to answer the following: Calculate the ∆Hrxn for the following: 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)

  10. Specific Heat – quantity of energy needed to increase the temperature of one gram of a substance by one degree Celsius • Every substance has a unique specific heat • Specific heat of liquid water is 4.18 J/g∙∘C • Specific Heat • Specific heat =

  11. Units • Heat in cal or J • Mass in g • Temperature in ∘C or K • Specific heat in J/g∙∘C Equivalent amounts • 1 cal = 4.18 J • 1000 cal = 1 kcal (Cal ) • Use as conversion factors • 0 ∘C = 273 K Temperature conversions Ex) 25 ∘C + 273 = 298 K Ex) 290 K – 273 = 17 ∘C • Specific Heat

  12. Specific Heat • Write unknown and givens cp= ? J/g∙∘C Q = 750. J m = 50. g Tinitial = 25 ∘C Tfinal = 100. ∘C • Identify the formula and rearrange, if needed Q = m cp∆T cp= Ex) An unknown substance, with a mass of 50. g, is heated from 25 ∘C to 100. ∘C. During the heating process the substance absorbs 750. J of energy. What is the specific heat of the unknown substance in J/g∙∘C? • Convert units and find intermediates, if needed ∆T = Tfinal – Tinitial ∆T = 100. ∘C – 25 ∘C = 75 ∘C • Plug in and solve cp = cp = 0.20 J/g∙∘C • Make sure the answer is reasonable

  13. Ex) A 21.0 g sample of a substance with a specific heat of 0.449 J/g∙∘C is heated from 288 K to 434 K. How much heat, in joules, was absorbed by the substance? • Specific Heat • Convert units and find intermediates, if needed • Tinital = 288 K – 273 = 15 ∘C • Tfinal = 434 K – 273 = 161 ∘C • ∆T = Tfinal – Tinitial • ∆T = 161 ∘C – 15 ∘C = 146 ∘C • Plug in and solve Q = (21.0 g)(0.449 J/g∙∘C )(146 ∘C) Q = 1380 J • Make sure the answer is reasonable • Write unknown and givens cp = 0.449 J/g∙∘C Q = ? J m = 21.0 g Tinitial = 288 K Tfinal= 434 K • Identify the formula and rearrange, if needed • Q = m cp∆T

  14. Lesson Objective (11E) • Calculate the heat of a chemical process using calorimetry • Calorimetry

  15. Calorimetry – measurement of the heat that is transferred during a reaction or physical process • Calorimeter – equipment used to measure the heat that is transferred during a reaction or physical process • Calorimetry Basic Calorimeter Setup

  16. Simple Calorimeter Bomb Calorimeter • Calorimetry

  17. Law of conservation of energy – energy is always conserved • Exothermic process • Heat lost by the system is gained by the water • Endothermic process • Heat gained by the system is lost by the water • Heat gained = heatlost • At constant pressure Q = ∆H • Exothermic Qwater = –Qprocess • Endothermic –Qwater= Qprocess • Calorimetry

  18. Calorimetry Calculation • Write unknown and givens mw = 1000. g cw = 4.18 J/g∙∘C Tinitial = 25.0 ∘C Tfinal= 56.5 ∘C Qw = ? Qp= ? mp = 2.80 g • Identify the formula and rearrange, if needed Qw= –Qp Qw = mwcw ∆Tw Ex) A calorimeter contains 1000. g of water. When 2.80 g of gasoline are combusted, it changes the temperature of the water in the calorimeter from 25.0 ∘C to 56.5 ∘C. How much heat is released by this chemical process? What is the heat of combustion of gasoline in kJ/g? • Convert units and find intermediates, if needed ∆Tw = Tfinal – Tinitial ∆Tw = 56.5 ∘C – 25.0 ∘C = 31.5 ∘C • Plug in and solve Qw= (1000. g)(4.18 J/g∙∘C)(31.5 ∘C) Qw = 132,000 J –Qp = 132,000 J Qp = – 132,000 J • Make sure the answer is reasonable × = –132 kJ Qp = ∆H = = –47.1 kJ/g

More Related