Moderators: Chem_Mod, Chem_Admin

Ryan Fang 1D
Posts: 53
Joined: Sat Jul 22, 2017 3:00 am


Postby Ryan Fang 1D » Tue Feb 13, 2018 11:15 am

(a) Calculate the reaction Gibbs free energy of I2(g) -> 2 I(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.

For this question, why does the solutions manual calculate ∆G° using 1200. K instead of 298 K? Since ∆G° is the Gibbs Free Energy of the reaction at standard state, shouldn't we use 298 K?
Last edited by Ryan Fang 1D on Tue Feb 13, 2018 11:31 am, edited 1 time in total.

Kelly Kiremidjian 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

Re: 11.15

Postby Kelly Kiremidjian 1C » Tue Feb 13, 2018 11:19 am

the question states that the reaction is at 1200K

Britney Alvey 1B
Posts: 31
Joined: Fri Jun 23, 2017 11:39 am

Re: 11.15

Postby Britney Alvey 1B » Tue Feb 13, 2018 11:19 am

The question states that the reaction is happening at 1200K so you would use this value to calculate deltaG. The solutions manual calculates delta G not . This is an important distinction because would be calculated at 298K by definition.

Return to “Gibbs Free Energy Concepts and Calculations”

Who is online

Users browsing this forum: No registered users and 2 guests