11.15






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Wilson Yeh 1L
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Joined: Fri Sep 29, 2017 7:06 am

11.15

Postby Wilson Yeh 1L » Thu Feb 08, 2018 5:42 pm

(a) Calculate the reaction Gibbs free energy of I2(g) --> 2I(g) at 1200. K (K = 6.8) when the partial pressures of I2 and I are 0.13 bar and 0.98 bar, respectively. (b) What is the spontaneous direction of the reaction? Explain briefly.

Can someone explain how to approach this? I'm so lost.

Lena Nguyen 2H
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Re: 11.15

Postby Lena Nguyen 2H » Thu Feb 08, 2018 5:52 pm

Start with the equations and
Everything is given except Q, which can be calculated using the partial pressures given. The Gibbs free energy of the reaction can then be calculated.
Hope this helps!

Wilson Yeh 1L
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Joined: Fri Sep 29, 2017 7:06 am

Re: 11.15

Postby Wilson Yeh 1L » Thu Feb 08, 2018 5:59 pm

Lena Nguyen 2H wrote:Start with the equations and
Everything is given except Q, which can be calculated using the partial pressures given. The Gibbs free energy of the reaction can then be calculated.
Hope this helps!

Thanks! I did it but ended up with 8.3 not the answer of 8.3 x 10^-1 in the back of the book. I feel like it might be because of the value of R I'm using. I used 8.314 x 10^-2, is that right?

EDIT: Yeah I used the wrong R. 8.314 works to get the right answer!

Essly Mendoza 1J
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Re: 11.15

Postby Essly Mendoza 1J » Thu Feb 08, 2018 11:04 pm

Hi, can someone explain why we use this Gibbs free energy equation specifically, thank you.

Jessica Benitez 1K
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Re: 11.15

Postby Jessica Benitez 1K » Fri Feb 09, 2018 6:08 pm

We need to use this Gibbs Free Energy equation because it includes K (the equilibrium constant) which is given in the question. We can also use the partial pressures given in the question to calculate Q which is also needed to solve for delta G. This equation contains values that we were either given or are able to find given the information in the question which is why it is the best to use in this situation.

zanekoch1A
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Joined: Thu Jul 13, 2017 3:00 am

Re: 11.15

Postby zanekoch1A » Sat Feb 10, 2018 4:19 pm

I'm not sure what I am doing wrong but I keep getting an answer around 1.5kj. I am using in the Gnot part of the equation a T of 273 and a k of 6.8 and and in the G side a T of 1200 and ln(.13/.98). What am I doing wrong?

torieoishi1A
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Re: 11.15

Postby torieoishi1A » Sun Feb 11, 2018 10:09 am

If deltaG is positive, how is the reaction spontaneous?

torieoishi1A
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Re: 11.15

Postby torieoishi1A » Sun Feb 11, 2018 10:20 am

Nevermind, I reread the question

Tim Foster 2A
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Re: 11.15

Postby Tim Foster 2A » Tue Feb 13, 2018 3:15 pm

zanekoch1A wrote:I'm not sure what I am doing wrong but I keep getting an answer around 1.5kj. I am using in the Gnot part of the equation a T of 273 and a k of 6.8 and and in the G side a T of 1200 and ln(.13/.98). What am I doing wrong?


Hey Zane, I think your natural log should be of (.98)^2/(.13)

AKatukota
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Re: 11.15

Postby AKatukota » Tue Feb 25, 2020 5:57 pm

How do you know delta G is 0 again?

AKatukota
Posts: 100
Joined: Thu Jul 25, 2019 12:18 am

Re: 11.15

Postby AKatukota » Tue Feb 25, 2020 6:01 pm

Im also a bit confused as to why we are calculating Q again?

AKatukota
Posts: 100
Joined: Thu Jul 25, 2019 12:18 am

Re: 11.15

Postby AKatukota » Tue Feb 25, 2020 6:18 pm

Also can you use Barr in calculating Q? I thought it had to be atm

Hannah_1G
Posts: 100
Joined: Wed Sep 18, 2019 12:22 am

Re: 11.15

Postby Hannah_1G » Thu Mar 05, 2020 12:02 am

How do we know this reaction is at equilibrium? Shouldn't only Q be used?


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