Answer : The equilibrium concentration of NO is, 0.0092 M.
Solution :
First we have to calculate the concentration of NO.
![\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20NO%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DNO%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D%3D%5Cfrac%7B0.3152mol%7D%7B2.0L%7D%3D0.1576M)
The given equilibrium reaction is,
![N_2(g)+O_2(g)\rightleftharpoons 2NO(g)](https://tex.z-dn.net/?f=N_2%28g%29%2BO_2%28g%29%5Crightleftharpoons%202NO%28g%29)
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)
![0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}](https://tex.z-dn.net/?f=0.0153%3D%5Cfrac%7B%280.1576-2x%29%5E2%7D%7B%28x%29%5Ctimes%20%28x%29%7D)
By solving the term, we get:
![x=0.0742,0.0839](https://tex.z-dn.net/?f=x%3D0.0742%2C0.0839)
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.