## Gibbs Free naught

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

605110118
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### Gibbs Free naught

Why is it that we can include Q and K in the equation? How do you know when to you which?

Ian Morris 3C
Posts: 101
Joined: Wed Sep 18, 2019 12:18 am

### Re: Gibbs Free naught

Q is quotient at any point in the reaction, and K is constant at equilibrium. You will always use both with this equation.

Ryan Narisma 4G
Posts: 104
Joined: Fri Aug 30, 2019 12:18 am

### Re: Gibbs Free naught

Hi! To answer your question, the $\Delta$G = $\Delta$G$^{\circ}$ + RT ln Q equation represents the gibbs free energy dependence on the concentrations/partial pressures of the products and the reactants at a certain instance in the reaction. If the reaction quotient is greater than the value of K at a given temperature, this indicates that there are too many products created, so the forward reaction will be nonspontaneous but the reverse reaction will be spontaneous because the reaction will shift to form more reactants (le'Chatlier). If Q is greater than K, then the value of $\Delta$G > 0 indicating a nonspontaneous reaction.

To put this into perspective let's say that for a given reaction at T=500K the value of K = 2 and the reaction quotient, Q = 5. If you substitute $\Delta$G$^{\circ}$ for -RTlnK (from our derivation in class), then you'd have -R(500) ln 2.
If you substitute that in the equation, then the equation will be: $\Delta$G = -R(500) ln 2 + R(500) ln 5.
If you were to isolate the log functions you'd get: R(500) (-ln 2 + ln 5) = $\Delta$G
Rearranging logs gives us: R(500)(ln 5 - ln 2) = $\Delta$G
Using the laws of logs: R(500(ln (5/2)) = $\Delta$G
Because 5/2 is greater than 1, the value of ln (5/2) will be positive.
If the reaction quotient was say 0.5, then the equation would become R(500(ln (0.5/2)) = $\Delta$G
In this instance, the value of 0.5/2 is less than 1, therefore the value of ln (0.5/2) will be negative

As expected, if Q<K, reaction is spontaneous in the forward direction and if Q>K, reaction is nonspontaneous in the forward direction

I hope this helps!