Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Frederick95 on March 06, 2008, 05:17:31 PM

1. Use tabulated standard enthalpies of formation to calculate a theoretical value for the molar enthalpy of combustion of acetone.
Here is what I have so far....
Write a balanced equation: C3H6O(l) + 4O2(g) > 3CO2(g) + 3H2O(l)
Under each compound, write out the enthalpy of formation, ignoring the number of moles.
C3H6O(l) + 4O2(g) > 3CO2(g) + 3H2O(l)
 0.0 393.5 285.8
I know i need to use this formula but can someone tell me how to sub in my values in to this formula?
∆H = ∑n∆H°f(prod.)  ∑n∆H°f(react.)
= [

Basically, all you have to do is take the products away from the reactants.
Example:
Calculate the enthaply of formation of methane, CH4 (g).
Equation: C(gr) + 2H2(g) > CH4(g)
Given table:
C(gr) = 394 Kjmol1
H2(g) = 286 Kjmol1
CH4(g) = 890 Kjmol1
C2H6(g) = 1560 Kjmol1
C3H8(g) = 2220 Kjmol1
So you have the equation again: C(gr) + 2H2(g) > CH4(g)
You know that >> C(gr) = 394 Kjmol1 + 2(H2(g) = 286 Kjmol1) > CH4(g) = 890 Kjmol1
SO:
According to the equation= products  reactants
Products: (394) + (2 x 286) = 966
Reactants: (890) = 890
(966)  (890) = 76 kjmol1
HOPE THIS HELPED.

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