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### Topic: Gibbs energy calc. (Q7a Atkins 8th ed.)  (Read 4904 times)

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#### zerokun

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• Mole Snacks: +0/-0 ##### Gibbs energy calc. (Q7a Atkins 8th ed.)
« on: June 11, 2008, 11:07:23 PM »
Q7.1(a) pg 235 - At 2257 K and 1.00 atm total pressure , water is 1.77 % dissociated at equilibrium by the way of the reaction 2H2O -> 2H2 + O 2 . Calculate K, molar G , del.(delta)G at this temperature.

This is my way of working it .
molar G = -2 X molar H2O = -2 X -228.57 = 457.14 J/mol = 457.14 X 103 kJ/mol

Since is equilibrium (dissociation ) , therefore del.G = 0
del.G = molar G + RT ln K
del.G =0, -molar G = RT ln K
457.14 X 103  = 8.314X 2257 ln K
K = 2.63 X 10-11

Which is wrong,
the answer given in book is  K =2.85 X 10-6
I don't know what is wrong with my calculation

#### Borek ##### Re: Gibbs energy calc. (Q7a Atkins 8th ed.)
« Reply #1 on: June 12, 2008, 03:52:45 AM »
Looks to me like K can be calculated from simple stoichiometry of the reaction equation.
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