# Bell Ringer May 11th

Bell Ringer 11th May The law of conservation of energy: energy cannot be ________ or _______. It can only be ________ or __________. Transfer of energy always takes place from a substance at a _______ temperature to a substance at a _______ temperature Three methods of heat energy transfer: _______ , __________, and __________. 1 Thermochemical Equations 2 Thermochemical Equations A Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, H.H.

Enthalpy (H) is the transfer of energy in a reaction (for chemical reactions it is in the form of heat) and H.H is the change in enthalpy. By definition, H.H = Hproducts Hreactants Hproducts < Hreactants, H.H is negative Hproducts > Hreactants, H.H is positive 3 Thermochemical & Endothermic/ Exothermic equations Depending on the sign of H.H, the reaction can either be exothermic or endothermic. Exothermic reactions release heat from the system to the surroundings so the temperature will rise. H.H will be negative because the reaction loses heat. Endothermic reactions absorb heat from the surroundings into the system so the temperature will decrease. H.H will be positive because the reaction absorbs heat.

4 Classify the following as endothermic or exothermic Ice melting Endothermic 2 C4H10(g) + 13 O2(g) 10 H2O(g) + 8 CO2(g) H.Hrx = -5506.2 kJ/mol Exothermic 2 HCl (g) + 184.6 kJ H2 (g) + Cl2 (g) Endothermic Water vapor condensing Exothermic 5 Exothermic vs. Endothermic EXOTHERMIC

ENDOTHERMIC A change in a chemical A change in chemical energy where energy/heat EXITS the chemical system Results in a decrease in chemical potential energy H is negativeH is negative energy where energy/heat ENTERS the chemical system Results in an increase in chemical potential energy H is negativeH is positive 6 Thermochemical equations using Standard Heat of Formations - Solving for change in H (H)H)H)

C2H2(g) + 2 H2(g) C2H6(g) Substance DHf (kJ/mol) C2H2(g) C2H6(g) 226.7 -84.7 Standard Heat of Formations = DHf Remember By definition, H.H = H products H reactants 7 Thermochemical equations using Standard Heat of Formations DHf Write the equation for the heat of formation Substance (kJ/mol)

of C2H6(g) C2H2(g) 226.7 1st: Using our balanced chemical C2H6(g) -84.7 equation, we see how many moles of each compound we have. C2H2(g) + 2 H2(g) C2H6(g) [(H2) does not have a DHf ] 1 mol of C2H2(g) and 1 mol C2H6(g) 2nd: We plug in the Hf for each of our compounds, remembering that H = [Hf products] [Hf reactants] rdH = [C2H6(g)] [C2H2(g)] = 3 : We solve for H H = [-84.7] [226.7] = -331.4 kJ/mol 8 Practice Problems

Substance Solve for the H.Hrx and write the following thermochemical equations. DHf (kJ/mol) CaCO3(s) -1207.6 CaO(s) -634.9 C 4H10 (g) -30.0 H2O (g) -241.82

CO2 (g) -393.5 1. What is the H.Hrx for the process used to make lime (CaO)? CaCO3(s) CaO(s) + CO2(g) 2. What is the H.Hrx for the combustion of C4H10(g)? 2 C4H10 (g) + 13 O2 (g) 10 H2O (g) + 8 CO2 (g) 9 Practice Problems Solve for the H.Hrx and write the following thermochemical equations. 1. What is the H.Hrx for the process used to make lime (CaO)? CaCO3(s) CaO(s) + CO2(g) Substance DHf (kJ/mol)

CaCO3(s) -1207.6 CaO(s) -634.9 C 4H10 (g) -30.0 H2O (g) -241.82 CO2 (g) -393.5 H.Hrx = [H.Hf (CaO) + H.Hf (CO2)] [H.Hf (CaCO3)] H.Hrx = [(-634.9)+(-393.5)] [(-1207.6)] H.Hrx = [ -1028.4] [-1207.6] = +179.2 kJ

CaCO3(s) CaO(s) + CO2(g) H.Hrx = 179.2 kJ/mol 10 Practice Problems Solve for the H.Hrx and write the following thermochemical equations. 2. What is the H.Hrx for the combustion of C4H10(g)? 2 C4H10 (g) + 13 O2 (g) 10 H2O (g) + 8 CO2 (g) H.Hrx = [H.Hf (H2O) + H.Hf (CO2)] [H.Hf (C4H10)] (We do not include O2 because its H.Hf is 0.) Substance DHf (kJ/mol) CaCO3(s) -1207.6

CaO(s) -634.9 C 4H10 (g) -30.0 H2O (g) -241.82 CO2 (g) -393.5 H.Hrx = [10(-241.82)+8(-393.5)] [2(-30.0)] H.Hrx = [ -5566.2] [-60.0] = -5506.2 kJ 2 C4H10 (g) + 13 O2 (g) 10 H2O (g) + 8 CO2 (g) H.Hrx = -5506.2 kJ/mol 11

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