In the thermochemistry section:
He says we have to know that delta G = delta G0 + RT ln (Keq)
But in the next slide, the equation given is delta G rxn = delta G0 = RT (Q/ln eq)
Which is it?
Lesson 10 video

 Posts: 466
 Joined: Sat Mar 30, 2019 8:39 pm
Re: Lesson 10 video
Can you tell me in which of the 10 thermochemistry subsections you found this, and maybe a timestamp? I see the subject you're talking about mentioned in part 10, but not the exact problem you're describing.
Re: Lesson 10 video
Yes, Thank you, these are in Part 7, the first one is in slide 9/11, Gibbs Equation, and the second one is in the next slide, slide 10/11

 Posts: 466
 Joined: Sat Mar 30, 2019 8:39 pm
Re: Lesson 10 video
Okay, I got you.
The problem that you're having is that the first equation presented, dG = dG_nought + RTln(Keq) is wrong. That should be dG = dG_nought + RTln(Q).
That change alone makes a massive difference. Now it is just saying that the favorability of a reaction depends on some constant (dG_nought) and the concentration of products and reactants. One would hope that is true!
The second one luckily is better, and this is just some use of the properties of logarithms: ln(A)ln(B) = ln(A/B)
so you had dGrxn = RTln(Keq) + RTln(Q) = RTln(Keq/Q)
This is totally true and actually just a very minor rearrangement of the first. If I had to pick one to memorize, it would definitely be dG = dG_nought + RTln(Q). But you do want to understand how these other relationships are derived and what they imply.
For example, since we said that dG_nought_rxn = RTln(Keq), you would want to notice that this is actually a little bit of a stupid statement in it's obviousness: It says that you can tell how favorable a reaction is under standard conditions (dG_nought_rxn) by just letting it go to equilibrium, then measuring the concentrations of products and reactants at equilibrium (and taking the log of that ratio, adjusted for coefficient exponents, at the end). Or simply put: You find out that if you have a lot of product and few reactants, ln(lots/little) = a positive number and your dG_nought_rxn ends up negative. If you have lots of reactants left and only very little product, then ln(little/lots) = a negative number and you end up with a positive dG_nought_rxn.
For all the intimidatinglooking math, don't be fooled: If you take some time out to rearrange and resolve these terms by yourself, you will find that they can become very intuitive.
The problem that you're having is that the first equation presented, dG = dG_nought + RTln(Keq) is wrong. That should be dG = dG_nought + RTln(Q).
That change alone makes a massive difference. Now it is just saying that the favorability of a reaction depends on some constant (dG_nought) and the concentration of products and reactants. One would hope that is true!
The second one luckily is better, and this is just some use of the properties of logarithms: ln(A)ln(B) = ln(A/B)
so you had dGrxn = RTln(Keq) + RTln(Q) = RTln(Keq/Q)
This is totally true and actually just a very minor rearrangement of the first. If I had to pick one to memorize, it would definitely be dG = dG_nought + RTln(Q). But you do want to understand how these other relationships are derived and what they imply.
For example, since we said that dG_nought_rxn = RTln(Keq), you would want to notice that this is actually a little bit of a stupid statement in it's obviousness: It says that you can tell how favorable a reaction is under standard conditions (dG_nought_rxn) by just letting it go to equilibrium, then measuring the concentrations of products and reactants at equilibrium (and taking the log of that ratio, adjusted for coefficient exponents, at the end). Or simply put: You find out that if you have a lot of product and few reactants, ln(lots/little) = a positive number and your dG_nought_rxn ends up negative. If you have lots of reactants left and only very little product, then ln(little/lots) = a negative number and you end up with a positive dG_nought_rxn.
For all the intimidatinglooking math, don't be fooled: If you take some time out to rearrange and resolve these terms by yourself, you will find that they can become very intuitive.