Degree of Dissociation from Vapour Density
In gaseous reversible reactions involving a change in the number of moles (e.g., dissociation), the observed molar mass of the mixture at equilibrium differs from the theoretical molar mass of the reactant. By measuring the Vapour Density, we can calculate the Degree of Dissociation ($\alpha$).
1. Basic Definitions
Vapour Density (VD): It is defined relative to Hydrogen gas.
- Theoretical Vapour Density ($D$): Calculated from the molecular formula of the reactant (before dissociation).
- Observed Vapour Density ($d$): Measured experimentally at equilibrium (average molar mass of the mixture).
2. Why does Vapour Density change?
When a gas dissociates, the number of moles increases. Since the total mass remains constant (Law of Conservation of Mass), the average molar mass (and thus the vapour density) decreases.
$$ \text{Average Molar Mass} \propto \frac{1}{\text{Total No. of Moles}} $$
If dissociation occurs ($\alpha > 0$), Total Moles $\uparrow$, Average Molar Mass $\downarrow$, Observed VD ($d$) < Theoretical VD ($D$).
3. Derivation of the Formula
Consider a general dissociation reaction where 1 mole of reactant $A$ gives $n$ moles of product(s).
| Reactant | Product | |
| Initial Moles: | 1 | 0 |
| At Equilibrium: | $1 - \alpha$ | $n\alpha$ |
Total Moles at Equilibrium: $$ (1 - \alpha) + n\alpha = 1 + (n - 1)\alpha $$
Since Mass is conserved:
$$ \text{Initial Moles} \times \text{Theoretical Mass} = \text{Eq. Moles} \times \text{Observed Mass} $$ $$ 1 \times M_{theoretical} = [1 + (n-1)\alpha] \times M_{observed} $$Substituting $M = 2 \times VD$:
$$ 1 \times 2D = [1 + (n-1)\alpha] \times 2d $$ $$ \frac{D}{d} = 1 + (n-1)\alpha $$ $$ \frac{D}{d} - 1 = (n-1)\alpha $$ $$ \frac{D - d}{d} = (n-1)\alpha $$Final Formula for $\alpha$
Where:
$D$ = Theoretical Vapour Density ($\frac{\text{Molar Mass}}{2}$)
$d$ = Observed/Experimental Vapour Density
$n$ = Number of moles of products formed from 1 mole of reactant.
4. Application Examples
Case 1: Dissociation of $PCl_5$
Reaction: $PCl_5 \rightleftharpoons PCl_3 + Cl_2$
- Value of n: $1 \text{ mole } \to 1 + 1 = 2 \text{ moles}$. So, $n = 2$.
- Formula: $\alpha = \frac{D-d}{d(2-1)} = \frac{D-d}{d}$
Case 2: Dissociation of $N_2O_4$
Reaction: $N_2O_4 \rightleftharpoons 2NO_2$
- Value of n: $1 \text{ mole } \to 2 \text{ moles}$. So, $n = 2$.
- Formula: $\alpha = \frac{D-d}{d}$
Case 3: Polymerization/Association (Reverse Case)
If molecules associate (e.g., $2A \rightleftharpoons A_2$), the number of moles decreases. Here $d > D$. The formula is modified or $n$ is taken as fraction (e.g., $n=0.5$).
5. Special Note on $\Delta n_g = 0$
For reactions where the number of moles does not change (e.g., $2HI \rightleftharpoons H_2 + I_2$), there is no change in volume or density at constant pressure.
Vapour Density Quiz
Test your concepts on Equilibrium calculations. 10 MCQs with explanations.
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