6th edition, 11.7?

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Tyra Nguyen 4H
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Joined: Thu Sep 27, 2018 11:25 pm

6th edition, 11.7?

Postby Tyra Nguyen 4H » Thu Jan 10, 2019 2:15 pm

The question comes with an image of 4 flasks illustrating the 4 stages towards the dissociation of a diatomic molecule, X2, over time until the reaction reaches equilibrium. Part c of the question asks, "Assuming that the initial pressure of X2 was 0.10 bar, calculate the value of K for the decomposition."

I am confused by this question, as we are only given one pressure, while the equilibrium constant equation has at least 3 terms. What am I missing? I saw that this question was answered earlier on Chem Community but also did not understand.

Chem_Mod
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Re: 6th edition, 11.7?

Postby Chem_Mod » Thu Jan 10, 2019 2:44 pm

You need to set up a I.C.E table, only one pressure was given but you can find the percent of dissociation in the picture which points to the change during the reaction.

Ashley Zhu 1A
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Re: 6th edition, 11.7?

Postby Ashley Zhu 1A » Thu Jan 10, 2019 2:56 pm

What's going on in the flask is the reaction X2 --> 2X, and normally, you'd just use the equation K = (Px)^2/(Px2). However, for this, I used the partial pressures of each gas since it only gives you the pressure of X2 at the very beginning, so you modify the equation to be K = (nx*Pi)^2/(nx2*Pi) where nx is the mole fraction of X present in the third (or fourth - just need one of the ones at equilibrium) flask, nx2 is the mole fraction of X2 in that flask, and Pi is the initial pressure given to you, 0.1 bar.
When you count the number of molecules in the flask, you see that there are 5 X2 molecules and 12 X molecules for a total of 17 molecules in total. With this, you find that nx = 12/17 and nx2 = 5/17. You then multiply these numbers each by the initial pressure because you want their partial pressures. Then, after everything is plugged into the equation, you get about 0.17 for K. Hope this helps.

RachelCheung1A
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Re: 6th edition, 11.7?

Postby RachelCheung1A » Thu Jan 10, 2019 4:41 pm

Thanks that's really helpful. Just wanted to know why would 5/17 be used for the calculation of partial pressure instead of 6/11 from the earlier part B?

Jeannine 1I
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Re: 6th edition, 11.7?

Postby Jeannine 1I » Thu Jan 10, 2019 4:56 pm

RachelCheung1A wrote:Thanks that's really helpful. Just wanted to know why would 5/17 be used for the calculation of partial pressure instead of 6/11 from the earlier part B?


It's because you're looking for the equilibrium constant K, so you use the flask that's at equilibrium! The flask that's before the one with 5/17 is still in the process of reaching that equilibrium state, and the 3rd flask is where equilibrium has been reached.


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