Degree of dissociation

The concept of the degree of dissociation was developed by the Swedish chemist Svante Arrhenius in the late 19th century. Arrhenius introduced this idea as part of his broader work on ionic theory, which was crucial in explaining the behavior of electrolytes in solutions. His theories were fundamental in developing the modern understanding of chemical reactions in solution and were critical in the advancement of physical chemistry.

He came to this discovery because Arrhenius studied how substances like acids, bases, and salts dissociate into ions when dissolved in water. He noted that these substances conducted electricity, which he attributed to the formation of ions in solution. He proposed the ionic theory, which suggested that electrolytes dissociate into charged particles (ions) in solution.

Degree Of Dissociation

Degree of dissociation: It is the extent to which any reactant gets dissociated. It is denoted by $\alpha$..

$\alpha=\frac{\text { number of molecules dissociated }}{\text { total number of molecules }}$

If "a" is the initial number of moles and the number of moles dissociated is "x" then

$\alpha=\frac{\mathrm{X}}{\mathrm{a}}$

Relation between degree of dissociation and Observed Molar mass/Vapor density

Due to dissociation, the total number of moles at equilibrium can be determined. Knowing the number of moles at equilibrium, the observed molar mass can be calculated.

Conversely, if the degree of dissociation is not known but the Observed Vapor density is available, then we can calculate the degree of dissociation.

We know about the law of conservation of mass

Mass initially taken $=$ Mass after dissociation at equilibrium
$\therefore$ Theoretical moles $\times$ Theoretical molar mass $=$ Observed Moles $\times$ Observed Molar mass

Using the above equation, we can calculate the unknown term required to be calculated.

Application Of Degree Of Dissociation

The degree of dissociation is a measure of how much a compound dissociates into its ions in a solution. It's particularly important in the context of weak electrolytes. Here are some applications:
Chemical Equilibria: It helps in understanding the extent to which weak acids or bases dissociate in solution, which is crucial for predicting the concentrations of ions at equilibrium

Buffer Solutions: The degree of dissociation informs the pH calculations of buffer solutions, which are used to maintain stable pH levels in various chemical and biological processes.

Temperature Dependence of Equilibrium Constant: Vant Hoff's Equation

$\begin{aligned} \ln \left[\frac{\mathrm{K}_{\mathrm{T}_2}}{\mathrm{~K}_{\mathrm{T}_1}}\right] & =\frac{\Delta \mathrm{H}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right] \\ \Rightarrow 2.303 \log _{10}\left[\frac{\mathrm{K}_{\mathrm{T}_2}}{\mathrm{~K}_{\mathrm{T}_1}}\right] & =\frac{\Delta \mathrm{H}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]\end{aligned}$

Using the above equation, the value of Keq at any unknown temperature can be calculated if the Keq value at a particular temperature $\Delta H$ is known.

Conversely, the above equation can also be used to calculate the value of of $\Delta \mathrm{H}$ if the values of Keq at two different temperatures are known.

Recommended topic video on (Degree of Dissociation )

Some Solved Examples

Example 1. The equilibrium constants $K_{p_1}$ and $K_{p_2}$ for the reaction $X \rightleftharpoons 2 Y$ and $Z \rightleftharpoons P+Q$,respectively are in the ratio of 1 : 9 . If the degree of dissociation of $X$ and $Z$ be equal then the ratio of total pressures at these equilibria is

1)1 $1: 9$
2) (correct) $1: 36$
3) $1: 1$
4) $1: 3$

Solution
$\begin{array}{lcclcc} & X \rightleftharpoons 2 Y ; & Z \rightleftharpoons P+Q \\ \text { Initial mol. } & 1 & 0 & 1 & 0 & 0 \\ \text { At equilibrium } & 1-a & 2 a & 1-a & a & a\end{array}$
$
\begin{aligned}
& K_{p_1}=\frac{p_y^2}{P_X}=\frac{\left(\frac{2 \alpha}{1+\alpha} P_1\right)^2}{\left(\frac{1-\alpha}{1+\alpha} P_1\right)} \\
& K_{p_2}=\frac{P_p P_Q}{P_Z}=\frac{\left(\frac{\alpha}{1+\alpha} P_2\right)\left(\frac{\alpha}{1+\alpha} P_2\right)}{\left(\frac{1-\alpha}{1+\alpha} P_2\right)} \\
& \Rightarrow \quad K_{p_1}=\frac{4 \alpha^2 P_1}{1-\alpha^2} \ldots \ldots(i) \\
& \Rightarrow \quad K_{p_1}=\frac{4 \alpha^2 P_1}{1-\alpha^2} \ldots \ldots(\text { ii })
\end{aligned}
$

Given is $\frac{K_{p_1}}{K_{p_2}}=\frac{1}{9}$
Substituting values of from equation $(i)$ \& (ii) into $($ iii $)$, we get
$
\frac{\frac{4 \alpha^2 P_1}{1-\alpha^2}}{\frac{\alpha^2 P_2}{1-\alpha^2}}=\frac{1}{9} \Rightarrow \frac{4 P_1}{P_2}=\frac{1}{9} \Rightarrow \frac{P_1}{P_2}=\frac{1}{36}
$

Example 2. Phosphorus pentachloride dissociates as follows in a closed reaction vessel,

$\mathrm{PCl}_{5(g)} \rightleftharpoons \mathrm{PCl}_{3(g)}+C l_{2(g)}$

If the total pressure at the equilibrium of the reaction mixture is P and the degree of dissociation of PCl₅ is x, the partial pressure of PCl₃ will be

$\begin{aligned} & \text { 1) }\left(\text { correct) }\left(\frac{x}{x+1}\right) P\right. \\ & \text { 2) }\left(\frac{2 x}{1-x}\right) P \\ & \text { 3) }\left(\frac{x}{x-1}\right) P \\ & \text { 4) }\left(\frac{x}{1-x}\right) P\end{aligned}$

Solution

Given,

$\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$

t = 0 1 0 0

t_eq 1-x x x

Total number of moles = 1 - x + x + x = 1 + x

Total number of moles = 1 + x

at t_eq, number of moles of PCL₃ = x

Thus partial pressure of $\mathrm{PCl}_3=\left(\frac{x}{1+x}\right) P$

Hence, the answer is the option (1).

