In: Chemistry
At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.500 M.
N2(g) + O2 (g) <----> 2NO(g)
If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re-established?
_____M
The balanced equilibrium reaction is:
N2(g) + O2(g) ---> 2NO(g)
Kc = [NO]2 / [N2] [O2] = (0.5 )2 / 0.1 X 0.1 = 25
Now the new concentration of NO in the system is 0.8 M
Let us see the changes through ICE table
N2(g) + O2(g) ---> 2NO(g)
initial 0.1 0.1 0.8
Change +x +x -2x
Equilibrium 0.1+x 0.1 +x 0.8-2x
Kc = [NO]2 / [N2] [O2] = 25
25 = (0.8-2x)2 / (0.1+x)(0.1+x)
5 = 0.8-2x / 0.1+x
0.5 + 5x = 0.8 -2x
7x = 0.3
x = 0.0429 M
Therefore [NO] after equilibrium = 0.8-2x = 0.8- 2(0.0429) = 0.7142 M