Chemistry Equillibrium Concentrations

[derek358]

Can someone help me solve this problem and explain the procedure taken to solve it?

A short explanation is fine because i will realize what to do but right now im just stuck?

In chemistry we were given a problem and told that only a few people will be able to solve it since it is an AP question.

the problem is as follows:

For the equillibrium 2A (gas) + B (gas) <> C (gas) K= 2.5 * 10^3 or (2500)
Assume 1 mole of A and 1 mole of B are placed into a 1 liter flask.
What are the concentrations of A,B,C at equillibrium?

Please help me answer this. Thanks guys.

 

[derek358]

come on guys can u please help me

i read something about it right now and it helped a little but im still not sure what the molarity of the 2 moles would be

 

[CycloWizard]

The general solution for an equilibrium constant is the activity of products over the activity of reactants. The 'activity' for your purposes will be nothing but the concentration of each species raised to the power of the stoichiometric coefficient.

Remember, this forum isn't for homework, though you can PM me with chemistry questions and I can probably help.

 

[BigPoppa]

K = [C]/([A]^2*[B.]) = 2.5x10^3

Use stoichiometry, pay special attention to the coefficient of gas A.

Addition:

Make an ICE (initial, change, equilibrium table):

.....2A + B --> C
I..1mol | 1mol | 0 mol
C.-2x | -x | +x
E. 1-2x | 1-x | x

The change line is your stoichiometric ratio line, you gain on C because of Le Chatelier's principle.

Plug the E line values into the K equation I gave you above, solve for x. Then plug x back into the E line for the final moles of each gas. Don't forget to square the A value.

 

[BigPoppa]

Man, not even a thank you for doing 90% of the work?

 

[CycloWizard]

Originally posted by: BigPoppa
Man, not even a thank you for doing 90% of the work?

Haha.... I already took care of it for him via PM. I guess he hasn't checked back here since.

 

[imported_booradley]

Woohoo.. this post caught my attention since the AP test was today at 8 A.M. Hope this kid did alrite.....

EDIT: (First post )