Kc=x^2/(.1-x/2)(.4-3x/2)
avoiding fractions, mutiply right hand side by 4/4
Kx=4x^2/(.2-x)(.8-6x)
.153(.16-1.2x-.8x+6x^2)=x^2
at this point, gather terms, put in quadratic form, and solve for x to get product concentration, and
.1-x/2 for concentration of N2
.4-3x/2 for concentration of H2
check my thinking.
What are the moles of substances present at equilibrium at 450C if 1 mol N2 and 4 mol H2 in a 10 L vessel react according to the following equation?
N2 (g) + 3H_2(g) ⇌ 2NH_3 (g) The equilibrium constant Kc is 0.153 at 450C.
Thanks a lot.
2 answers
I believe my friend made a couple of typos. I think the Kc expression should be 4x^2/(0.2-2x)(.8-6x)^3
where I have bolded the numbers I think it should be.
where I have bolded the numbers I think it should be.
2 answers