Question

If $a$ is the degree of dissociation of $PC{l}_5$ at equilibrium for the reaction, $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$
The total number of moles of the reaction Consider intial moles of $P$ to be 1:

• A. $1-a$
• B. $1+a$
• C. 3
• D. $1$

Answer 1

The correct option is B $1+a$

Given: degree of dissociation of $PC{l}_5$ = $a$
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$
Initial moles $1$ $0$ $0$
After diss. $1-a$ $a$ $a$

Total moles at equilibrium = $1-a+a+a$ = $1+a$