[CHM 112] Burdge Problems

[Andrew Pounds]

When I sent this out earlier the equations didn't come through -- 
hopefully this makes more sense..

On 07/10/13 16:05, wrote:
>
> Hey Dr. Pounds,
>
> I was working on problem #46 from Chapter 18 (part2). I had a few 
> questions.
>
> 1. Why do we use (delta G=deltaG^o +RTlnK) rather than just finding 
> the delta H and delta S and using the other formula to find the values?
>
In this part of the problem they have given you the K value.  Since K is 
at the equilibrium condition, $\Delta G$ has to be zero.  Because you 
know that you can compute the value of $\Delta G^{\circ}$.

> 2. When do we know when to find just delta G vs. delta G^o?
>
If you are looking for an equilibrium constant, you will need $\Delta 
G^{\circ}$.

> 3. Also, on the second part of that question, we used this formula 
> again(delta G=deltaG^o +RTlnK), BUT rather than the K, there is a Q? 
> Why is that? When do we know it is a K or a Q?
>

So here they want to know the actual value of the Gibbs Free Energy 
function when it is NOT at equilibrium, but rather at its initial 
state.  Using the value of $\Delta G^{\circ}$ you just calculated and 
the values of the non-equilibrium condition (Q), you can determine the 
value for $\Delta G$.

> 4. Last, why do we leave delta G as joules when I thought it is 
> supposed to be in kilojoules?
>

It's purely a matter of convenience.  If you are looking for the value 
of K then you will be dividing by the gas constant R in the units of J/K 
mol.

> Thank you so much!
>


-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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