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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 loge. Once you do this, you just solve for deltaG.