Thermochemical Equations

Thermochemical Equations Chemistry 11(C)

Lesson Objectives • Classify reactions as endothermic or exothermic • Complete calculations using thermochemical equations • Thermochemical Equations

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

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

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

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

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

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

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

Hess’s Law – the sum of the enthalpy changes for each step of a reaction is equal to the overall enthalpy change • Manipulate the equations so the sum of the given equations equals the overall equation • If the reaction is reversed, the sign of ∆H must be reversed • Multiply given equations by coefficients • Hess’s Law Ex) 4Fe(s) + 3O2(g)2Fe2O3(s) ∆H= –1625 kJ 2Fe2O3(s)4Fe(s) + 3O2(g) ∆H= 1625 kJ 3[ 2Fe2O3(s)4Fe(s) + 3O2(g) ∆H= 1625 kJ ] 6Fe2O3(s)12Fe(s) + 9O2(g) ∆H= 4875 kJ

Hess’s Law • 376 kJ • 188 kJ 2H2O2(l ) → 2H2(g) + 2O2(g) H2O2(l) → H2(g) + O2(g) Ex) 2H2O2(l ) → 2H2O(l ) + O2(g) ∆H=? + H2(g) + O2(g) → H2O2(l ) ∆H1= –188 kJ ∆H= –196 kJ Rearrange formulas so each chemical is on the same side of the reaction as the goal formula 2H2(g) + O2(g) → 2H2O(l ) ∆H2= –572 kJ • Multiply by coefficients so the number of moles of each chemical match the goal formula • Keep in mind that matching chemicals on opposite sides of the combined reaction will cancel each other out Find the sum of the thermochemical equations

Lesson Objectives • Classify reactions as endothermic or exothermic • Complete calculations using thermochemical equations • Solve for heat of reaction when given heats of formation • Solve problems using Hess’s law • Thermochemical Equations

Thermochemical Equations