Answer : The equilibrium concentration of NO is, 0.0092 M.
Solution :
First we have to calculate the concentration of NO.

The given equilibrium reaction is,

Initially conc. 0 0 0.1576
At eqm. (x) (x) (0.1576-2x)
The expression of
will be,
![K_c=\frac{[NO]^2}{[N_2][O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%7D)

By solving the term, we get:

Neglecting the 0.0839 value of x because it can not be more than initial value.
Thus, the value of 'x' will be, 0.0742 M
Now we have to calculate the equilibrium concentration of NO.
Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M
Therefore, the equilibrium concentration of NO is, 0.0092 M.