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How would I solve this? (Chem question with ICE table)?
kc=1.2 N2 + 3H2 -> 2Nh3
initial concenctrations:
[H2] = 0.75 mol/L
[N2] = 0.55 mol/L
[NH3] = 0.5 mol/L
(predict the cange in concentration that each gas will undergo in reaching equilibrium
so how would I do this? I'd use kc = [products]/ [reactancts] I guess, but It would end up having a cubic in there right? (because of the 3 H2)?
How would you do this equation? (Idk how to work with cubics?)
1 Answer
- HPVLv 710 years agoFavorite Answer
Molarity . . . . . . .N2 + 3H2 <==> 2NH3
initial . . . . . . . . 0.55 . . 0.75 . . . . . .0.50
change . . . . . . . .-x . . . .-3x . . . . . . .2x
at equilibrium . .0.55-x . 0.75-3x . . .0.50+2x
Kc = ([NH3]^2 / [N2][H2]^3) = ((0.50+2x)^2 / (0.55-x)(0.75-3x)^3)) = 1.2
(4x^2 + 2x + 0.25) / (27x^4 - 35.1x^3 + 16.2x^2 - 3.21x + 0.23) = 1.2
4x^2 + 2x + 0.25 = (1.2)(27x^4 - 35.1x^3 + 16.2x^2 - 3.21x + 0.23)
4x^2 + 2x + 0.25 = 32.4x^4 - 42.1x^3 + 19.4x^2 - 3.8x + 0.28
32.4x^4 - 42.1x^3 + 15.4x^2 - 5.8x + 0.03 = 0
That's pretty UGLY! Using the quartic equation solver referenced below, the only usable root is x = 0.005.
[N2] = 0.55 - x = 0.55 - 0.005 = 0.545 M
[H2] = 0.75 - 3x = 0.75 - 0.015 = 0.735 M
[NH3] = 0.50 + 2x = 0.50 + 0.01 = 0.510 M
