Question

For the following reaction:
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$0.4mol$ of $PC{l}_5\left(g\right)$, $0.2mol$ of $PC{l}_3\left(g\right)$ and $0.6mol$ of $C{l}_2\left(g\right)$ are taken in a $1L$ flask.
If $K_c=0.2$, then the direction in which reaction proceeds is :

• A. Towards reactant side
• B. Towards products side
• C. Reaction is at equilibrium
• D. None of the above

Answer 1

The correct option is A Towards reactant side

Given Reaction is:
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$Concentration=\frac{Moles}{Volume}$
$\left[PC{l}_5\right]=\frac{0.4}{1}=0.4M\left[PC{l}_3\right]=\frac{0.2}{1}=0.2M\left[C{l}_2\right]=\frac{0.6}{1}=0.6M$
Finding $Q_c$
$Q_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$

puttting the values,
$Q_c=\frac{0.2\times 0.6}{0.4}=0.3M>K_c(=0.2)$
Since, $Q>K_c$ the reaction proceeds in backward direction i.e. reactant $\left(PC{l}_5\right)$ side.