Hmwrk 11.45 6th edition

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Lina Petrossian 1D
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Joined: Fri Oct 05, 2018 12:16 am

Hmwrk 11.45 6th edition

Postby Lina Petrossian 1D » Tue Jan 15, 2019 6:06 pm

11.45 (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.

Can someone please go through this problem step by step for me?

Andrea Zheng 1H
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

Re: Hmwrk 11.45 6th edition

Postby Andrea Zheng 1H » Tue Jan 15, 2019 11:25 pm

a) To solve this problem, you would use the reaction Cl2(g) <-> 2Cl(g). From table 2, you see that the K for this reaction is 1.2*10^-7.

ICE TABLE: Because it asks for the equilibrium composition, you can use the ICE table (initial, change, equilibrium). To set this up, you would need to find the initial concentration of Cl2. It is given that it is 2.0 mmol in a 2L vessel, so converting to mol/L, you would get 0.001 mol/L of Cl2. This means the initial concentration of Cl2 is 0.001 M. Because you start with Cl2, there would be 0M Cl initially. For change, you would lose 1 Cl2 as it dissociates to 2 Cl, so you have -1x for Cl2 and +2x from Cl. This means that in the equilibrium line, you would get 0.001-x for Cl2 and 2x for Cl.

SET EQUAL TO K AND SOLVE FOR X:Because these are the values at equilibrium, you can set them equivalent to the value of K from the table, so 1.2*10^-7 = [products]/[reactants] = (2x)^2/(0.001-x). If you multiply this out, you get 4x^2+(1.2*10^-7)x-(1.2*10^-10)=0. You would then use the quadratic formula to get the value of x. The result is x=-5.5*10^-6 or 5.5*10^-6. A negative value of x doesn't make sense, as you can't have a negative concentration, so you would use x=5.5*10^-6.

PLUG X INTO EQUATION FOR EACH SUBSTANCE:
Because you know the equilibrium concentration for Cl2 is 0.001-x, then 0.001-(5.5*10^-6) is around 0.001 M, the equilibrium concentration of Cl2.
Because you know the equilibrium concentration for Cl is 2x, then 2*(5.5*10^-6) is around 1.1*10^-5 M, the equilibrium concentration of Cl.

b) you would use the same procedure for (b), except with a K value of 1.2*10^-4, which comes from Table 2.

c) the more stable one is the Cl2. When comparing the ratio for Cl2:Cl and F2:F, Cl2:Cl is greater. This shows that Cl2 is more stable because at equilibrium, the ratio of Cl2:Cl is greater than the ratio of F2:F.

204929947
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Joined: Fri Apr 06, 2018 11:03 am

Re: Hmwrk 11.45 6th edition

Postby 204929947 » Fri Feb 08, 2019 10:00 pm

im not sure where you got 1.2*10^-7 if in the table it says the equilibrium constant is 1.0*10^-5

Andrea Zheng 1H
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

Re: Hmwrk 11.45 6th edition

Postby Andrea Zheng 1H » Sat Feb 09, 2019 2:22 pm

Because this problem is giving you the molar concentrations (given mmol and L), you would use the column Kc, as the book says that the Kc column is the eq constant for the molar concentration of gases.


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