Question

$1.6mole$ of $P{Cl}_5\left(g\right)$ is placed in $4{dm}^3$ closed vessel. When the temperature is raised to $500K$, it decomposes and at equilibrium $1.2mole$ of $P{Cl}_5\left(g\right)$ remains. What is the $K_c$ value for the decomposition of $P{Cl}_5\left(g\right)$ to $P{Cl}_3\left(g\right)$ and ${Cl}_2\left(g\right)$ at $500K$?

• A. $0.013$
• B. $0.050$
• C. $0.033$
• D. $0.067$

Answer 1

The correct option is C $0.033$

The given reaction is :-

${PCl}_5\left(g\right)\rightleftharpoons {PCl}_3\left(g\right)+{Cl}_2\left(g\right)$

Initial moles : $1.6$ $0$ $0$ (given)

At eqm : $1.2$ $0.4$ $0.4$ (given)

Now, conc. of ${PCl}_5$ at equation $=\frac{1.6}{4}mole/{dm}^3$

conc. of ${PCl}_3$ at equation $=\frac{0.4}{4}mole/{dm}^3$

conc. of ${Cl}_2$ at equation $=\frac{0.4}{4}mole/{dm}^3$

$\therefore K_C=\frac{\left[{PCl}_3\right]\left[{Cl}_2\right]}{\left[{PCl}_5\right]}=\frac{0.1\times 0.1}{0.4}=0.025$