Equilibrium and Disassociation of a Diatomic Molecule (11.7)

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Saira Shahid 3K
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Equilibrium and Disassociation of a Diatomic Molecule (11.7)

Postby Saira Shahid 3K » Wed Nov 11, 2015 9:29 pm

I'm having trouble with this question (11.7):

The following flasks show the dissociation of a diatomic molecule, X2, over time (picture attached).
(a) Which flask represents the point in time at which the reaction has reached equilibrium?
(b) What percentage of the X2 molecules have decomposed at equilibrium?
(c) Assuming that the initial pressure of X2 was 0.10 bar, calculate the value of K for the decomposition.

Can someone explain the picture to me and help me on starting the question? Thank you!
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Chem_Mod
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Re: Equilibrium and Disassociation of a Diatomic Molecule (1

Postby Chem_Mod » Wed Nov 11, 2015 9:54 pm

Count the number of diatomics left in each flask: 11, 8, 5, 5

Since no change occurred after the third flask, equilibrium is reached here.
6 out of 11 diatomics have disocciated so about 55% have disocciated

Hint for part c) : The equilibrium expression is the partial pressures of the products over the partial pressures of the reactants. Each partial pressure is equal to the mole fraction of that species times the total pressure (0.10 bar).

Mole fraction of A = # of molecules of A / total # of separate molecules

Helen Shi 1J
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Re: Equilibrium and Disassociation of a Diatomic Molecule (11.7)

Postby Helen Shi 1J » Sat Nov 25, 2017 2:27 pm

can someone explain part c in detail, please?

lauren chung 2f
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Re: Equilibrium and Disassociation of a Diatomic Molecule (11.7)

Postby lauren chung 2f » Wed Nov 29, 2017 2:54 am

For part c, K = (partial pressure of X)^2/(partial pressure of X2)

You find the partial pressure of X by using the formula listed above, (12 molecules/17 molecules)x0.10 bar= 0.071
The partial pressure of X2 is (5 molecules/17 molecules)x0.10 bar= 0.030

K= (0.071)^2/(0.030)
= 0.017

Ashley P 4I
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Re: Equilibrium and Disassociation of a Diatomic Molecule (11.7)

Postby Ashley P 4I » Sun Jan 20, 2019 7:56 pm

So x^2 is derived from the ice table? Is it like a short hand? Or is it a different formula?


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