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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
$\text{K}_{\text{c}}\text{ for }\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g})+\text{Cl}_2(\text{g})\text{ is }0.04\text{ at }25^\circ\text{C}.$
How much moles of $PCl_5$ must be added to 3L flask to obtain a chlorine concentration of $0.15M$.
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Answer

$\begin{matrix}&\text{PCl}_5(\text{g})&\rightleftharpoons&\text{PCl}_3(\text{g})&+&\text{Cl}_2(\text{g})\\\text{Initial Conc. }&\text{x}&&0&&0\\\text{Final Conc.}&\text{x}-0.15&&0.15&&0.15\end{matrix}$
$\text{K}_{\text{c}}=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}$
$0.04=\frac{0.15\times0.15}{\text{x}-0.15}$
$\text{x}-0.15=\frac{0.15\times0.15}{0.04}$
$=\frac{225}{1000}\times\frac{100}{4}$
$=\frac{225}{400}=\frac{9}{16}$
$\text{x}=\frac{9}{16}+\frac{15}{100}=\frac{9}{16}+\frac{3}{20}$
$=\frac{180+48}{320}=\frac{228}{320}=0.7\text{mol L}^{-1}$
Number of moles of $PCl_5$ per litre = 0.7mol
Number of moles of $PCl_5$ in 3L flask = 0.7 × 3 = 2.1mol.

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