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Old Nov 23rd 2016, 07:21 AM   #1
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Oxygen pressure for oxidizing gold problem

Hi Guys! I'm not sure what to do, I don't epect anyone to complete this for me (not that i'd be adverse to it) but can anyone give me some guidance please?

The question is linked: http://imgur.com/0cXlnkB

2AU(solid) + 3/2O2(gas) <--> Au203(solid)

@25degrees.

Gibbs free energy change = +78.7kj/kmol.

Estimate the value of oxygen pressure required oxidizing gold Is the reaction spontaneous in air?

Do I use aA + bB -> cC + dD? if so I'm not sure what to use for each value?

Thanks!

Last edited by Sarahchemmy; Nov 23rd 2016 at 07:45 AM. Reason: Broken link
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Old Nov 23rd 2016, 10:30 AM   #2
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The main thrust of the problem posed seems to be altering the balance point of the reversible reaction by increasing the concentration of one of the reactants (the oxygen).

I can vaguely remember (from my brief foray into chemistry many moons ago) that the Gibbs free energy is important in the determination of this sort of thing.
But that is about as far as I can go.

You might get a better response from:
Chemistry Help Forum
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Old Nov 24th 2016, 05:56 AM   #3
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Thanks I've sent a message over there too.

So so far my method has been;

0 = deltaG + RTlnP/Po

To eventually get a pressure of 726.75mmhg

To which I divided by the stoichiometry (3/2 02) to get 484.5mmhg
Converting to bar to get 0.653bar

This yields the partial pressure for 02?
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