Textbook Question

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Riya Shah 4H
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Joined: Wed May 02, 2018 3:00 am

Textbook Question

Postby Riya Shah 4H » Thu Jan 17, 2019 4:05 pm

(a) A sample of 2.0 mmol Cl2(g) was sealed into a 2.0-L reaction vessel and heated to 1000. K to study its dissociation into Cl atoms. Use the information in Table 11.2 to calculate the equilibrium composition of the mixture. (b) If 2.0 mmol F2 was placed into the reaction vessel instead of the chlorine, what would be its equilibrium composition at 1000. K? (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.

Could someone please help me out with this question?

Seohyun Park 1L
Posts: 73
Joined: Fri Sep 28, 2018 12:29 am

Re: Textbook Question

Postby Seohyun Park 1L » Thu Jan 17, 2019 5:17 pm

You would use ICE tables to find the appropiate equilbirum concentrations of Cl2 and F2 and the K values are in the textbooks. For part C, you can compare K values or compare concentration values to see whether Cl2 or F2 separates less, leading for it to be more "stable."

Sean_Rodriguez_1J
Posts: 62
Joined: Fri Sep 28, 2018 12:17 am

Re: Textbook Question

Postby Sean_Rodriguez_1J » Thu Jan 17, 2019 6:47 pm

For a), you would start out with balancing the equation of Cl2 dissociating into Cl atoms (Cl2 --> 2CL). We'll use this later. Then, we are given the amount of mols and the volume, so we can calculate the initial concentration of Cl2, which is (2.0x10^-3 mol) / (2.0L) = 0.0010 mol*L^-1. From here, we can make a simple ICE chart. Note the +2x due to the 2 mols of Cl(g).
Cl2(g) -----> 2Cl(g)
I 0.0010 0
C -x +2x
E 0.0010-x 2x
With the rest of the info we have, we can then make an equation for K. 1.2*10^-7 = (2x)^2/(0.0010-x). Our K here is small, so we can approximate this to 1.2*10^-7 = 4x^2/0.0010. Solving for x gives you the answer 5.5 * 10^-6. That will leave the amount of Cl2(g) virtually unchanged, and there will be 2*(5.5*10^-6) = 1.1*10^-5 mol*L^-1.
b) The process here would be exactly the same, but solving for fluorine instead of Chlorine. Replace the K in a) with 1.2*10^-4, and you should be on your way.
For c), We can look at the two equilibrium constants at 1000. K. Cl2's is 1.2*10^-7, and F2's is 1.2 x10^-4. This means that at equilibrium, there is more Cl2 than F2 ie Chlorine favors its reactants more. Therefore, Cl2 is more stable.


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