### 11.111

Posted:

**Sun Feb 18, 2018 7:46 am**I was wondering if someone could explain to how they solved for DeltaG2, I understand the setup of the problem but I'm not sure how they got rid of e in one of the very last steps.

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=135&t=27939

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Posted: **Sun Feb 18, 2018 7:46 am**

I was wondering if someone could explain to how they solved for DeltaG2, I understand the setup of the problem but I'm not sure how they got rid of e in one of the very last steps.

Posted: **Sun Feb 18, 2018 9:44 am**

To get rid of e, take the natural log of both sides to get ln 10 = (2.00 x 10^5 + deltaG2)/(8.3145 x 298.15) and solve for deltaG2 from there.

Posted: **Fri Feb 23, 2018 2:32 pm**

Since ΔG(1) is given, you can solve for k(1). The relationship between k(1) and k(2) is given, with k(2)=k(1)/10. You can solve for ΔG(2) using k(2). For this problem, you should use ΔG=-RT ln(k).

Posted: **Fri Feb 23, 2018 2:37 pm**

When you take the natural log (ln) of both sides, you get rid of e because ln is equal to log_{e}. Once you do this, you just solve for deltaG.