## delta G naught when K<1

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

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

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

Posts: 60
Joined: Fri Sep 28, 2018 12:16 am

### delta G naught when K<1

Why is delta G naught positive when K<1? Wouldn't the forward reaction be favored and therefore spontaneous, making it negative?

tierra parker 1J
Posts: 61
Joined: Fri Sep 28, 2018 12:17 am

### Re: delta G naught when K<1

i think it has to do with the fact that if K < 1 then there are more reactants than products at equilibrium so the reaction sits to the left making the reverse reaction favorable

Jeannine 1I
Posts: 73
Joined: Fri Sep 28, 2018 12:27 am

### Re: delta G naught when K<1

Since K is the concentration of products over the concentration of reactants ($\frac{[P]}{[R]}$) at equilibrium, when K<1, this means there is a higher concentration of reactants compared to products. This means that the *reverse* reaction is favored, which in turn means the forward reaction is NOT favored. Thus, when K<1, the forward reaction is NOT spontaneous, resulting in a positive delta G nought.

Hope that helps(:

Brandon Mo 4K
Posts: 70
Joined: Fri Sep 28, 2018 12:15 am

### Re: delta G naught when K<1

When K<1, there is a higher concentration of reactants than products. This indicates that products are not being favored in equilibrium. This would lead to how delta G would be positive, or nonspontaneous.

You can also look at the equation delta G = -RTln(K). Having a K<1 would result in a negative value in the natural log, resulting in a positive delta G.