HW 1: Question #9

[Shaniya Kerns 4D]

At equilibrium, the concentrations in this system were found to be [N2] = [O2] = 0.300 M and [NO] = 0.600 M.
N2(g) + O2(g) ---- 2 NO(g)
If more NO is added, bringing its concentration to 0.900 M, what will the final concentration of NO be after equilibrium is re-established?
I need help with finding the correct answer.
I found X = 0.0750 M and need help with the the next step in order to find NO final.
[NO]final= 0.075

[Spencer T]

First calculate K_c, then use an ice table on the equilibrium reaction, including the added concentration of [NO].

Set K_c equal to the ice table result.

Notice that the expression is a perfect square, so you can take the sqrt of both sides to solve for x.

Finally, refer back to the ice table to calculate [NO]_final.

[Hansika_Sakshi_2k]

calculate Kc first. Then write the Kc equation in terms of x and then solve for x using Kc. Then find final concentration , 0.900 - 2x.

[ivy knabe 3i]

i checked my achieve and i got a different x than you, but the next step should be [NO]final = 0.900-2x, since the ice table calculates initial change, and the NO is added after