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11.45 Part C

Posted: Sun Jul 30, 2017 2:51 pm
by Jacinda Wollenweber 1D
C) Use your results from parts a and b to determine which is thermodynamically more stable relative to its atoms at 1000 K, Cl2 or F2.

Results: Cl2= .001mol/L; 2Cl=1.1x10^-5mol/L
F2=8.4x10^-3mol/L; 2F=3.2x10^-3mol/L

How exactly do we know which one is the most stable based on these results?

Re: 11.45 Part C

Posted: Sun Jul 30, 2017 9:56 pm
by Davalanya 1F
Fluorine is small and has more lone pairs than Cl. Eventhough fluorine may have a high electronegativity, it loses its electrons a lot easier than Cl due to being an almost completed octet.

Re: 11.45 Part C

Posted: Sun Jul 30, 2017 9:58 pm
by Davalanya 1F
Also, you may want to check your calculations because it's supposed to be Cl2 = 8.2*10^4 and F2= 3.2*10^4. Additionally based on these results it is seen that Cl2 has the largest number making it more stable.

Re: 11.45 Part C

Posted: Tue Aug 01, 2017 8:10 pm
by derek1d
Thank you, but how based on the calculations of concentration are we able to tell? The solutions manual says because of the larger equilibrium concentration, but how does the equilibrium concentration relate back to thermodynamic stability?