## 11.83

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

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

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

Michelle Dong 1F
Posts: 110
Joined: Fri Sep 29, 2017 7:04 am

### 11.83

Calculate the equilibrium constant at 25 degrees celsius and at 150 degrees celsius for each of the following reactions, using data available in appendix 2A.

Why can't we just use the standard Gibbs free energy of formation to calculate the standard Gibbs free energy and use that for both temperatures?

Kayla Danesh 1F
Posts: 38
Joined: Thu Jul 13, 2017 3:00 am

### Re: 11.83

The ΔGf° of a substance has to do with the reaction where the substance is formed from the elements that exist in their most stable forms at a pressure of 1 atm and, most of the time, 298K. Once we have a different temperature than the "standard" one for the specific element, we can no longer use this value.

Thuy-Anh Bui 1I
Posts: 56
Joined: Sat Jul 22, 2017 3:00 am
Been upvoted: 2 times

### Re: 11.83

That is also a viable method to solve this problem because you're solving for K at 25°C (standard state). If you look up the standard Gibbs free energy of formation for the products and reactants, you can use the difference of sums method to calculate ΔG°. Set ΔG°=-RTlnK to solve for K at 25°C. Then for K at 150°C, you can either use ΔG°=-RTlnK again or the Van't Hoff equation.