5I.13

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Luyan Zhang - 2D
Posts: 103
Joined: Sat Jul 20, 2019 12:16 am

5I.13

Postby Luyan Zhang - 2D » Wed Jan 08, 2020 9:28 pm

In the answer for part b, it says that the concentration of F2 is 8*10^-4 M. But since 2.0 mmol of F2 was added into the vessel containing 2.0L, shouldn't the concentration b e 0.001M?

And can somebody also explain why Cl2 would be more stable? Is it because it disassociated less?

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

Re: 5I.13

Postby JamieVu_2C » Wed Jan 08, 2020 9:40 pm

.001 M is the initial concentration of F2 before equilibrium is reached, but at equilibrium, the concentration of F2 is 8.3 x 10^-4M.

Cl2 is more stable because its K value, the equilibrium constant, is smaller. This means that at equilibrium there are more reactants than products, so there is more of Cl2(g) than its product.

JasonLiu_2J
Posts: 109
Joined: Sat Aug 24, 2019 12:17 am

Re: 5I.13

Postby JasonLiu_2J » Wed Jan 08, 2020 9:43 pm

When 2.0mmol of F2 is added to the reaction vessel with 2.0L, you find that the initial concentration of F2 is indeed 0.001 M. However, this does not give the equilibrium concentration, which is what you want to find. In this case, you would need to set up the dissociation equation of F2<->2F and use its Kc value (1.2 x 10^-4) to calculate the equilibrium concentration of F2 using an ICE table. Doing this will give you an equilibrium concentration of 8.4 x 10^-4 for F2.
Cl2 is more stable because the reaction is strongly reactant favored. Since the Kc for Cl2<-> 2Cl is 1.2 x 10^-7, you know that the equilibrium concentration of reactants is much higher than that of the products, indicating that the reaction strongly favors the reactants. This shows that the Cl2 is more stable because it is favored in the reaction.
Hope this helps!


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