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  • <img alt="\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{3} \quad \mathrm{~K}_{\mathrm{c}}=1.844" src="/latex-image/?%5Cmathrm%7BPCl%7D_%7B5%7D%20%5Crightleftharpoons%20%5Cmathrm%7BPCl%7D_%7B3%7D&plus;%5Cmathrm%7BCl%7D_%7B3%7D%20%5C

\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{3} \quad \mathrm{~K}_{\mathrm{c}}=1.844 3.0 moles of \mathrm{PCl}_{5} is introduced in a 1 \mathrm{~L} closed reaction vessel at 380 \mathrm{~K}. The number of moles of \mathrm{PCl}_{5} at equilibrium is _________\times 10^{-3}.
(Round off to the Nearest Integer)
 

Answers (1)

best_answer

\begin{matrix} & \mathrm{PCl_5} & \rightleftharpoons &\mathrm{PCl_3} &+ &\mathrm{Cl_2} &\quad \mathrm{,K_c=1.844} \\\mathrm{t =0}& 3& & -& &- & \\ \mathrm{t =t_{eq}}& (3-x)& &(x) & &(x) & \end{matrix}

 

\mathrm{ \therefore K_{C}=\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]} }

\begin{aligned} \Rightarrow 1.844 &=\frac{x \times x}{3-x} \end{aligned}

\begin{aligned} &\Rightarrow x^{2}=5.532-1.844 x \\ &\Rightarrow x^{2}+1.844 x-5.532=0 \end{aligned}

On solving,

\begin{aligned} x &=1.6 \\ \therefore\left[\mathrm{PCl}_{5}\right] &=3-1.6 \\ &=1.4 \\ &=1400 \times 10^{-3} \mathrm{M} . \end{aligned}

Hence, the correct answer is 1400.

Posted by

sudhir.kumar

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