State vs Path Functions

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Laura Gu 2K
Posts: 12
Joined: Fri Sep 26, 2014 2:02 pm

State vs Path Functions

Postby Laura Gu 2K » Mon Feb 09, 2015 7:26 pm

On the practice midterm for winter 2014, for question 5C, why is E a path function and not a state function?

Sarah H Brown 1L
Posts: 56
Joined: Fri Sep 26, 2014 2:02 pm

Re: State vs Path Functions

Postby Sarah H Brown 1L » Mon Feb 09, 2015 8:36 pm

Potential is equal to work over charge.

E = w/q (you can check this because volts = joules per coulomb.) **IN THIS CASE "q" IS NOT HEAT, but CHARGE.

Because E can be related to q in this way, which is a path function, E is also a path function.

Also, remember that to find the Eo of a half reaction that is a combination of two known half reactions, you can't simply add together the Eo's of each half reaction (and a quality of a state function is linear addition). If you have:

Half Rxn A Eo= EoA
Half Rxn B Eo = EoB

Half Rxn A + Half Rxn B = Half Rxn C

And you want EoC, you have to solve using free energy Go = -nFEo:

-nFEoC = -nFEoA + -nFEoB

and solve for EoC. Moles of electrons vary depending on the half reaction.

#13.27 and 33 demonstrates this concept which helps explain that E is not a state function.
Last edited by Sarah H Brown 1L on Wed Feb 11, 2015 9:24 am, edited 1 time in total.

Hannah Owens- Lecture 1 section 1E
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Joined: Fri Sep 26, 2014 2:02 pm

Re: State vs Path Functions

Postby Hannah Owens- Lecture 1 section 1E » Tue Feb 10, 2015 10:05 pm

why does e= w/q ?

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: State vs Path Functions

Postby Niharika Reddy 1D » Tue Feb 10, 2015 10:22 pm

Potential difference (E, or max potential) is measured as work(in J)/charge(in C). I don't think the q Sarah used in E=w/q is referring to q as in heat. I would say since work is a path function, potential difference, which is calculated using work, is a path function as well.

Sarah H Brown 1L
Posts: 56
Joined: Fri Sep 26, 2014 2:02 pm

Re: State vs Path Functions

Postby Sarah H Brown 1L » Wed Feb 11, 2015 9:25 am

Thanks for clarifying! Yeah, that is confusing. "q" denotes charge in a physics context when relating potential, work, and charge.


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