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Topic: molar enthalpy of combustion of acetone next step?  (Read 31880 times)

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Offline mo_1984

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Re: molar enthalpy of combustion of acetone next step?
« Reply #30 on: March 07, 2008, 10:16:01 PM »
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%


Offline ARGOS++

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Re: molar enthalpy of combustion of acetone next step?
« Reply #31 on: March 08, 2008, 11:00:12 AM »

Dear Mo_1984;                 

Question 1Well done!

Question 2Frederick95 and I got a different value as you may see above!
                  You did Reply # 9 not correct!  ( mAl ).
                  You did also same Mistake as Frederick95 first: Tooo less significant figures!

Question 3:  Needs to be recalculated, because Question 2!


Good Luck!
                    ARGOS++


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