What is wrong with this Chemistry..

[FishyTheFish]

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What is wrong with the Chemistry I'm doing?

Consider the Haber Process occurring in a rigid container:

N2 + 3H2 <--> 2NH3

Decreasing the volume would increase the pressure and because there are 4 moles of gas on the left side and only 2 on the right side, the products would be favored, right? Well mathematically, no.

Consider Q , the reaction quotient, for this reaction: Q = [NH3]²/[H2]³[N2].
We can rewrite the concentrations as moles/L (so [NH3]² becomes (moles NH3)²/L² where L is the volume of the container)

Doing this for all the concentrations,Q = (moles NH3)²/[(moles H2)³*(moles N2)] * L²

By this expression for Q, if we were to decrease the volume of the container, L² would decrease and thus Q would decrease, meaning there would be less product and more reactant.

So why is this and which one is actually correct?

 

[Double Trouble]

What is wrong with the Chemistry I'm doing?

Consider the Haber Process occurring in a rigid container:

N2 + 3H2 <--> 2NH3

Decreasing the volume would increase the pressure and because there are 4 moles of gas on the left side and only 2 on the right side, the products would be favored, right? Well mathematically, no.

Consider Q , the reaction quotient, for this reaction: Q = [NH3]²/[H2]³[N2].
We can rewrite the concentrations as moles/L (so [NH3]² becomes (moles NH3)²/L² where L is the volume of the container)

Doing this for all the concentrations,Q = (moles NH3)²/[(moles H2)³*(moles N2)] * L²

By this expression for Q, if we were to decrease the volume of the container, L² would decrease and thus Q would decrease, meaning there would be less product and more reactant.

So why is this and which one is actually correct?

"SDN isn't for homework help"

that being said, if you divide Q by a smaller number (L) it will be larger. For the larger Q to equal the right side, the top of the fraction (products) has to be larger.

 

[ReptarBar]

weird.

i know why you are wrong, though. K is always the same, bro. thus, if we decrease L^2 on top, [NH3] must increase to keep K same or [H2] / [N2] must decrease.

/thread.

 

[kpcrew]

"By this expression for Q, if we were to decrease the volume of the container, L² would decrease and thus Q would decrease, meaning there would be less product and more reactant."

I didn't really think about this for too long so I might be wrong but it seems that since Q is decreased, to reach equilibrium you want Q to go back to K. So the reaction would have go towards the right to the products to restore equilibrium.

 

[aSagacious]

Sorry, SDN is not for homework help. Thread closed.