i am so lost plz check if this is right and if not help me fix it. PLZ PLZ PLZ
1) Using the table of standard enthalpies of formation, I calculated the theoretical value for the molar enthalpy of combustion of acetone by:
Writing a balanced equation: C3H6O (l) + 4O2 (g) --> 3CO2 (g) + 3H2O (l)
Then each compound, I wrote out the enthalpy of formation.
C3H6O (l) + 4O2 (g) --> 3CO2 (g) + 3H2O (l)
-248.1 0.0 -393.5 -285.8
After I put these enthalpy values into the equation and now I used the number of moles to find the theoretical value:
∆H = ∑n ∆H°f(prod.) - ∑n ∆H°f(react.)
= [3(-393.5) + 3(-285.8] – [(-248.1) + 4(0.0)] kJ/mol
= [-2037.9 – (-248.1)] kJ/mol
= -1789.8 kJ/mol
2) Calculate the molar enthalpy of combustion of acetone, using the experimental evidence.
∆T = t2 – t1
= 25.0oC – 20.0oC
= 5.0oC
QTotal = QWater + QAl = mWcWΔT + mAlcAlΔT
= (100.0g)(4.18J/(g*oC)(5 oC) + (100.0g)(0.91 /(g*oC)(5 oC)
= 2090J + 455J
= 2545 J
Molar Mass of Acetone:
C3H6O
C3 = 3 X 12 = 36
H6 = 6 X 1 = 6
O = 1X 16 = 16
= 58 g/mol
Moles = Mass / molar mass
= 0.092g / 58g/mol
= 1.59 X 10-3 mol
H = - Q
= -2545J (it’s an exothermic reaction therefore it needs to have a negative value to get the correct percent error/enthalpy change)
n ∆H°f = ∆H
∆H°f = ∆H/ n
∆H°f = -2545J / 1.59 X 10-3 mol
= -1600.63 kJ/ mol
3) Calculate the percentage error for this experiment.
Error % = [the accepted answer - your observed answer] / the accepted answer (then multiply by 100)
= [(-1789.8kJ/mol) - (-1600.63kJ/mol)] / (-1789.8kJ/mol) X 100%
= 10.6%