No one can figure out this question! NOT EVEN PAID PH.D ONLINE PROFESSIONALS, PLEASE HELP!?
At a particular temperature 8.0 mol NO2 gas is placed in a 1.0-L container. Over time the NO2 decomposes to NO and O2:
2NO2(g) 2NO(g) + O2(g)
At equilibrium the concentration of NO(g) was found to be 3.8 mol/L. Calculate the value of K for this reaction (using units of mol/L for the concentrations).
I ASKED YAHOO ANSWERS B4 AND THEY COULDN’T GET IT RIGHT, I PAID ONLINE PROFESSIONALS $21.75 FOR AN ANSWER/EXPLANATION FROM A PH.D HOLDER IN CHEMISTRY AND THEY GOT IT WRONG TOO.
My Reasoning:
NO is given at 3.8.
O2 would be half of NO at 1.9.
I don’t know how to find NO2 though.
K = [O2][NO]^2 / [NO2]^2
K = [1.9][3.8]^2 / [?]^2
Instructor told me 3.8 since it is at equilibrium, but that does not work either. In my course, the system automatically checks the homework before we can submit it. Each time it says my answer is wrong, but the system isn’t wrong so it’s something flawed. I CANT SUBMIT THE HOMEWORK UNLESS IT’S RIGHT!! I AM GOING CRAZY WITH THIS STUPID PROBLEM.
PLEASE HELP ME, I PAID A CHEM INSTRUCTOR WITH A PH.D IN CANADA AND THEY GOT THE ANSWER WRONG. I AM ENTIRELY FRUSTRATED WITH THIS PROBLEM AND HAVE BEEN TRYING FOR DAYS. PLEASE HELP.
That the concentration NO2 is 3.8 that is. But that is wrong anyway. I will give you 10 pts plus thumbs up plus thumbs up from my sister plus thumbs up from my friend(s), pleaseeeee help.
THANK YOU VERY MUCH PISGAHCHEMIST. I will get as many people as I can to thumbs you up when I give you best answer.
Seriously, thank you. You’re amazing. Amazingly simple response that so many could not figure out. You are loads better than any teacher I have ever had. Thank you again!
This is probably wrong but try 0.475 mol/L….are you making sure to put the “mol/L” part on right because sometimes if you don’t add the units it could say you are wrong.
[NO] = 3.8M, [O2] = 1.9M, [NO2] = 8.0 – 3.8 = 4.2M
K = (1.9)(14.4)/(17.6) = 15.5
You’re almost there:
…. 2NO2 < ==> 2NO + O2
I……8.0 …………..0…….0
E .. 4.2 ………….3.8….1.9
K = [NO]^2 [O2] / [NO2]^2
K = 3.8^2 x 1.9 / 4.2^2
K = 1.56
Since you started with 8.0 mol/L of NO2 and at equilibrium you have 3.8 mol/L, then 3.8 mol/L of NO2 must have reacted. 8.0 – 3.8 = 4.2 mol/L which is the equilibrium concentration of the remaining NO2.