Chemical Energetics & Thermodynamics Q5

Chemical Energetics & Thermodynamics Q5

Question Summary

Part a(i) • Temperature correction Plot a best fit line using the points after 3 minutes and extrapolate it to intersect the line Time = 2.5 min Maximum temperature assuming no heat loss = 22.0oC Extrapolate the temperature recorded at Time = 2.5min upwards Maximum temperature rise = 22.0oC - 15.0oC = 7.0oC

Part a(ii) • Total volume of solution = 2 x 50.0 cm3 • Assuming all solutions to have a density of 1.0 g/cm3 • Total mass of solution = 100 cm3x 1.0 g/cm3 • Specific heat capacity = 4.2 J g-1 K-1 • Change in temperature = 7.0 oC = 100 cm3 = 100 g

Part a(ii) • q = m c ΔT • q = 100g x 4.2 J g-1 K-1 x 7.0oC Energy transferred Specific heat capacity = 4.2 J g-1 K-1 Mass of solution = 100g Temperature change = 7.0 oC = 2940 J = 2.94 kJ

Part a(ii) • ΔHneuto= = = -58.8 kJ/mol

Part b(i) • Original setup (HCl) 0.0500 mol H2O 0.0500 mol NaCl 0.0500 mol HCl (0.0500 mol H+) 0.0500 mol NaOH (0.0500 mol OH-) + +

Part b(i) • New setup (H2SO4) 0.0500 mol H2SO4 (0.100 mol H+) 0.0500 mol NaCl 0.0500 mol H2O 0.0250 mol H2SO4 (0.0500 mol H+) 0.0500 mol NaOH (0.0500 mol OH-) + + +

Part b(i)

Part b(ii)

Thank you Comments?

Chemical Energetics & Thermodynamics Q5