Thermochemistry Calculations: Heat of Reactions and Formation, Assignments of Chemistry

Download Thermochemistry Calculations: Heat of Reactions and Formation and more Assignments Chemistry in PDF only on Docsity! 1) work, cost, tire wear, distance, gasoline are not state functions while change in location, elevation, longitude and latitude are. 2) In a coffee-cup calorimeter (with a cup that has no heat capacity) (I'm assuming there are 200. g) q = mc∆T q = 200 g× 4.184 J/gK× 2.31 K = 1930 J So 1930 J of heat was transferred to the surroundings. If we know that 2.17 × 10-3 mol of methane were reacted, we determine the enthalpy change for the chemical equation CH4(g) + 2 O2(g) −→ CO2(g) + 2 H2O(l) qrxn = −1930 J ∆Hrxn = q n = −1930 J 2.17×10−3 mol = −891 kJ/mol 3) The final temperature of the "hot" water and the "cold" ice must be the same, 273 K. The heat lost by the hot water must be gained by the cold ice to achieve that. qhot = mC∆T = −qcold = −(mC∆T + m∆H) 150 g× 4.184 J/gK× (273− 298) K = − (mice × 4.184 J/gK× (273− 273) K + mice × 333 J/g) mice = 47.1g 4) It takes 100 kJ to walk a kilometer, ∆Hwalk ∼ −100 kJ/km For a car, gasoline's enthalpy of combustion is - 48 kJ/g. With a car that gets 8.0 km/L (density is 0.68 g/cm3) we can expect to burn ∆Hcar ∼ −48 kJ/g× 0.68 g/mL× 1000 mL/8.0 km = −4080 kJ You save about 4000 kJ/km by walking. I ignored the 30% efficiency, many answers are acceptable. 5) We want to know the enthalpy associated with the process Ca+2(g) + 2 Cl−(g) −→ CaCl2(s) We know that Ca(s) + Cl2(g) −→ CaCl2(s) ∆Hf(CaCl2(s)) = −795.8 kJ/mol Ca(s) −→ Ca(g) ∆Hf(Ca(g)) = 179.3 kJ/mol Ca(g) −→ Ca+(g) + e− IE1(Ca) = 590 kJ/mol Ca+(g) −→ Ca+2(g) + e− IE2(Ca) = 1145 kJ/mol 1/2 Cl2(g) −→ Cl(g) ∆Hf(Cl(g)) = 121.7 kJ/mol Cl(g) + e− −→ Cl−(g) EA(Cl) = −349 kJ/mol Rearranging, we get... Ca+2(g) + e− −→ Ca+(g) − IE2(Ca) = −1145 kJ/mol Ca+(g) + e− −→ Ca(g) − IE1(Ca) = −590 kJ/mol Ca(g) −→ Ca(s) −∆Hf(Ca(g)) = −179.3 kJ/mol Ca(s) + Cl2(g) −→ CaCl2(s) ∆Hf(CaCl2(s)) = −795.8 kJ/mol 2 Cl(g) −→ Cl2(g) − 2∆Hf(Cl(g)) = −2(121.7) kJ/mol 2 Cl−(g) −→ 2 Cl(g) + e− − 2EA(Cl) = −2(−349) kJ/mol _________________________________________________________________________ Ca+2(g) + 2 Cl−(g) −→ CaCl2(s) ∆Hlatt = −2260 kJ/mol The lattice formation energy of NaCl is -788 kJ/mol so this is almost three times as favorable. Since we are doubling the charge on the cation, we are bringing together more highly charged species and it is more stabilized. 6) At constant pressure, the heat absorbed from the environment by the formation of 1 mole of calcium carbonate is q = 38.95 kJ. This is equal to the enthalpy. The energy change associated with this reaction doesn't pay the p-V penalty (2.47 kJ/mol) of releasing carbon dioxide so it should be smaller. ∆H = ∆E + P∆V ∆E = (38.95− 2.47) kJ/mol = 36.48 kJ/mol 7) The benzoic acid experiment is used to calibrate the calorimeter and determine its heat capacity. The caffeine experiment is then useful to to determine the heat of combustion of caffeine at constant volume (i.e., not the enthalpy of combustion). The 0.235 g sample of benzoic acid should produce 26.38 kJ/g which raised the temperature of the calorimeter 1.642 K. The heat capacity of the calorimeter is q = Ccal∆T 0.235 g× 26.38 kJ/g = Ccal × 1.642 K Ccal = 3.78 kJ/K The 0.265 g sample of caffeine raised the temperature 1.525 K. That corresponds to a heat transfer of q = 3.78 kJ/K× 1.525 K = 5.76 kJ The molar enthalpy of combustion of caffeine is ∆Hc(caff) = 5.76 kJ 0.265 g × 194.2 g caff1 mol caff = 4220 kJ/mol If there is uncertainty of 0.002 K in the temperatures and 0.001 g in the masses, the estimated uncertainty in the final molar enthalpy of combustion would be: The fractional uncertainty in the temperature of the first calibration step is 0.002/1.642 = 0.00122 The fractional uncertainty in the mass of the first calibration step is 0.001/0.235 = 0.00426