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Transcript
Chapter 5 Problems 1. How many joules must be added to a 63.43-g sample of Fe(s) to raise the sample’s temperature from 19.7 to 54.2° C? Assume that the specific heat of iron is constant over this temperature range. Answer: 9.8 x 10^2 J 2. How many joules must be added to a 7.92-g sample of CH3CH2OH(l) to raise the sample’s temperature from 6.8 to 75.2 ° C? Answer: 1.31 x 10^3 J 3. How many joules must be removed from a 287-g sample of CH3CH2OH(l) to lower the sample’s temperature from 56.2 to 19.8 ° C? Answer: 2.53 x 10^4 J 4. A 1.00 M aqueous solution of NaOH, a 0.50 M aqueous solution of H2SO4, and a coffee-cup calorimeter were allowed to stand at a room temperature of 25.4 °C until the temperature of all three reached 25.4 C. A 50.0-mL sample of the 1.00 M NaOH was then placed in the calorimeter. 50.0 mL if the 0.50 M H2SO4 was added as rapidly as possible and the two solutions were mixed thoroughly, the temperature rose to 31.9° C. From this experiment, what is the heat of neutralization of one mole of H2SO4? For simplicity, assume that the densities of the NaOH and H2SO4 solutions were 1.00 g/mL and that the specific heat of the solution after reaction was 4.18 J/g °C. Answer: -112 kJ/mol 5. A 1.00 M solution of NaOH, a 1.00 M solution of HCl, and a calorimeter were allowed to stand at a room temperature of 19.5°C until the temperature of all three reached 19.5°C. A 200.0-mL sample of the 1.00 M NaOH was then placed in the calorimeter, 200.0 mL of the 1.00 M HCl was added as rapidly as possible, and the two solutions were mixed thoroughly.The temperature rose to 25.8°C. a) From this experiment, what is the heat of neutralization of one mole of HCl? Assume densities of NaOH and HCl solutions were 1.00 g/mL and that the specific heat of solution after reaction was 4.18 J/g°C. b) Compare the heat of neutralization of one mole of HCl with the heat of neutralization of one mole of sulfuric acid in Problem 4 and explain the difference. Answer: a) –53 kJ/mol b) molar heat of neutralization HCl = ½(molar heat of neutralization H2SO4) 6. For the burning of methane gas, CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) Heat of rxn = -890.32 kJ. How many kilojoules will be given off by the burning of 451 g of methane? Answer: 2.51 x 10^4 kJ 7. For burning of isooctane, 2 (CH3)2CHCH2C(CH3)3 (l) + 25O2(g) -> 16CO2(g) + 18H2 (l). Heat of rxn = -10930.9 kJ. How many kilojoules will be given off by the burning of 369 g of isooctane? Answer: 1.77 x 10^4 kJ 8. Calculate deltaH(rxn) for the reaction 2Ca(s) + 2C(graphite) + 3O2(g) -> 2CaCO3(s) from the following information: 2Ca(s) + O2(g) -> 2CaO(s) ΔH = -1270.18 kJ C(graphite) + O2(g) -> CO2(g) ΔH = -393.51 kJ CaO(s) + CO2(g) -> CaCO3(s) ΔH = -178.32 kJ Answer: -2413.84 kJ 9. Given the thermochemical equation 2Hg(l) + O2(g) -> 2HgO(s) deltaH = -181.66 kJ a) Write the thermochemical equation for the decomposition of 2 mol HgO(s) to Hg(l) and O2(g). b) Write the thermochemical equation that shows the formation of 1 mol HgO from Hg(l) and O2(g). Answer: a) 2HgO(s) -> 2Hg(l) + O2(g) ΔH = +181.66 kJ b) Hg(l) + 1/2O2(g) ->HgO(s) ΔH = -90.83 kJ