# The Burning of Primary and Secondary Alcohols Extended Experimental Investigation

2737 WordsSep 12, 201111 Pages
The Burning of Primary and Secondary Alcohols Year 11A Chemistry By Jarrod Ahern Abstract The aim of the investigation was to determine whether primary alcohols use less energy than secondary alcohols for fuel. The hypothesis is if primary alcohols are heated and results are taken, they produce a lesser heat of reaction then secondary alcohols. The method used was to find the average bond energies of three relating primary and secondary alcohols and compare it with the average theoretical values. Hence the primary alcohols produce less energy most of the time and this was tested by using the correct methods. Contents Pg 1 - Aim Pg 1-2 - Theory Pg 2 - Hypothesis Pg 2-3 - Variables Pg 3 - Apparatus Pg 3 - Procedure Pg…show more content…
6. Light burner wick and center the flame. 7. Make sure there are no variables affecting the flame. 8. When temperature rises to the desired level of 70 degrees Celsius, extinguish flame by putting the lid back on the burner. 9. Reweigh the burner without the lid and record the final weight, also taking ‘tare’ into consideration. Analysis Table 7: Energy Released in Reactions (kJ/mole) Alcohol | Trial 1 | Trial 2 | Trial 3 | Average | Propan-1-ol | 387 | 420 | 374 | 393.67 | Propan-2-ol | 509 | 454 | 235 | 399.33 | Butan-1-ol | 773 | 657 | 573 | 667.67 | Butan-2-ol | 861 | 524 | 434 | 606.33 | Pentan-1-ol | 526 | 896 | 588 | 670 | Pentan-2-ol | 877 | 877 | 655 | 803 | Ɛ = c x m x ∆T This equation is used to find the number of Joules, then J/g = Joules ÷ Initial weight of burner – Final weight of burner will find the joules per gram. Then J/mol = Joules per gram (J/g) x Molar mass of the alcohol will find the joules per mole. Full Working out for Propan-1-ol: Ɛ = c x m x ∆T = 4.2 J/g x 30mL x 46°C = 5796J J/g = Joules ÷ Initial weight of burner – Final weight of burner = 5796 ÷ 237.44 – 236.54 = 5796 ÷ 0.9 = 6440 J/g J/mol = Joules per gram (J/g) x Molar mass of the alcohol = 6440 J/g x 60.1 = 387044 J/mol ≈ 387 kJ/mol Average = (387 + 420 + 374) ÷ 3 = 1181 ÷ 3 = 393.67 kJ/mol This process is repeated for all primary and secondary alcohols. (See appendices) Graph 1: Comparing Bond