1. Jeremy Koo 1 Chemical Energetics & Thermodynamics Q5

2. Question Summary 2 Aim of experiment: to determine Δ Hneut o of reaction Aqueous HCl + Aqueous NaOH Volume: 50cm 3 Concentration:1 mol/dm 3 Stand for 1 hour to reach lab temperature Temperature taken every 30s for 2.5min At 3min, solutions were mixed Temperature taken again

3. Part a(i) 3 Temperature correction Extrapolate the temperature recorded at Time = 2.5min upwards 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 = 22.0 o C Maximum temperature rise = 22.0 o C - 15.0 o C = 7.0 o C

4. Part a(ii) 4 Total volume of solution = 2 x 50.0 cm 3 Assuming all solutions to have a density of 1.0 g/cm 3 Total mass of solution = 100 cm 3 x 1.0 g/cm 3 Specific heat capacity = 4.2 J g -1 K -1 Change in temperature = 7.0 o C = 100 cm 3 = 100 g

5. Part a(ii) 5 q = m c Δ T q = 100g x 4.2 J g -1 K -1 x 7.0 o C Energy transferred Specific heat capacity = 4.2 J g -1 K -1 Mass of solution = 100g Temperature change = 7.0 o C = 2940 J = 2.94 kJ

6. Part a(ii) 6 = -58.8 kJ/mol

7. Part b(i) 7 Original setup (HCl) 0.0500 mol HCl (0.0500 mol H + ) 0.0500 mol NaOH (0.0500 mol OH - ) 0.0500 mol NaCl 0.0500 mol H 2 O

8. Part b(i) 8 New setup (H 2 SO 4 ) 0.0500 mol H 2 SO 4 (0.100 mol H + ) 0.0250 mol H 2 SO 4 (0.0500 mol H + ) 0.0500 mol NaOH (0.0500 mol OH - ) 0.0500 mol NaCl 0.0500 mol H 2 O

9. Part b(i) 9 Temperature rise would still be 7.0 o C 1 mol H2SO4 dissociates to form 2 mol H+ Only half the volume of acid is required to neutralize NaOH However, the total number of moles of H + and OH - that react are still the same Δ Hneut o remains the sameTemperature rise remains the same

10. Part b(ii) 10 Original experiment HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) Δ H neut o = -58.8 kJ/mol New experiment HCl (aq) + NaOH (s) NaCl (aq) + H 2 O (l) Δ H = -99.8kJ/mol NaOH (s) Na + (aq) + OH - (aq) Δ H soln o = -44.4 kJ/mol Energy released to dissolve NaOH pellets Reaction more exothermic

11. Comments? 11 Thank you