How much heat energy is required (at constant pressure) to convert 50g of ice at 100K to liquid water at 315K given the following data: Cwater = 4.184 J/gK, Cice = 2.1 J/gK, Hfusion = 6.01 kJ/mol. 315 K Temperature (K) 273 K 100 K Energy (J) Heating ice up to melting point: q = cice  mice  Tice = (2.1 J/gK)(50g)(273K-100K) = 18165 J Converting ice to liquid water: qp = n  Hfus,water = (50g)(1 mol/18g)(6.01 kJ/mol) = 16.69 kJ Heating up liquid water: q = cwater  mwater  Twater = (4.184J/gK)(50g)(315K-273K) = 8786 J Total Heat Required = q1 + q2 + q3 = 18.165 kJ + 16.69 kJ + 8.786 kJ = 43.64 kJ

A certain piece of metal with a mass of 140. 0 g is heated to 135° C A certain piece of metal with a mass of 140.0 g is heated to 135° C. The metal is then placed into a coffee cup calorimeter containing 100.0 g of water initially at 25 ° C. Determine the specific heat of the metal if the calorimeter and metal reach thermal equilibrium at 37° C. q metal = q water 140.0g ·c (37° C - 135 ° C = 100.0 g ·4.18 J/gC · (37 ° C - 25 ° C ) c= 5016 J = .366 J/gC 140 g · 98 ° C

Enthalpy- heat evolved or absorbed by the reaction Thermochemical equation – balanced chemical equations that show the associated enthalpy change. Heat of reaction- enthalpy change that accompanies a reaction Standard heat of formation- the change in enthalpy for the reaction that forms 1 mol of the compound (usually reported at 298 K)

Guidelines for Thermochemical Equations : Enthalpy is an extensive property. The magnitude of ΔH is directly proportional to the amount of reactant consumed by the process.

How much heat is released when 4 How much heat is released when 4.50 g of methane gas is burned in a constant – pressure system according to the equation below? CH4 (g) + 2O2(g)  CO2 (g) + 2 H2O (l) ΔH= -890 kJ

Guidelines for Thermochemical Equations : 2. The enthalpy change for a reaction is equal in magnitude but opposite in sign to ΔH for the reverse reaction. CH4 (g) + 2O2(g)  CO2 (g) + 2 H2O (l) ΔH= -890 kJ CO2 (g) + 2 H2O (l)  CH4 (g) + 2O2(g) ΔH= 890 kJ

Guidelines for Thermochemical Equations : 3. The enthalpy change for a reaction depends on the state of the reactants and products.

Hess’s Law Hess's Law of Constant Heat Summation states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.  Germain Henri Hess

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ 2H2 + O2  2 H2O -571.66 kJ Step #3: Multiply reaction #2 by 2

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ 2H2 + O2  2 H2O -571.66 kJ CH4 + 2O2  CO2 + 2H2O -890.36 kJ Step #4: Sum up reaction and H

Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

Calculation of Heat of Reaction Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O Hrxn =  n Hf(products) -  n  Hf(reactants)     Substance Hf   CH4 -74.80 kJ O2 0 kJ CO2 -393.50 kJ H2O -285.83 kJ Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ] Hrxn = -890.36 kJ