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Delta G vs Delta G0

Posted: Thu Feb 09, 2017 12:38 pm
When asked for delta G, being given Delta H0 and Delta S0, should we just plug in the given temperature (Delta G0 = DeltaH0 - T DeltaS0) or
calculate Delta G using -

Delta G = DeltaG0 + RT ln(Q) <Where DeltaG0 = DeltaH0 - (298)DeltaS0>

????

Re: Delta G vs Delta G0

Posted: Thu Feb 09, 2017 1:10 pm
I think you just plug it into the first one because you have the 0 subscript on all the givens, so you'd be solving for delta G0.

Re: Delta G vs Delta G0

Posted: Fri Feb 10, 2017 2:38 am
You would just solve using the first equation. To solve the second equation, you would need the reaction quotient Q.

Re: Delta G vs Delta G0

Posted: Sun Feb 12, 2017 9:53 pm
But what is the actual difference between calculating a $\Delta G versus a \Delta G^{\circ}$ ?

Re: Delta G vs Delta G0

Posted: Sun Feb 12, 2017 11:37 pm
You use the same equations for both. The difference is that deltaG0 is with standard conditions whereas deltaG is with nonstandard conditions.

Re: Delta G vs Delta G0

Posted: Sun Feb 12, 2017 11:43 pm
The naught sign means that the substance is in standard state so delta g naught would be calculated using standard enthalpy of formation and standard entropy values. You can find the standard enthalpy of formation of the reaction by doing sum of the standard enthalpy of formation of products - reactants (values usually given in a chart). And you can do the same with the standard entropy values then plug the values into the equation delta G0= deltaH0 -TdeltaS0. Another method would be using deltaG0= -RTlnK whereas delta G= delta G0 + RTlnQ