Calculate the equilibrium concentrations of N2, H2, and NH3 present when a mixture that was initially 0.10M N2, 0.10 M H2, and 0.10 M NH3 comes to equilibrium at 500 degrees C.

Question

N2(g) + 3H2(g) <--> 2NH3(g)
Kc = 0.040 (at 500 degrees C)

I'm getting confused with the math.

Answer 1

Why don't you show the math so we will know what is confusing you instead of us guessing at the problem. Here is the chemistry.
.........N2 + 3H2 ==> 2NH3
I.......0.1...0.1......0.1
C.......-x....-3x......2x
E......0.1-x..0.1-3x...2x

Kc = 0.04 = (NH3)^2/(N2)(H2)^3

Answer 2

0.040 = (.1+2x)^2 / (.1-x)(.1-3x)^3
-.00996-.4012x-3.9712x^2+1.08x^3

I'm not sure how to solve for x. I got the chemistry correct

Answer 3

the second line equals 0