Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Frederick95 on March 06, 2008, 05:17:31 PM
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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.)
= [
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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 Kjmol-1
H2(g) = -286 Kjmol-1
CH4(g) = -890 Kjmol-1
C2H6(g) = -1560 Kjmol-1
C3H8(g) = -2220 Kjmol-1
So you have the equation again: C(gr) + 2H2(g) --> CH4(g)
You know that >> C(gr) = -394 Kjmol-1 + 2(H2(g) = -286 Kjmol-1) --> CH4(g) = -890 Kjmol-1
SO:
According to the equation= products - reactants
Products: (-394) + (2 x -286) = -966
Reactants: (-890) = -890
(-966) - (-890) = -76 kjmol-1
HOPE THIS HELPED.
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