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Topic: Thermochemistry and Sponteneity (free energy change)  (Read 4376 times)

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Offline Shinigami_lover

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Thermochemistry and Sponteneity (free energy change)
« on: May 01, 2008, 11:10:14 PM »
ok, so i encountered this problem on homework and i am really struggling with things like these. It's dealing with free energy. here's the question,

Determine the free energy change for the reaction: CuO(s) +H2(g) --> Cu(s) + H2O(l)
▲H (enthalpy: Hf-Hi) = -128.5 kJ/mol
▲S (entropy: Sf-Si) = -70.1 J/mol (would u change this to KJ/mol?)

I think in order to get this you need to use the formula ▲G (free energy)=▲H-T▲S. In the back of my book, it says that CuO = -128.3, H2 = 0, Cu = 0 and H2O = -237.1 (all in terms of ▲Gf). This is what i did so far,

Cu + H2O --> CuO + H2
(0 + - 237.1) - (-128.3 - 0) = -108.7 KJ/mol.

I think i got this totally wrong. can anyone help me and explain the right way to use what's listed and not listed for this problem? thanx a lot!

-Shinigami_Lover

Offline Shinigami_lover

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Re: Thermochemistry and Sponteneity (free energy change)
« Reply #1 on: May 01, 2008, 11:18:07 PM »
oops... forgot to mention, the temp. is 25 degrees C --> 298 K

Offline enahs

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Re: Thermochemistry and Sponteneity (free energy change)
« Reply #2 on: May 02, 2008, 12:16:12 AM »
How you worked it is fine, if the tables in the back of your are for 25 C. And if they are reasonably close to 25 C there will not be too much error.

But the other way to work it is to use the values given.
You typed up the equation!

ΔG = ΔH - TΔS

Yes, you would change units.
Just change to correct units, and plug into this equation (you are given the values).

Is it in agreement with your other method?

Is your answer correct?



Offline Shinigami_lover

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Re: Thermochemistry and Sponteneity (free energy change)
« Reply #3 on: May 03, 2008, 11:12:29 PM »
ok, i think i got it thanx ^ ^

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