Asked by Anonymous
5.88 moles of nitrogen and 16.2 moles of oxygen are mixed and heated at 2000C° until equilibrium is established ,11,28 moles of nitric oxide are formed . Calculate the value of equilibrium constant?
Answers
Answered by
DrBob222
I'm assuming a closed system of some volume I'll call L.
.................N2 + O2 ==> 2NO
I.............5.88.....16.2...........0
C..............?.........?...............?
E..............?..........?..............11.28
So ? for 2NO = 11.28. Then 11.28/2 = 5.64 mols O2 and 5.64 mols N2 were used and the table can be redone this way.
.................N2 + O2 ==> 2NO
I.............5.88.....16.2...........0
C.........-.5.64.....-5.64........11.28
E..........0.240.....10.56.......11.28
Do you want Kc. Note that no volume is available so mols/L can't be calculated; however, the volume, whatever it is, cancels in the calculation. Therefore, write the Kc expression, substitute thre E values in mols, and solve for Kc. To be complete, I would substitute 11.28/L, 0.240/L, and 10.56/L for the molarities of each component. Post your work if you get stuck.
.................N2 + O2 ==> 2NO
I.............5.88.....16.2...........0
C..............?.........?...............?
E..............?..........?..............11.28
So ? for 2NO = 11.28. Then 11.28/2 = 5.64 mols O2 and 5.64 mols N2 were used and the table can be redone this way.
.................N2 + O2 ==> 2NO
I.............5.88.....16.2...........0
C.........-.5.64.....-5.64........11.28
E..........0.240.....10.56.......11.28
Do you want Kc. Note that no volume is available so mols/L can't be calculated; however, the volume, whatever it is, cancels in the calculation. Therefore, write the Kc expression, substitute thre E values in mols, and solve for Kc. To be complete, I would substitute 11.28/L, 0.240/L, and 10.56/L for the molarities of each component. Post your work if you get stuck.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.