Moderators: Chem_Mod, Chem_Admin

Jesus Rodriguez 1J
Posts: 53
Joined: Sat Jul 22, 2017 3:00 am


Postby Jesus Rodriguez 1J » 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.

Lena Nguyen 2H
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am
Been upvoted: 1 time

Re: 11.111

Postby Lena Nguyen 2H » 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.

Jenny Cheng 2K
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 1 time

Re: 11.111

Postby Jenny Cheng 2K » 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).

Ivy Lu 1C
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am

Re: 11.111

Postby Ivy Lu 1C » 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.

Return to “Gibbs Free Energy Concepts and Calculations”

Who is online

Users browsing this forum: No registered users and 0 guests