Question

The equation, $ \alpha = \dfrac{{D - d}}{{\left( {n - 1} \right)d}} $ is correctly matched for which of the following reactions?
A) $ A \rightleftarrows {\raise0.7ex\hbox{ $ n $ } \!\mathord{\left/
 {\vphantom {n 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}B + {\raise0.7ex\hbox{ $ n $ } \!\mathord{\left/
 {\vphantom {n 3}}\right.}
\!\lower0.7ex\hbox{ $ 3 $ }}C $
B) $ A \rightleftarrows {\raise0.7ex\hbox{ $ n $ } \!\mathord{\left/
 {\vphantom {n 3}}\right.}
\!\lower0.7ex\hbox{ $ 3 $ }}B + {\raise0.7ex\hbox{ $ {2n} $ } \!\mathord{\left/
 {\vphantom {{2n} 3}}\right.}
\!\lower0.7ex\hbox{ $ 3 $ }}C $
C) $ A \to {\raise0.7ex\hbox{ $ n $ } \!\mathord{\left/
 {\vphantom {n 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}B + {\raise0.7ex\hbox{ $ n $ } \!\mathord{\left/
 {\vphantom {n 4}}\right.}
\!\lower0.7ex\hbox{ $ 4 $ }}C $
D) $ A \rightleftarrows {\raise0.7ex\hbox{ $ n $ } \!\mathord{\left/
 {\vphantom {n 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}B + C $

Answer 1

Hint :Degree of dissociation of an equilibrium reaction is defined as the fraction of a mole of reactant/reactants undergoing dissociation. The relation between vapor density and number of moles of a reaction is given as $ \dfrac{D}{d} = \dfrac{m}{M} $ where D and M are density and molecular weight at equilibrium, d and m are initial density and molecular weight.

Complete Step By Step Answer:
The relation between degree of dissociation and vapor density is given as $ \alpha = \dfrac{{D - d}}{{\left( {n - 1} \right)d}} $
We can write a relation between vapor density and number of moles as $ \dfrac{D}{d} = \dfrac{m}{M} $
Let us solve this relation for every reaction separately.
A) Given reaction: $ A \rightleftarrows {\raise0.7ex\hbox{ $ n $ } \!\mathord{\left/
 {\vphantom {n 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}B + {\raise0.7ex\hbox{ $ n $ } \!\mathord{\left/
 {\vphantom {n 3}}\right.}
\!\lower0.7ex\hbox{ $ 3 $ }}C $
Let us assume that the concentration of A initially is ‘c’ and now we can write the concentrations of the products and reactants as follows.

Time	A	B	C
$ t = 0 $	$ c $	$ 0 $	$ 0 $
$ T = t $	$ c(1 - \alpha ) $	$ \dfrac{n}{2}c\alpha $	$ \dfrac{{2n}}{3}c\alpha $

Total number of moles initially: $ c + 0 + 0 = c $
Total number of moles at equilibrium: $ c\left( {1 - \alpha } \right) + {\raise0.7ex\hbox{ $ n $ } \!\mathord{\left/
 {\vphantom {n 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}c\alpha + {\raise0.7ex\hbox{ $ n $ } \!\mathord{\left/
 {\vphantom {n 3}}\right.}
\!\lower0.7ex\hbox{ $ 3 $ }}c\alpha = c\left( {1 + \alpha \left( {{\raise0.7ex\hbox{ $ {5n} $ } \!\mathord{\left/
 {\vphantom {{5n} 6}}\right.}
\!\lower0.7ex\hbox{ $ 6 $ }} - 1} \right)} \right) $
By substituting this in the above equation, we get $ \dfrac{D}{d} = \dfrac{{c\left( {1 + \alpha \left( {{\raise0.7ex\hbox{ $ {5n} $ } \!\mathord{\left/
 {\vphantom {{5n} 6}}\right.}
\!\lower0.7ex\hbox{ $ 6 $ }} - 1} \right)} \right)}}{c} = 1 + \alpha \left( {{\raise0.7ex\hbox{ $ {5n} $ } \!\mathord{\left/
 {\vphantom {{5n} 6}}\right.}
\!\lower0.7ex\hbox{ $ 6 $ }} - 1} \right) $
By solving, we get $ \dfrac{D}{d} - 1 = \alpha \left( {{\raise0.7ex\hbox{ $ {5n} $ } \!\mathord{\left/
 {\vphantom {{5n} 6}}\right.}
\!\lower0.7ex\hbox{ $ 6 $ }} - 1} \right) \Rightarrow \dfrac{{D - d}}{{d\left( {{\raise0.7ex\hbox{ $ {5n} $ } \!\mathord{\left/
 {\vphantom {{5n} 6}}\right.}
\!\lower0.7ex\hbox{ $ 6 $ }} - 1} \right)}} = \alpha $
Clearly, this equation is not the correct match.
B) Similarly for this reaction, we write the table as:

Time	A	B	C
$ t = 0 $	$ c $	$ 0 $	$ 0 $
$ T = t $	$ c(1 - \alpha ) $	$ \dfrac{n}{2}c\alpha $	$ \dfrac{{2n}}{3}c\alpha $

Total number of moles initially: $ c + 0 + 0 = c $
Total number of moles at equilibrium: $ c\left( {1 - \alpha } \right) + {\raise0.7ex\hbox{ $ n $ } \!\mathord{\left/
 {\vphantom {n 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}c\alpha + {\raise0.7ex\hbox{ $ {2n} $ } \!\mathord{\left/
 {\vphantom {{2n} 3}}\right.}
\!\lower0.7ex\hbox{ $ 3 $ }}c\alpha = c\left( {1 + \alpha \left( {n - 1} \right)} \right) $
By substituting this in the above equation, we get $ \dfrac{D}{d} = \dfrac{{c\left( {1 + \alpha \left( {n - 1} \right)} \right)}}{c} \Rightarrow \dfrac{D}{d} = 1 + \alpha \left( {n - 1} \right) $
By solving, we get $ \dfrac{D}{d} - 1 = \alpha \left( {n - 1} \right) \Rightarrow \dfrac{{D - d}}{{d\left( {n - 1} \right)}} = \alpha $
This equation is a correct match for the given equation.
C) Since degree of dissociation is applicable for only equilibrium reaction, this option can be ruled out.
D) We can calculate the same for this reaction and the resultant equation would not be a match for the given equation.
Therefore, the answer is option B.

Note :
For a reaction to be correctly matched to the given equation, the total number of moles of products should be ‘n’. So calculate the number of moles of products in every equation given. Number of moles of products for option B is $ \dfrac{n}{3} + \dfrac{{2n}}{3} = n $ and hence this would be a correct match.