At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M
and [NO]=0.600 M.
N2(g)+O2(g)↽−−⇀2NO(g)
N
2
(
g
)
+
O
2
(
g
)
↽
−
−
⇀
2
NO
(
g
)
If more NO
NO
is added, bringing its concentration to 0.900 M,
0.900
M,
what will the final concentration of NO
NO
be after equilibrium is re‑established?
[NO]final=
What's with the funny post. You really know how to make it difficult to read and understand the question.
.........................N2(g) + O2(g) <--> 2NO(g)
at Equil...........0.100....0.100..........0.600
Keq = (NO)^2/(N2)(O2) = (0.6)^2/(0.1)(0.1) = 36
Qeq with NO = 0.900, then (0.9)^2/(0.1)(0.1) = 81
Since Qeq > Keq, the NO is too large and the reaction will re-establish equilibrium by shifting to the left.
.....................N2(g) + O2(g) ==> 2NO(g)
I.....................0.100....0.100...........0.900
C....................+x............+x...............-2x
E..................0.100+x....0.100+x.......0.900-2x
Keq = (NO)^2/(N2)(O2) = (0.900-2x)^2/(0.100-x)(0.100-x) = 36
Solve for x, evaluate 0.900 - 2x and that's you answer for (NO) after equilibrium has been re-established.
Post your work if you get stuck.