Question

3 Moles of ${PCl}_5$ kept in 1L closed vessel was allowed to attain equilibrium at 380K .Calculate the composition of the mixture at equilibrium,$K_c$:

• A. $\left[{PCl}_5\right]=1.41M,\left[{PCl}_3\right]=\left[{Cl}_2\right]=2.59M$
• B. $\left[{PCl}_5\right]=1.41M,,\left[{PCl}_3\right]=\left[{Cl}_2\right]=1.59M$
• C. $\left[{PCl}_5\right]=1.41M,,\left[{PCl}_3\right]=\left[{Cl}_2\right]=5.59M$
• D. $\left[{PCl}_5\right]=1.41M,,\left[{PCl}_3\right]=\left[{Cl}_2\right]=9.59M$

Answer 1

Answer: (B)

$\left[{PCl}_5\right]=1.41M,,\left[{PCl}_3\right]=\left[{Cl}_2\right]=1.59M$

${PC{l}_5}_{\left(g\right)}\rightleftharpoons {PC{l}_3}_{\left(g\right)}+{C{l}_2}_{\left(g\right)}$

Initial $3$ mole $0$ $0$

At eqm $3-x$ $x$ $x$

Eqm conc. $\frac{3x}{1}$ $\frac{x}{1}$ $\frac{x}{1}$

$\therefore K_C=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}=\frac{x\times x}{3-x}$

now, $\frac{x^2}{3-x}=1.8\Rightarrow x^2+1.8x-5.4=0$

$\therefore x=1.59$

So the $\left[PC{l}_5\right]=3-x=3-1.59=1.41M$

$\left[PC{l}_3\right]=\left[C{l}_2\right]=x=1.59M$ .