## 11.81

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

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

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

Wilson Yeh 1L
Posts: 42
Joined: Fri Sep 29, 2017 7:06 am

### 11.81

A gaseous mixture consisting of 2.23 mmol N2 and 6.69 mmol H2 in a 500.-mL container was heated to 600. K and allowed to reach equilibrium. Will more ammonia be formed if that equilibrium mixture is then heated to 700. K? For N2(g) + 3 H2(g) <--> 2 NH3(g), K = 1.7 x 10-3 at 600. K and 7.8 x 10-5 at 700. K.

How would I approach this problem? I'm thinking I basically need to show whether it's exothermic or endothermic then I can pretty much answer the question, but how would I do so? Can I just say that the K value at 700. K is less thus the reaction at the higher temperature favors the reactants even more, producing less ammonia? Or do I actually have to like show work and derive this mathematically? Thanks!

David Minasyan 1C
Posts: 54
Joined: Thu Jul 13, 2017 3:00 am

### Re: 11.81

I'm pretty sure you can just look at the K value changing and come to a conclusion based off that. I don't really know how to solve this mathematically, so using K sufficed for me.

skalvakota2H
Posts: 52
Joined: Sat Jul 22, 2017 3:01 am

### Re: 11.81

I believe a description of the changes in the reaction due to increase in temperature is acceptable. As mentioned, by observing that the equilibrium constant decreases, this implies increased reactant formation. As a result, this deems that there will be less ammonia formed upon heating.

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