Easy chem questions...

[Svart Aske]

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In a 973K, 12L flask.
2H2 (g) + S2 (g) < -> 2H2S (g)
At equilibrium we have 2.5 moles of H2, 1,35x10^-5 moles of S2, 8.7 moles of H2S.

The first two parts ask you to calculate the Keq and Kp; those are easy. But now it says 2.53 moles of Ar are added into the flask and asks you to calculate the new Keq and reaction quotient Q. Since Ar is pretty unreactive, why would be it even be part of the system? I feel like I'm missing some fundamentals.

 

[doc toothache]

http://forums.studentdoctor.net/showthread.php?t=575417

 

[Svart Aske]

^I don't see how that answers my question. The system volume is fixed, so how does adding Ar affect the equilibrium? I understand that Ar changes the partial pressures of the gases, but that doesn't help with finding Keq ad Q.

 

[Dencology]

http://forums.studentdoctor.net/showthread.php?t=575417

yes, i don't get it either.

 

[curelightwounds]

I haven't worked out the problem yet, but adding an inert gas into the closed system would increase pressure.

Increasing the pressure would push equilibrium in the direction of the side with fewer mols of gas.

 

[Svart Aske]

Ah I should've seen that! But I don't remember how to work out the new K (and Q).

 

[jay47]

In a 973K, 12L flask.
2H2 (g) + S2 (g) < -> 2H2S (g)
At equilibrium we have 2.5 moles of H2, 1,35x10^-5 moles of S2, 8.7 moles of H2S.

The first two parts ask you to calculate the Keq and Kp; those are easy. But now it says 2.53 moles of Ar are added into the flask and asks you to calculate the new Keq and reaction quotient Q. Since Ar is pretty unreactive, why would be it even be part of the system? I feel like I'm missing some fundamentals.

Not sure if this is trying to trick you or not...
I think if you added Ar to the mix, you have to assume it will eventually come to equilibrium, and Keq is defined as

Keq= (C^c x D^d)/(A^a x B^b)​

for the reaction

aA(g) + bB(g) = cC(g) + dD(g)​

However, we don't know how Ar will react, if at all, with the reactants, Therefore we do no know the concentrations or molar ratios of any hypothetical products.​

Either one of two things has occurred:
1. They want you to assume that Ar does NOT react and simply changes the partial pressures the gases exert. This would not change Keq because of how Keq is defined. This would negligibly affect the volume and therefore negligibly affect the concentrations, not changing any calculations you could do (because we make "Ideal Gas" assumptions and gases have "no volume"). However, as pointed out earlier, the reaction would shift to the right which would increase Keq by an unknown amount) because there are fewer moles of gas on the right (Le Chatlier's Principle). We cannot know how much though because we don't know how it will react. Just as with any reaction, we cannot assume a mechanism therefore we cannot assume which products would be made.​

2. OR, this is just a dumb question that is trying to trick you or is not well thought out. ​

My guess is that the latter of these two is the case.​

 

[doc toothache]

http://www.shodor.org/UNChem/advanced/equ/index.html

 

[Svart Aske]

🙄, please, there's no need to be insulting by suggesting that I need to brush up on equilibrium. This problem clearly presents an unusual scenario.

@jay47, thanks for attempting.

 

[doc toothache]

🙄, please, there's no need to be insulting by suggesting that I need to brush up on equilibrium. This problem clearly presents an unusual scenario.

@jay47, thanks for attempting.

Perhaps you missed a section.

"changing the volume/pressure (only gases):
Increasing the volume has the same effect as decreasing the pressure and vice versa so we are only going to talk about changing the pressure. When you increase the pressure, the system will shift so the least number of gas molecules are formed because the more gas molecules there are, the more collisions there are. These collisions and the presence of gas molecules are what cause the pressure to increase. Likewise, when you decrease the pressure, the system will shift so the highest number of gas molecules are produced. For example, in the equation:

N2 (g) + 3H2 (g) <---> 2NH3 (g)

if the pressure is increased, the system will shift to the right because fewer gas molecules are produced in the forward reaction than in the reverse reaction."

 

[Svart Aske]

I've already acknowledged that the pressure will change and hence affect the direction of the system. But how do you calculate Keq and Q based on the information given? How do you utilize the information that 2.53 moles of Ar were added?

 

[doc toothache]

I've already acknowledged that the pressure will change and hence affect the direction of the system. But how do you calculate Keq and Q based on the information given? How do you utilize the information that 2.53 moles of Ar were added?

See your pm.