Quiz 1 Prep #8

Moderators: Chem_Mod, Chem_Admin

Jasmine Holloway 1G
Posts: 39
Joined: Fri Sep 25, 2015 3:00 am

Quiz 1 Prep #8

Postby Jasmine Holloway 1G » Tue Jan 26, 2016 8:45 am

Cl2(g) 2Cl(g)

Can someone explain why this reaction is spontaneous at high temperatures?

Posts: 62
Joined: Fri Sep 25, 2015 3:00 am

Re: Quiz 1 Prep #8

Postby 704578485 » Tue Jan 26, 2016 1:07 pm

This reaction involves simply breaking the Cl-Cl bond into plain unbonded Cl. Whenever you break a bond, heat must be added. Since only the Cl-Cl bonds are broken and no bonds are formed, we know the reaction is endothermic, taking in heat. This means that deltaH is positive and deltaS is positive too since it is equal to qp/T. Since deltaG=deltaH-TdeltaS we can see that delta G will be negative(meaning the reaction is spontaneous) when T is a high value.

Posts: 6
Joined: Fri Sep 25, 2015 3:00 am

Re: Quiz 1 Prep #8

Postby haileymoto » Tue Jan 26, 2016 1:33 pm

Since the bond between the Cl2(g) molecule is broken to form 2Cl(g), the process is endothermic and therefore delta H is positive. The way I figured out whether delta S is positive or negative is by thinking about how the reactant Cl2 (g) is merely one gas molecule in a flask, whereas the product 2Cl (g) is two molecules and therefore delta S is positive. From the equation: delta G = delta H-T delta S, in order to have a negative delta G with a positive delta H and a positive delta S, temperature has to be a high value.

Posts: 18878
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 714 times

Re: Quiz 1 Prep #8

Postby Chem_Mod » Tue Jan 26, 2016 1:41 pm

The above discussions are correct. Well done! Since the entropy is increasing due to an increase in the number of particles, the reaction will be spontaneous at high temperature (entropy driven)

Return to “Gibbs Free Energy Concepts and Calculations”

Who is online

Users browsing this forum: No registered users and 2 guests