A hypothetical gas dissociates as . At a certain temperature, the observed equilibrium vapour densit — Chemical Equilibrium Chemistry Question
Question
A hypothetical gas $A$ dissociates as $A(g) \rightleftharpoons 3B(g)$. At a certain temperature, the observed equilibrium vapour density is experimentally found to be exactly half of its theoretical vapour density. Calculate the value of $(100 \times \alpha)$, where $\alpha$ is the degree of dissociation.
Answer: 50
💡 Solution & Explanation
For the reaction $A(g) \rightleftharpoons 3B(g)$, $1\text{ mole}$ of reactant yields $n=3$ moles of product. The formula is $\alpha = \frac{D - d}{(n - 1)d}$. We are given that $d = D/2$, meaning $D = 2d$. Substituting these values into the formula: $\alpha = \frac{2d - d}{(3 - 1)d} = \frac{d}{2d} = 0.5$. The question asks for $100 \times \alpha = 100 \times 0.5 = 50$.
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