WebAssign #14 q=c∙m∙ΔT ΔHchange 693 kJ How much heat is required to change 213 g of ice from -46.5ᵒC to 0.0ᵒC, melt the ice, warm the water from 0.0ᵒC to 100ᵒC, boil the water, and heat the steam to 173.0ᵒC? 100ᵒ to 173ᵒC q=c∙m∙ΔT boil ΔHchange melting 0.0ᵒC to 100ᵒC -46.5ᵒC to 0.0ᵒC 693 kJ

Using ∆Hf to calculate enthalpy changes or ∆H⁰rxn Energy and Chemical Change: Basic Concepts There are two ways that we will calculate the standard enthalpy of reaction: Hess’ Law (adding equations) Comparing products to reactants both use enthalpies of formation Using ∆Hf to calculate enthalpy changes or ∆H⁰rxn The standard enthalpy (heat) of formation (∆H°f) of a compound is the change in enthalpy for the formation of one mole of a compound

Calculating Enthalpy Change Energy and Chemical Change: Basic Concepts Calculating Enthalpy Change Hess’s law: If you can add two equations, you can add their enthalpies Used to find the enthalpy of the overall reaction. Think of it like you’re doing a puzzle

Applying Hess’ Law Identify products and reactants Hess’ Law Steps: Identify products and reactants Match coefficients to overall equation Add equations (cancel spectators) Add enthalpies Whatever you do to the equation, you must also do to its enthalpy Some problems have 3 equations – follow the same principles

Applying Hess’ Law Ex13) Use the following chemical equations to determine ΔH for this equation: 2H2O2(l) → 2H2O(l) +O2(g) 2H2(g) +O2(g) → 2H2O(l) ΔH = -572 kJ H2(g) +O2(g) → H2O2(l) ΔH = -188 kJ ΔH = -196 kJ

Determine ΔH for each reaction Ex14) 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) ∆H = ? 2CO(g) + O2(g) → 2CO2(g) ∆H = -556.0 kJ N2(g) + O2(g) → 2NO(g) ∆H = 180.6 kJ Ex15) 4Al(s) + 3MnO2(s) → 2Al2O3(s) + 3Mn(s) ∆H = ? 4Al(s) + 3O2(g) → 2Al2O3(s) ∆H = -3352 kJ Mn(s) + O2(g) → MnO2(s) ∆H = -521kJ -746.6 kJ -1789 kJ

Uses the standard enthalpies of formation for each individual molecule Energy and Chemical Change: Basic Concepts A simpler way: Uses the standard enthalpies of formation for each individual molecule f reactants f products The standard heat of formation of an element in its standard state is zero. Ex) Au, O2; ΔH°f = 0 kJ

Standard Enthalpies of formation Ex 16) Use standard enthalpies of formation to calculate ∆H°rxn for CH4(g) + 2O2(g) → CO2(g) +2H2O(l) 1. Find ∆H°f for known elements ∆H°f (CH4(g)) = -75 kJ ∆H°f (O2(g)) = 0.0 kJ ∆H°f (CO2(g)) = -394 kJ ∆H°f (H2O(g)) = -286 kJ

Standard Enthalpies of formation Ex16) Use standard enthalpies of formation to calculate ∆H°rxn for CH4(g) + 2O2(g) → CO2(g) +2H2O(l) 2. Use formula ∆H°rxn = ∆H°f(prod) - ∆H°f(react) ∆H°rxn = (∆H°f(CO2) + ∆H°f(H2O)) – (∆H°f(CH4) + ∆H°f(O2)) ∆H°rxn = [(-394 kJ + 2(-286 kJ)] – [(-75 kJ + 2(0.0 kJ)] = [-966 kJ] – [-75 kJ] = -966 kJ + 75 kJ = -891 kJ The combustion of 1 mol CH4 releases 891 kJ

Practice ∆H°rxn ∆H°rxn = 178.3 kJ ∆H°rxn = 66.36 kJ Use standard enthalpies of formation to calculate ∆H°rxn for each of the following reactions. Ex17) CaCO3(s) → CaO(s) + CO2(g) Ex18) N2(g) + 2O2(g) → 2NO2(g) Ex19) 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l) ∆H°rxn = 178.3 kJ ∆H°rxn = 66.36 kJ ∆H°rxn = -1660.3 kJ