11.15

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

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

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

Joslyn_Santana_2B
Posts: 51
Joined: Wed Sep 21, 2016 2:58 pm

11.15

1. Why does solution's manual use positive RTlnQ instead of negative? How do you know which equations to use?

2. The solution states, "Because deltaG is positive, the reaction will be spontaneous to produce I_2. I thought positive deltaG meant non-spontaneous ?
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Chem_Mod
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Re: 11.15

1. Because we know that the equation is not at equilibrium, we will instead use this equation.$\Delta G=\Delta G^{\circ} +RTln(Q)$

$\Delta G=\Delta G^{\circ} +RTln(Q)$

If we were solving for $\Delta G^{\circ}$, then we would use the following formula.

$\Delta G^{\circ}=-RTlnK$

2. You are right, a positive delta G should be non-spontaneous.

David Julfayan 1F
Posts: 16
Joined: Wed Sep 21, 2016 2:59 pm

Re: 11.15

The way I understood it was that ∆G is positive so the reaction is spontaneous, favoring the reactants (going to the left to I_2). Its not spontaneous going to the right.