Equilibrium and total pressure

[howtomedicine]

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Reading through general chemistry when I encountered this passage on gasses and equilibrium.

1 CS2(g) + 4 H2(g) <-> 1 CH4(g) + 2 H2S(g)

At equilibrium the total pressure is
Ptotal = Pcs2+Ph2+Pch4+Ph2s = 1.0atm -2x

I'm not understanding why it's 1.0 - 2x

I understand that the pressure does decrease at equilibrium if you start with all reactants because you're going from more molecules of gas -> less molecules of gas, but why would it be decreased by 2x?


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[howtomedicine]

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[5words]

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Can you please post the whole question? And i am not following you... So far what i am getting is that two gas were placed in separate containers . The H2 in the reaction flask at 1 atm and the Cs in Flask 1 at 1 atm. And then the stocks were opened allowed both compounds to mix, and we somehow ended up with 0.8 atm of H2 in the reaction flask, meaning that we have 0.2 atm of Cs in the reaction flask. So the whole reaction becomes

0.2 atm of (1) Cs + 0.8 atm of (4) H2 --------- 0 atm of CH4 and 0 atm of 2 H2S. at t=0 .. so i need to know what it the question?

OH, why is P at equilibruim 1- 2x.. I just check the explanation and it;s making sense.

So we know that the system will shift to the side with less number of moles so the pressure has to decrease. So Treat the reaction as you would treat a PH problem.

                            1 Cs + 4 H2 ------------> Ch4  + 2 Hs2    Since they are all gases , it follows that

at t= 0                     0.2      0.8                     0            0
The change                - 1*x     - 4*x                     x            2*x
Concentration at K        0.2 - x    0.8-4x                x           2x
Pressure of eahc become   PCs          PH2                Pch4       PHS2

Pt = Pcs + Ph2 + PCh4 + PHS2 = 0.2-x   + 0.8 -4x   + x + 2x = 1 - 2x

This is what their explanation is showing .. I wonder if there is a shortcut though.. ill check this question later.. idk what they meant by "the decrease of H2 partial pressure is twice that of the internal pressure"

 

[BerkReviewTeach]

This passage appears in an older (kind of outdated) version of the book. You shouldn't worry about this question any longer, which is why we removed the passage from circulation.