Example 3. Ammonia under a pressure of 15 atm, at 27^oC is heated to 327^oC in a closed vessel in the presence of catalyst. Under these conditions, NH₃ partially decomposes to H₂ and N₂.The vessel is such that the volume remains effectively constant, whereas the pressure increases to 50 atm. Calculate the percentage of NH₃ actually decomposed.

1)50%

2)33.33%

3) (correct)66.7%

4)89%

Solution

Given that the temperature is doubled, this means that if NH₃ was not dissociated then the pressure would have been 30 atm. But after dissociation, the total pressure is 50 atm.

Let $\alpha$ be the degree of dissociation
$
\begin{aligned}
& \text { Moles } \quad 2 \mathrm{NH}_3(\mathrm{~g}) \rightleftharpoons 3 \mathrm{H}_2(\mathrm{~g})+\mathrm{N}_2(\mathrm{~g}) \\
& \begin{array}{llll}
\text { Initial } & a & 0 & 0
\end{array} \\
& \text { at equilibrium } \quad a-a \alpha \quad 3 a \alpha / 2 \quad a \alpha / 2 \\
& \text { Total moles }=a+a \alpha \\
& \frac{\text { Initial moles }}{\text { Final moles }}=\frac{\text { Pressure of } \mathrm{NH}_3 \text { at } 600 \mathrm{~K}}{\text { Final pressure after dissociation at } 600 \mathrm{~K}} \\
& \Rightarrow \frac{a}{a+a \alpha}=\frac{30}{50} \\
& \Rightarrow \alpha=\frac{20}{30} \\
& \% \text { dissociation }=\alpha \\
& \% \text { dissociation }=\frac{20}{30} \times 100=66.7 \% \\
&
\end{aligned}
$

Hence, the answer is the option (3).

Example 4.Gaseous N₂O₄ dissociates into gaseous NO₂ according to the reaction

$\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})$

At 300 K and 1 atm pressure, the degree of dissociation of N₂O₄ is 0.2.

If one mole of N₂O₄ gas is contained in a vessel, then the density of the equilibrium mixture is :

1)1.56 g/L

2) (correct)3.11 g/L

3)4.56 g/L

4)6.22 g/L

Solution

$\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})$

Moles at equilibrium
$
(1-0.2)
$$
$$
=0.8
$

Molar mass of mixture $\left(\mathrm{M}_{\mathrm{eff}}\right)=\chi_{\mathrm{N}_2 \mathrm{O}_4} \times 92+\chi_{\mathrm{NO}_2} \times 46$
$
\begin{aligned}
& \therefore \mathrm{M}_{\mathrm{eff}}=\frac{0.8 \times 92+0.4 \times 46}{1.2}=76.67 \\
& \therefore \mathrm{d}=\frac{\mathrm{P} \times \mathrm{M}_{\mathrm{eff}}}{\mathrm{RT}}=\frac{1 \times 76.67}{0.0821 \times 300}=3.11 \frac{\mathrm{g}}{\mathrm{L}}
\end{aligned}
$

Hence, the answer is the option (2).

Example 5. The self-ionization constant for pure formic acid $\mathrm{K}=\left[\mathrm{HCOOH}_2^{+}\right]\left[\mathrm{HCOO}^{-}\right]_{\text {has been estimated as }} 10^{-7}$ room temperature. What percentage of formic acid molecules in pure formic acid is converted to formate ions? The density of formic acid is $1.25 \mathrm{~g} / \mathrm{cm}^3$1)

1)0.01%

2) (correct)0.001%

3)0.02%

4)0.002%

Solution

The density of HCOOH is given as $1.25 \mathrm{~g} / \mathrm{cm}^3$
$\therefore$ Mass of HCOOH in 1 litre solution $=1.25 \times 10^3 \mathrm{~g}$
The concentration of HCOOH,
$
\mathrm{c}=\frac{1.25 \times 10^3}{46}=27.17 \mathrm{M}
$

As it is given the HCOOH goes to auto-ionization, so $\left[\mathrm{HCOOH}_2^{+}\right]=\left[\mathrm{HCOO}^{-}\right]$
Also, $\left[\mathrm{HCOOH}_2^{+}\right]\left[\mathrm{HCOO}^{-}\right]=10^{-7}$
$\therefore\left[\mathrm{HCOO}^{-}\right]=\sqrt{10^{-7}}=3.16 \times 10^{-4}$
$\%$ dissociation of $\mathrm{HCOOH}=\frac{\left[\mathrm{HCOO}^{-}\right] \times 100}{[\mathrm{HCOOH}]}$
$
\Rightarrow \alpha=\frac{3.16 \times 10^{-4} \times 100}{27.17}=0.001 \%
$

Hence, the answer is the option(2).

Hence, the answer is the option(2).

Summary

The degree of dissociation is the amount of electrolytes dissociating into its ions. Basically, it occurs in case of weak electrolytes because the Strong electrolytes dissociate completely. The applications of degree of dissociation involve chemical equilibrium which gives us the idea of Understanding weak acid and bases dissociation.