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Topic: Ka From Thermodynamic Data  (Read 2894 times)

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Offline Boxxxed

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Ka From Thermodynamic Data
« on: April 25, 2011, 01:57:19 PM »
 a.  Write a balanced chemical equation for the acid dissociation of acetic acid, CH3COOH,  in water. (1 point)

CH3COOH + H2O --> CH3COO- + H3O+

b.  Using the thermodynamic data in appendix B, calculate ΔrxnGo and Ka for this reaction at 298 K.   Do not include water as a reactant in your equation.

                         G0f (kJ mol-1)
CH3COOH (aq)   -396.46
CH3COO- (aq)   -369.31
H3O+ (aq)            0
H2O(l)                -285.8

Grxn = Gproducts - GReactants = (-369.31 + 0)-(-396.46) = 27.15

First of all, shouldn't Grxn be negative if this is spontaneous?

Subbing into lnK = -G/RT

lnK = -27.15/(8.314)(298) =
lnK = -0.01095
K = 0.989

Ka for acetic acid is 1.8x10-5

What am I doing wrong?

Offline Boxxxed

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Re: Ka From Thermodynamic Data
« Reply #1 on: April 25, 2011, 02:00:30 PM »
Figured it out, I did not keep energy units constant first time through.


Why does gibbs have a positive value then?

I guess looking at the small ka value that it is reactant favoured. Understood now

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