## 5G 17

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

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

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

LBacker_2E
Posts: 104
Joined: Wed Sep 18, 2019 12:21 am
Been upvoted: 1 time

### 5G 17

I am confused on why the answer in the book said that the I was the reactant and the I2 was the product in the graph, when the equation we were given said the reverse. Is this just a typo, or can someone explain why it should be this way?

Daniel Honeychurch1C
Posts: 109
Joined: Thu Jul 11, 2019 12:15 am

### Re: 5G 17

I believe that it is a typo, because the reaction in 5G.13 (which is what you graph in 5G.17) is I2 (aq) --> 2I (g).