Question

If 1.00 mol each of carbon dioxide and hydrogen is initially injected into a 10.0-L reaction chamber at 986 degrees Celsius, what would be the concentrations of each entity at equilibrium?

CO2(g) + H2(g) „²„³ CO(g) + H2O(g) k = 1.60 for 986 degrees Celsius
Heres my work:

C = n/v
= 0.1 mol/L
CO2(g) + H2(g) „²„³ CO(g) + H2O(g)
I 0.1 0.1 0 0
C -x -x +x +x
E 0.1-x 0.1-x x x

Keq = [CO][H2O]/[CO2][H2]

1.60 = x(x)/(0.1 -x)(0.1 ¡V x)

This is the part I am stuck with please help thanks a lotļ

Answer 1

ok. You have
x*x/(0.1-x)(0.1-x) = 1.60. That is
X^2/(0.1-X)^2 =1.60
Take the square root of both sides.
X/(0.1-X) = sqrt 1.60
X/0.1-X)=1.265
X = 1.265(0.1-X)
X = 0.1265 - 1.265X. Rearrange to
X+1.265X = 0.1265
X = 0.1265/2.265 = 0.0558 which rounds to 0.056 M for CO and H2O.
0.1 - X = 0.1 - 0.0558 = 0.0442 which rounds to 0.044 for CO2 and H2.
I don't know how picky your prof is about significant figures so check these out. Check my work. Check my arithmetic. check it all.

Answer 2

It is incorrect, where , sadly I cannot say:(

Answer 3

Check the s.f. and key in the right number for s.f. The problem is solved correctly.