Equilibrium Partial Pressure: Question 11.7

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Lauren Worley 3C
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Joined: Fri Sep 25, 2015 3:00 am

Equilibrium Partial Pressure: Question 11.7

Postby Lauren Worley 3C » Sun Nov 08, 2015 12:34 pm

The book gives us the decomposition equation of X2(g)=2X(g)
At equilibrium the mole fraction of X2 is 5/17 and the fraction of X is 12/17
We are also given that the initial pressure of X2 is 0.1bar

In the answer book, to find the partial pressures of X and X2 at equilibrium the mole fraction and initial pressure multiplied. Why can we still use the initial pressure? Wouldn't the pressure have increased when it reached equilibrium because each X2 molecule breaks down into two X molecules making the "n"(the number of molecules) in PV=nRT increase?

SubparChemist
Posts: 27
Joined: Fri Sep 25, 2015 3:00 am

Re: Equilibrium Partial Pressure: Question 11.7

Postby SubparChemist » Sun Nov 08, 2015 6:37 pm

There is no actual increase in matter here though, the overall total number of atoms remains the same.

This analogy might help:

If I have 3 apples and I cut two of them in half, I haven't created more apples, I still have the same amount of apples. I still have 3 apples, one of them is just in a different "form" per se.


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