Question

In $A_3\left(g\right)\rightleftharpoons 3A\left(g\right)$ reaction, the initial concentration of $A_3$ is $a$ mol $L^{-1}$. If $x$ is degree of dissociation of $A_3$. The total number of moles at equilibrium will be:

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

Answer 1

The correct option is D none of these

$A_3\left(g\right)\rightleftharpoons 3A\left(g\right)$

Initial concentration $a$ $b$

$x$ is the degree of dissociation

$1-x$ $3x$

$a(-x)$ $3x$

$\therefore \left[A_3\right]=a(1-x)$

$\left[A\right]=3x$

$K_c=\frac{[A{]}^3}{\left[A_3\right]}$......molarity=moles\litre

$=\frac{27x^3}{a(1-x)}=$ total numbers of moles.