Question

In a $ 0.25 $ litre tube disassociation of $ 4 $ moles of $ NO $ takes place. If its degree of dissociation is $ 10\% $ . The value of $ 2NO \rightleftharpoons N{}_2 + O{}_2 $ is;
 $ a)\dfrac{1}{{(18){}^2}} \\
  b)\dfrac{1}{{(8){}^2}} \\
  c)\dfrac{1}{{16}} \\
  d)\dfrac{1}{{32}} \\ $

Answer 1

Hint :One should know what is the degree of dissociation, and Le Chatelier's principle. The relationship between both of them in terms of gasses. Let us find out the relationship and solve the given problem.

Complete Step By Step Answer:
Let us solve the problem with the help of the given values;
We are given that the volume $ V = 0.25 $ , the initial molar mass of $ NO = 4 $ mol, $ \alpha = 0.1 $
 $ 2NO \rightleftharpoons N{}_2 + O{}_2 $
At time $ t = 4 $ is the given value of $ NO $ so at equilibrium it will be $ 4 - 4\alpha $ where $ \alpha $ is degree of dissociation which is given to us as $ 10\% $ ( $ 0.1 $ in terms of moles)
 $ 4(1 - 0.1)2 \times 0.1 \times 0.1 $
 $ Kp = \dfrac{{[N{}_2][O{}_2]}}{{[NO]{}^2}} $
 $ \dfrac{{[\dfrac{{4 \times 0.1}}{{0.25 \times 2}}][\dfrac{{4 \times 0.1}}{{0.25 \times 2}}]}}{{[\dfrac{{4 \times 0.9}}{{0.25}}]}} $
 $ Kp = \dfrac{1}{{(18){}^2}} $
The equation in terms of Le chatelier's principle $ Kp $ will be constant with temperature. When pressure is added to the system it increases the value of $ P{}_T $ so the degree of dissociation will decrease to keep $ Kp $ constant.
Hence the correct option is option A.

Additional Information:
Degree of dissociation is the fraction of original solute molecules that are dissociated. It is denoted by $ \alpha $ . More evidently the degree of dissociation refers to the amount of solute which are dissociated into ions or radicals per mole. If there is presence of very strong acids and bases, the degree of dissociation will approximately be $ 1 $ . Weak acids and bases will have lesser degree of dissociation. The relationship between degree of dissociation and van't hoff factor $ i $ is given when the solute dissociates into $ n $ ions then,
 $ i = 1 + \alpha (n - 1) $ .

Note :
In terms of gasses,the degree of dissociation refers to the percentage of gas molecule which dissociates. The relationship between $ Kp $ and $ \alpha $ exists depending on the stoichiometry of the equation.