11.7

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Melissa Per 2J
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Joined: Fri Sep 29, 2017 7:04 am

11.7

Postby Melissa Per 2J » Sun Nov 26, 2017 10:07 pm

In this question, we have four flasks demonstrating the disassociation of a molecule till it reaches equilibrium in flask 3. In part c it asks us to: Assuming that the initial pressure of X2 was 0.10 bar, calculate the value of K for the decomposition. With no other information aside from the pictures showing us this molecule reaching equilibrium in its environment, how am I supposed to find the value of K? I thought I would have to use the PV=nRT because the question gave us P. Please help. Thank you.

Andrea Grigsby 1I
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

Re: 11.7

Postby Andrea Grigsby 1I » Sun Nov 26, 2017 10:24 pm

use the method of how you would find the equilibrium constant
K=[X]^2 / [X2]
=[(12/17)*0.1]^2 / [(5/17)*0.1]
=0.17

the 12/17 and 5/17 are the number of molecules that have/ haven't dissociated, as given in the diagrams of the 4 containers

Michelle Nguyen 2L
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am

Re: 11.7

Postby Michelle Nguyen 2L » Sun Nov 26, 2017 10:45 pm

I thought this problem was confusing though in that the solution manual seems to assume that the pressure within the flask remains constant throughout the reaction. I am predisposed to believe this cannot actually be the case since PV= nRT, hence with volume and temperature staying the same, pressure depends on the number of moles of gas and as the reaction proceeds we get more moles of gas (one mole of reactant produces two moles of product). I am therefore confused as to why the solution manual proceeds with this problem in such a way; is there something I'm missing?

Robert Estrada
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Joined: Fri Sep 29, 2017 7:05 am

Re: 11.7

Postby Robert Estrada » Mon Nov 27, 2017 8:39 pm

How did you determine that flask 3 was the one at equilibrium?

Gurvardaan Bal1L
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Re: 11.7

Postby Gurvardaan Bal1L » Tue Nov 28, 2017 12:17 am

Why do you multiply the mole fractions by 0.1 bar? I understand doing that for X2, but for 2X why wouldn't it be 0.2 bar(seeing as the coefficient is now doubled so the pressure would be too)?

Hena Sihota 1L
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Re: 11.7

Postby Hena Sihota 1L » Tue Nov 28, 2017 1:59 am

@RobertEstrada At equilibrium the concentrations of the reactants and products remain constant. From flask 1 to 2 and from flask 2 to 3 the concentrations of the reactant and product change. However, from flask 3 to 4 the concentrations of the reactant and product remain the same; therefore, you know that the reaction reached equilibrium in flask 3.


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