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.
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
3 answers
It is incorrect, where , sadly I cannot say:(
Check the s.f. and key in the right number for s.f. The problem is solved correctly.