Constant-Pressure Calorimetry Atmospheric pressure is constant ΔH = qP qsystem = -qsurroundings System = reaction or process conducted in the water Surroundings = water in the calorimeter + calorimeter. Assume insulator prevents heat transfer to the room. qsystem =-qsurroundings = - (qwater + qcalorimeter) For most calculations, the qcalorimeter can be ignored qsystem = - qwater csystemmsystem ΔTsystem = - cwatermwater ΔTwater Example 3.88 g of ammonium nitrate is dissolved in 60.0 g of water in a coffee cup calorimeter, that is initially at 23.0 oC. As the ammonium nitrate dissolves the temperature goes to 18.4 oC. What is the molar ΔH for dissolving ammonium nitrate? NH4NO3(s) NH4+(aq) + NO3-(aq) ΔTsolution = 18.4oC – 23.0oC = -4.6oC mwater = 60.0g msolution = 60.0 g + 3.88 g = 63.88 g cwater = 4.184J/goC = cfinal solution (by assumption) qsolution = csolutionmsolution ΔTsolution qsolution = 4.184J/goC)(63.88g)(-4.6oC) = -1229J mNH4NO3 = 3.88g molesNH4NO3 = 3.88g/80.032g/mol = 0.04848mol molar enthalpy for dissolving = ΔH = qNH4NO3/molesNH4NO3 q NH4NO3 = -qsolution ΔH = +1229J/0.04848mol ΔH = 25.4 kJ/mol Bomb Calorimetry (Constant-Volume Calorimetry) - For combustion reactions - Substance placed in a sealed “bomb” and filled to a high pressure of O2 - Substance ignited by a spark - Heat from the combustion is transferred to the water - Must account for heat capacity of the calorimeter (includes heat capacity of water). Value of Ccalorimeter is found by calibration (usually by manufacturer) qrxn = -Ccalorimeter(ΔT) A 1.800g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/oC. The temperature of the calorimeter plus contents increases from 21.36oC to 28.78oC. What is the enthalpy of combustion per gram of octane? Per mole of octane? 2 C8H18 + 25O2 16 CO2 + 18 H2O ΔTwater = 28.78oC – 21.36oC = 7.42oC Ccal = 11.66kJ/oC msample = 1.80g qrxn = -Ccal (ΔTwater) qrxn = -11.66kJ/oC(7.42oC) = -86.52kJ ΔHcombustion(in kJ/g) ΔHcombustion = -86.52kJ/1.80g = -48.1 kJ/g ΔHcombustion(in kJ/mol) molesC8H18 = 1.80g(1mol/114.2302g)=0.01575 mol ΔHcombustion = -86.52kJ/0.01575mol = -5492 kJ/mol Hess’s law - If a reaction is carried out in a series of steps, H for the overall reaction is the sum of H’s for each individual step. Example: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) 2H2O(g) 2H2O(l) H = -802 kJ H = -88 kJ Sum: CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ To sum reactions: Sum things separately on each side of the arrows. Cancel out things that appear equally left and right of the arrow. It's just like summing math equations. Remember........enthalpies are extensive, so they refer to the reactions as written. H2O(g) H2O(l) H = -44 kJ compare to above Standard Enthalpy of Formation (ΔHf0 ) The enthalpy change for the formation of one mole of a substance from the most stable forms of its component elements at 298 Kelvin, 1 bar pressure, 1 mol/L The standard enthalpy of formation of the most stable form on any element is zero Many ΔHf0 values have been tabulated. Using ΔHf0’s to Calculate Enthalpies of Reaction For a reaction: o H rxn nH of products mH of reactants Where n and m are the stoichiometric coefficients of each reactant or product from the balanced chemical equation. Example Calculate the enthalpy change for the following reaction: N2O4(g) + 4 H2(g) N2(g) + 4 H2O(g) From a table of standard enthalpies of formation. N2O4(g) H2(g) N2(g) H2O(g) 9.66 kJ/mol 0 kJ/mol 0 kJ/mol -241.82 kJ/mol element element notice value for the gas is used ΔH = [(1 x ΔHN2)+ (4 x ΔHH2O)] – [(1 x ΔHN2O4) + (4 x ΔHH2)] ΔH = [(1x 0 kJ/mol) + (4 x -241.82kJ/mol)] – [(1x l9.66kJ/mol) + (4 x 0kJ/mol)] = -976 kJ PAY ATTENTION TO THE PARENTHESES AND BRACKETS TO KEEP THE ORDER OF ADDITIONS AND SUBTRACTIONS CORRECT! Example Calculate the enthalpy change for the following reaction: 2 KOH(s) + CO2(g) K2CO3(s) + H2O(g) From a table of standard enthalpies of formation. KOH(s) CO2(g) K2CO3(s) H2O(g) -424.7 kJ/mol -393.5 kJ/mol -1150.18 kJ/mol -241.82 kJ/mol ΔH = [(1 x ΔHK2CO3) + (1 x ΔHH2O)] – [(2 x ΔHKOH) + (1x ΔHCO2)] ΔH = [(1x (-1150.18kJ/mol)) + (1x(-241.82kJ/mol))] – [(2 x (-424.7kJ/mol)) + (1 x (-93.5kJ/mol))] = -149.1 kJ