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. What is the degree of dissociation $\alpha$?
Answer: B
💡 Solution & Explanation
For $A(g) \rightleftharpoons 3B(g)$, $1\text{ mole}$ yields $n=3$ moles of product. Formula: $\alpha = (D - d) / ((n - 1)d)$. Given $d = D/2$, or $D = 2d$. Substituting: $\alpha = (2d - d) / ((3 - 1)d) = d / (2d) = 0.5$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes