## 5I.13

Shanzey
Posts: 120
Joined: Wed Sep 18, 2019 12:20 am

### 5I.13

For part c, how are we able to tell which reaction is more thermodynamically stable? Does it have to do with whether we would get more monatomic molecules or diatomic molecules for the reactions, and them comparing the K values of each equation?

DHavo_1E
Posts: 118
Joined: Sat Aug 17, 2019 12:17 am

### Re: 5I.13

Hello,

I am also confused about this question. Could you also explain what it means to be thermodynamically stable? Thank you!

Angus Wu_4G
Posts: 102
Joined: Fri Aug 02, 2019 12:15 am

### Re: 5I.13

I think the reaction that is more thermodynamically stable is the reaction that produces more product. So for part C I think all you do is compare the 2 products and the one with more product is the correct answer.

KSong_1J
Posts: 101
Joined: Thu Jul 11, 2019 12:17 am

### Re: 5I.13

I’m not sure if this is the right way to do it but what I did was look at the Kc values for both and compared them. Since both are diatomic molecules, they’re more stable in the form F2/Cl2 instead of by themselves, so the one that produces more F2/Cl2 is more stable. Since the equation for Cl2 has a smaller K value (1.2x10^-7) that means it favors reactants (Cl2) more as compared to the equation for F2 (which has a K value of 1.2x10^-4). Therefore, Cl2 is more stable.

Megan Vu 1J
Posts: 101
Joined: Thu Jul 25, 2019 12:15 am

### Re: 5I.13

For this specific part C, I compared the two K values to each other with F2 and Cl2. With this comparison, I saw that Cl1 had a much smaller K value than F2. Because of this, it is evident that Cl2 is much more stable because they are less likely to react with a smaller K value in comparison.

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