Question

$A_{3\left(g\right)}\rightleftharpoons 3A_{\left(g\right)}$
In the above reaction, the initial moles of $A_3$ is "a". If $\alpha$ is degree of dissociation of $A_3$. The total number of moles at equilibrium will be:

• A. $a-\frac{a\alpha }{3}$
• B. $\frac{a}{3}-\alpha$
• C. $\left(\frac{a-a\alpha }{2}\right)$
• D. None of these

Answer 1

Answer: (D)

None of these

Given that $A_{3\left(g\right)}\rightleftharpoons 3A_{\left(g\right)}$

initially $a$ $0$

At equilibrium $a-x$ $3x$ $(x=a\alpha )$.

$a-a\alpha$ $3a\alpha$

Total number of moles at equilibrium $=a-a\alpha +3a\alpha =a(1+2\alpha )$.

Answer is none of those. (D).