# Chemical Forums

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

Title: Standard Enthalpies
Post 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.)
= [
Title: Re: Standard Enthalpies
Post by: scopski on March 06, 2008, 06:10:53 PM
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.
Title: Re: Standard Enthalpies
Post by: scopski on September 25, 2009, 01:07:02 PM
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