Combustion of Alcohols
Goal: The purpose of this kind of investigation is to use calorimetry to determine the molar enthalpy change in the combustion of every of a number of alcohols. Issue: How do the enthalpies of combustion change as the alcohol elements become greater? Prediction: I predict the larger the molecule the more the will be unveiled. Materials: About lab linen
Process: On laboratory sheet
Declaration: On different page
Analysis: b) i) Ethanol – q=(100g)(4. 18J/g/C)(25. 5C)
Propanol – q=(100g)(4. 18J/g/C)(21. 5C)
Butanol – q=(100g)(4. 18J/g/C)(23. 5C)
ii) Ethanol – 10659J/0. 985g= 12. 8kJ/g
Propanol – 8987J/0. 845g= 10. 6kJ/g
Butanol – 9823J/0. 75g = 13. 1 kJ/g
iii) Ethanol – n=0. 985g/46. 01g/mol = 0. 0214 mol
Propanol – n=0. 845g/60. 01g/mol sama dengan 0. 0141 mol
Butanol – n=0. 75g/74. 1g/mol = zero. 0101 mol
iv) Propanol – 8987J/0. 0141 mol = 637kJ/mol
Ethanol – 10659J/0. 0214 mol sama dengan 498kJ/mol
Butanol – 9823J/0. 0101mol sama dengan 973kJ/mol
C2H6O + 2O2 -> 2CO2 +3H2O + 498kJ
C2H6O + 2O2 -> 2CO2 +3H2O ΔH=498kJ
2C3H8O + 9O2 -> 6CO2 + 8H2O + 637kJ
2C3H8O + 9O2 -> 6CO2 & 8H2O ΔH=637kJ
2C4H10O + 12O2 -> 8CO2 + 10H2O & 973kJ
2C4H10O & 12O2 -> 8CO2 & 10H2O ΔH=973kJ
d) According to the scored data, because the liquor molecules turn into larger, the enthalpies from the combustion become larger also therefore , the greater the molecule, the more energy is produced. The prediction I manufactured was correct.
e) Sources of experimental error through this experiment can be if you failed to measure the liquor right after it was burned. In the event you spilled any alcohol. In the event you didn't record the temperature after you create the fire.
f) Ethanol 498kJ/mol as well as 1369kJ/mol = 36%
Propanol 637kJ/mol as well as 2008kJ/mol = 32%
Butanol 973kJ/mol as well as 3318 kJ/mol = 29%
The Fresh Design is not an suitable method to answer the question as the values are just 32% in the accepted principles.
g) Butanol, because it uses less alcohol per gram...