Q. At equilibrium total number of moles for the reaction $2 HI \rightleftharpoons H _{2}+ I _{2}$, if $\alpha$ is degree of dissociation, are
Equilibrium
Solution:
$HI$
$\rightleftharpoons +H _{2}$
$I _{2}$
Initial moles
$2$
$0$
$0$
Change
$- \alpha$
$\frac{\alpha}{2}$
$\frac{\alpha}{2}$
moles at equilibrium
$2- \alpha$
$\frac{\alpha}{2}$
$\frac{\alpha}{2}$
The reaction is $2 HI \rightleftharpoons H _{2}+ I _{2}$.
$\alpha$ is the degree of dissociation.
Total number of moles $=2-\alpha+\frac{a}{2}+\frac{a}{2}=2$.
| $HI$ | $\rightleftharpoons +H _{2}$ | $I _{2}$ | |
| Initial moles | $2$ | $0$ | $0$ |
| Change | $- \alpha$ | $\frac{\alpha}{2}$ | $\frac{\alpha}{2}$ |
| moles at equilibrium | $2- \alpha$ | $\frac{\alpha}{2}$ | $\frac{\alpha}{2}$ |