Question

At 100⁰C the vapour density of nitrogen peroxide $({{N}_{2}}{{O}_{2}})$ is 26.8. The percentage dissociation into $N{{O}_{2}}$ molecule is:
A. 71.4%
B. 61.57%
C. 83.56%
D. 67.39%

Answer 1

Hint: To calculate the percentage dissociation we have the formula of $\dfrac{D-d}{d}\times 100$ where 'd' is the observed vapour density and 'D' in the initial density and the formula to calculate the initial density is $\dfrac{Mwt}{2}$ where 'Mwt' is the molecular weight of the compound.

Complete step by step solution:
From your chemistry lessons you have learned about the vapour density, degree of dissociation and percent dissociation.
Vapour density is the density of vapour in the relation with hydrogen.
Let us take a chemical reaction and find the relation of degree of dissociation with vapour density.
Degree of dissociation is defined as fraction of no. of mole of reactant undergoing dissociation and it is denoted as $'\alpha '$
\[\alpha =\dfrac{no.\,of\,mole\,dissociated}{no.\,of\,moles\,present\,initially}\]
So, in the reaction
\[A\rightleftharpoons yB\]
Initial mole- for A-1
for B-0
At equilibrium $(1-\alpha )$ $y\alpha $ ( Where $\alpha $ is the degree of dissociation)
So, the no. of moles of A and B at equilibrium will be = $1+\alpha +y\alpha =1+\alpha (y-1)$
Initial if the volume of 1 mole is V then, the volume of mixture of A and B at equilibrium will be = $[1+\alpha (y-1)]V$
Initial molar density (D) = $\dfrac{molecular\,weight}{Volume}=\dfrac{m}{V}$
And, density after dissociation (d) = $\dfrac{molecular\,weight}{[1+\alpha (y-1)]V}$
So, the ratio of $\dfrac{D}{d}=[1+\alpha (y-1)]$
\[\therefore \alpha =\dfrac{D-d}{d(y-1)}\]
And if we have to find the percentage dissociation then the formula will be,
\[Percentage\,dissociation=\dfrac{D-d}{d(y-1)}\times 100\]…………. (10\[\]
Where y= no. of moles of products formed from one mole of reactant
And, to find the initial density (D) we also have a formula = $\dfrac{Mwt}{2}$
Here, in the question the reaction that will takes place is,
\[{{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)\]
In the question the value of final vapour density (d) is given as 26.8
We have to calculate the value of initial density(D) = $\dfrac{Mwt\,of\,{{N}_{2}}{{O}_{4}}}{2}=\dfrac{92}{2}=46$
And y= 2 (2 moles of $N{{O}_{2}}$)
By putting all the values in equation 1 we will get,
\[%\,dissociation=\dfrac{46-26.8}{26.8(2-1)}\times 100=71.4%\]

Thus the correct option will be (A).

Note: In any reaction the no. moles of product from one mole of reactant is two you can directly use the formula of $\alpha =\dfrac{D-d}{d}$ . And if we have to solve the question in terms of molecular mass then the formula will be $\alpha =\dfrac{M-m}{m}$ where 'M' is the initial molecular mass and 'm' is the final molecular mass after equilibrium.