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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 19 of 73
Marks: +1, -0
A sample of pure PCl5\mathrm{PCl}_5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5\mathrm{PCl}_5 was found to be 0.5 × 10−1 mol L−110^{-1}\,\text{mol L}^{-1}. If value of Kc is 8.3 × 10−310^{-3}, what are the concentrations of PCl3\mathrm{PCl}_3 and Cl2\mathrm{Cl}_2 at equilibrium?
PCl5(g)\mathrm{PCl}_{5(g)} ⇌ PCl3(g)+Cl2(g)\mathrm{PCl}_{3(g)} + \mathrm{Cl}_{2(g)}
Solution:  
PCl5(g)\mathrm{PCl}_{5(g)} ⇌ PCl3(g)+Cl2(g)\mathrm{PCl}_{3(g)} + \mathrm{Cl}_{2(g)}
Initial conc.y00Conc. at equilibriumy−xxx\begin{array}{c|c|c|c} \text{Initial conc.} & y & 0 & 0 \\ \text{Conc. at equilibrium} & y - x & x & x \end{array}
But y - x = 0.5 × 10−1 mol L−110^{-1}\,\text{mol L}^{-1}
∴ KcK_c = [PCl2][Cl2][PCl5]\frac{[\mathrm{PCl}_2][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]} = x.xy−x or \frac{x.x}{y-x} \text{ or }K_c$==x^2/{0.5×10^{-1}}=8.3×= 8.3 \times10^{-3}⇒x=2.04×\Rightarrow x = 2.04 \times10^{–2} \mol L^{–1}<br>< br>[PCl_3]==[Cl_2]=2.04×= 2.04 \times10^{–2} \mol L^{–1}$
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