destroyer gchem 146

[Tina324]

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N2 +3H2 <--> 2 NH3
if 8 moles of N2 and 8 moles of H2 are placed in a 2 liter flask and allowed to come to equilibrium, at equilibirum, 2 moles of NH3 are formed, calculate the Keq.

ans: Keq= 1^2/(3.5)(2.5)^2

helllp! if you have the answer key..the main part that i don't understand is why x=.5 =/

im in super freak out mode, so i really appreciate your help now more than ever guys =)

 

[Tina324]

bump bump bummmmmp

 

[keliao]

relating to 2 Liter flask at eqm?? .......am lost too..

 

[JBarr29]

ok..so when dealing with initial constant and eq questions, you always deal with Molarity, not just moles. I think you understand this. as the destroyer shows, you fill out the start change eq chart with molarity, and the changing concentrations. your equillibrium concentration of NH3 would be 2x correct? in the problem it states that you yield 2 moles of NH3 at eq, which means your molarity concentration is =1. set your 2x=1, and solve for x. x=.5 so now all you have to do is get the other eq concentrations using x=.5 and you should be on the right path. if you still do not understand let me know and il try and do a better job explaining it.

 

[ZenoVT]

At equilibrium 2 mols of NH3 exists. So according the the equation 3 mols of H2 and 1 mol of N2 is needed to make 2 mols of NH3; therefore 8N2 - 1N2 = 7N2, 8H2 -3H2 = 5H2.

Keq = [NH3]^2 / [N2][H2]^2

7 mol N2 / 2 liter = [N2] = 3.5 M
5 mol H2 / 2 liter = [H2] = 2.5 M
2 mol NH3 / 2 mliter = 1 M

Now it's just plug and chug: Keq = 1^2 / (3.5)(2.5)^2