## 9.55, 6th edition (part c)

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

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

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

Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

### 9.55, 6th edition (part c)

Question 9.55 asks you to write a balanced chemical equation for the formation reaction of various gases. In part C, it asks you to do so for CO(g). Why do you use C(s, graphite) for the chemical equation and not C(s, diamond) or C(g)? Is C(s, graphite) the most stable form of carbon?

Hovik Mike Mkryan 2I
Posts: 95
Joined: Fri Sep 28, 2018 12:25 am

### Re: 9.55, 6th edition (part c)

Hello,
I believe professor Lavelle mentioned in class that at room temperature the most stable is graphite and at higher temperatures diamond can become more stable. Hope this helped!

Steven Garcia 1H
Posts: 79
Joined: Fri Sep 28, 2018 12:16 am

### Re: 9.55, 6th edition (part c)

C(s, graphite) is the most stable element of carbon so that's why you would use C(s, graphite) as opposed to C(s, diamond). This can be confirmed if you look at the appendix of the textbook. The enthalpy of formation of C(s, graphite) is 0, which is what we expect for elements in their most stable form.