$\mathrm{K}_{\mathrm{c}}$ for $\mathrm{PCl}_{5}(\mathrm{~g}) ightleftharpoons…

$\mathrm{K}_{\mathrm{c}}$ for $\mathrm{PCl}_{5}(\mathrm{~g}) ightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$ is $0.04$ at $250^{\circ} \mathrm{C}$. How many moles of $\mathrm{PCl}_{5}$ must be added to a $3 \mathrm{~L}$ flask to obtain $\mathrm{a} \mathrm{Cl}_{2}$ concentration of $0.15 \mathrm{M}$
  1. $4.2$ moles
  2. $2.1$ moles
  3. $5.5$ moles
  4. $6.3$ moles

Solution

At equilibrium the moles of $\mathrm{Cl}_{2}$ must be
$=0.15 \times 3=0.45$
$\mathrm{PCl}_{5} ightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}$
Eqm. Conc. $\quad \frac{x-0.45}{3} \quad \frac{0.45}{3} \quad \frac{0.45}{3}$
$\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{PCl}_{3}ight]\left[\mathrm{Cl}_{2}ight]}{\left[\mathrm{PCl}_{5}ight]}$
$\therefore 0.04=\frac{0.15 \times 0.15}{(\mathrm{x}-0.45) / 3}$
$\therefore \mathrm{x}=2.1$ moles ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more EQUILIBRIUM questions on Aicharya