Chemistry really need help!
posted by Angelica on .
I am really struggling with these problems!!
A reaction mixture that consisted of 0.20 mol N2 and 0.20 mol H2 was introduced into a 25.0 L reactor and heated. At equilibrium, 5.0% of the nitrogen gas has reacted. What is the value of the equilibrium constant Kc for the reaction below at this temperature?
N2 + 3H2 <> 2NH3
I don't know how to approach it because Im' not sure how to factor in the 5% of N2 that reacted. I found the molarities of the reactants..and i know the Kc equation I just don't know how to set it up...someone please help!!

at equilibirum here are your moles of chems.
N2= .2*.95moles (.2*.05 reacted)
H2= .23*.2*.05
NH3=2*.2*.05
check those. 
Can you explain why (3*.3*.05) for H2 is subtracted from .2?

nevermind i think I get it now... thanks for your help

I was also wondering about this question. So once you get those moles all you should have to do is find the concentration by dividing by 25 L?
Then plug it into the equation for Kc? I keep getting 320, which isnt right.
Kc= (2*.2*.05/25L)^2/(.2*.95/25L)(.23*.2*.05/25L)^3
That right? 
That's right. Divide by 25 L to convert to concns.
You should end up with the following concns.
N2 = 7.6 x 10^3
H2 = 6.8 x 10^3
NH3 = 8 x 10^4
And if I didn't goof somewhere that gives a Kc of 267.8 which needs to be rounded to the correct number of s.f. I don't know what the correct s.f. are because I don't know if the 0.2 mole is 0.2 or 0.20 or 0.200 or what. And the volume may be 25 or 25.0 L. 
5 mol of ammonia were introduced into a 5 L rxn chamber in which it is partially decomposed at high temperatures. 2NH3(g)<> 3H2(g)+ N2(g) at equlilibrium at a particular temperature, 80% of the ammonia had reacted calculate kc for the rxn

13.5

Answer is 17.3