Can you solve this GCHEM question please

[PreDental88]

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The question reads:

N2 + 3H2 <--> 2NH3

8 moles of N2 and 8 moles of H2 are placed in a 2L flask and allowed to come to equilibrium. At equilibrium, 2 moles of NH3 are formed, calculate Keq.

I understand how to set up the Keq equation: [NH3]^2 / [N2] [H2]^3 and that concentration would have to be in molarity (moles over liter), but I don't understand how they calculated the following concentration of reactants and products.

The answer is:


Keq = (1)^2 / (3.5)(2.5)^3

If anyone can help with this problem, I would greatly appreciate it!

 

[mashinka]

haha this took a stupidly long time to solve but i got it!

Lets figure out the concentrations first. So we have 8 moles of N2 and H2 in 2L. This means that the molarity of each reactant is 4M.

We have 2 moles of out product in a 2L solution, so the molarity of NH3 is 1M.

Now, lets make a nice little ICE chart.

N2 + 3H2 <--> 2NH3
I 4 4 0
C -x -3x +2x
E (4-x) (4-3x) (2x)

So since we know that that concentration of our product is going to be 1M, we can say that 2x = 1M. So x=(1/2)
Now, we plug that into the rest of the equations.

Our final concentration of N2 is going to be [4-(1/2)] = 3.5M

Our final concentration of H2 is going to be [4-3(1/2)] = 2.5M

So now, we go back to the Ksp equation that you had and plug in the numbers

[NH3]^2 / [N2] [H2]^3

[1]^2 / [3.5][2.5]^3

ya dig?

 

[PreDental88]

Yes, I dig very much, thank you. lol, but really that was helpful thanks!

 

[MowgliR]

was this from Destroyer?

 

[theilleztpaki]

was this from Destroyer?

yes