posted by Jessica on .
I would also like to see Nathaniels work solved:P I have a similar 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 lotƒº
It is incorrect, where , sadly I cannot say:(
It looks ok to me.
and we can check it.
(0.0558)^2/(0.0442)^2 = 1.594 which is so close to 1.60 (Keq in the problem) it isn't funny. If you are keying into a computer data base, then I suspect there is a problem with s.f. You must decide how many s.f. to report and key those numbers in.
Hmm.. I will check over if I typed in the correct numbers and I will let you know
I will rewrite the equaiton making it clearer
CO2(g) + H2(g)<---> CO(g) + H2O(g) k = 1.60 for 986 degrees Celsius
Does this change anything?
For simplicity, all answers should be correct to two (2) significant figures and not expressed in scientific notation.
The above was also found next to the question
Then 0.056 for CO and H2O should be correct.
And 0.044 for CO2 and H2 should be correct.