Question

The degree of association can be defined as the number of moles of a particular substance associated per more of the substance taken.
For example: If out of 10 moles of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ ​, 3 moles of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$combine with $\text{ }{{\text{H}}_{\text{2}}}\text{ }$ ​ to form $\text{ N}{{\text{H}}_{\text{3}}}\text{ }$ ​, then the degree of association of $\text{ }{{\text{N}}_{\text{2}}}\text{ = 0}\text{.3 }$

Consider the equilibrium situation: $\text{ }{{\text{N}}_{\text{2}}}\left( \text{g} \right)\text{+2}{{\text{H}}_{\text{2}}}\left( \text{g} \right)\rightleftharpoons \text{2N}{{\text{H}}_{\text{3}}}\left( \text{g} \right)\text{ }\!\!~\!\!\text{ }$ initially $\text{ }{{\text{N}}_{\text{2}}}\text{ }$​ & $\text{ }{{\text{H}}_{\text{2}}}\text{ }$were mixed in $\text{ }1\text{ }:\text{ }3\text{ }$ molar ratio and after a long time the mean molar mass of the mixture was found to be $\text{ }\frac{34}{3}\text{ g }$. The degree of association of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$​ is________.
A)$\text{ 0}\text{.5 }$
B)$\text{ 0}\text{.4 }$
C)$\text{ 0}\text{.9 }$
D)$\text{ 0}\text{.25 }$

Answer 1

Hint: The degree of association is the fraction of the total number of molecules which combine to form a bigger molecule. For a general reaction, $\text{ nA }\rightleftharpoons \text{ (A}{{\text{)}}_{\text{n}}}\text{ }$ the number of effective moles is equal to the sum of the number of moles of the product and the reactant. The mean molar mass is equal to the calculated molar mass to the effective number of moles at equilibrium.

Complete Solution :
- There are many organic solutes which in non-aqueous solutions undergo association that is two or more molecules of the solute associated to form a bigger molecule. Thus, the number of effective molecules or particles decreases, and consequently the colligative properties are less than the calculated based on single molecules.
- Consider ammonia gas $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ reacts with hydrogen gas $\text{ }{{\text{H}}_{\text{2}}}\text{ }$ to form an ammonia molecule $\text{ N}{{\text{H}}_{\text{3}}}\text{ }$ . Initially, at a time ‘0’ one mole of nitrogen gas reacts with 3 moles of hydrogen gas.at zero time no ammonia gas is formed.
- At the time $\text{ }{{\text{t}}_{\text{eq}}}\text{ }$ the $\text{ (1}-\text{x) }$ moles of nitrogen gas reacts with $\left( \text{3}-3\text{x} \right)\text{ }$ hydrogen gas and forms $\text{ 2x }$ moles of ammonia gas. The association reaction of ammonia is as shown below,
$\text{ }\begin{matrix}
   {} & {{\text{N}}_{\text{2}}}\text{(g)} & \text{+} & \text{2}{{\text{H}}_{\text{2}}}\text{(g)} & \rightleftharpoons & \text{2N}{{\text{H}}_{\text{3}}}\text{(g)} \\
   \text{t=0} & 1 & {} & 3 & {} & 0 \\
   \text{t=}{{\text{t}}_{\text{eq}}} & \left( 1-\text{x} \right) & {} & \left( 3-3\text{x} \right) & {} & 2\text{x} \\
\end{matrix}\text{ }$
Let x be the degree of association. Then,
The number of effective moles is equal to the sum of the moles of at the equilibrium .the number of effective moles are calculated as follows,
$\text{ The number of effective moles = }\left( 1-\text{x} \right)+\left( 3-3\text{x} \right)+2\text{x = 4}-2\text{x }$

- Since the colligative properties are proportional to the number of moles .therefore the mean molar mass of the mixture is equal to the ratio of the sum of normal molar mass to the total number of moles at equilibrium.it is denoted as,
$\text{ }\,\text{mean molar mass = }\dfrac{2x\times 17+\left( 3-3\text{x} \right)\times 2+\left( 1-\text{x} \right)\times 28}{\left( 4-2\text{x} \right)}\text{ }$
We are given the molar mass of the mixture as $\text{ }\dfrac{34}{3}\text{ g }$ . Let's calculate the value of ‘x’ which is the degree of association for the reaction.it is,
$\begin{align}
  & \text{ }\dfrac{34}{3}\text{ = }\dfrac{2\text{x}\times 17+\left( 3-3\text{x} \right)\times 2+\left( 1-\text{x} \right)\times 28}{\left( 4-2\text{x} \right)}\text{ } \\
 & \Rightarrow 34\times \left( 4-2\text{x} \right)\text{ = 3}\times \left( 2\text{x}\times 17+\left( 3-3\text{x} \right)\times 2+\left( 1-\text{x} \right)\times 28 \right) \\
 & \Rightarrow 136-68\text{x = 102x + 18x }-18\text{x + 84}-\text{84x} \\
 & \Rightarrow \text{ 68x = 34 } \\
 & \therefore \text{x = }\dfrac{\text{34}}{\text{68}}\text{ = 0}\text{.5 } \\
\end{align}$
Thus the degree of dissociation of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ is $\text{ 0}\text{.5 }$ .
So, the correct answer is “Option A”.

Note: Note that generally the association of molecules to form an associated molecule is written as $\text{ nA }\rightleftharpoons \text{ (A}{{\text{)}}_{\text{n}}}\text{ }$ where n is the number of molecules combining. Thus the number of moles of the associated molecule is generally written as $\text{ }\dfrac{\alpha }{n}\text{ }$ . Here association reaction results in one molecule only thus n are takes like 1.