Question
Question: In a \( 0.25 \) litre tube disassociation of \( 4 \) moles of \( NO \) takes place. If its degree of...
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⇌N2+O2 is;
a)\dfrac{1}{{(18){}^2}} \\\
b)\dfrac{1}{{(8){}^2}} \\\
c)\dfrac{1}{{16}} \\\
d)\dfrac{1}{{32}} \\\
Solution
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, α=0.1
2NO⇌N2+O2
At time t=4 is the given value of NO so at equilibrium it will be 4−4α where α is degree of dissociation which is given to us as 10% ( 0.1 in terms of moles)
4(1−0.1)2×0.1×0.1
Kp=[NO]2[N2][O2]
[0.254×0.9][0.25×24×0.1][0.25×24×0.1]
Kp=(18)21
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 PT 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 α . 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+α(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 α exists depending on the stoichiometry of the equation.