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$\mathrm{PCl_{5}}$ , dissociates as

$\mathrm{PCl_{5}(g)\rightleftharpoons PCl_{3}(g) +Cl_{2}(g)}$

5 moles of $\mathrm{PCl_{5}}$ are placed in a 200 litre vessel which contains 2 moles of $\mathrm{N_{2}}$ , and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant $\mathrm{K_{p}}$ , for the dissociation of $\mathrm{PCl_{5}}$ is _______ $\mathrm{\times 10^{-3}}$ . (nearest integer)

(Given: $\mathrm{R=0.082 \;L \;atm \;K^{-1}mol^{-1}}$ : Assume ideal gas behaviour)
Option: 1
1107

Option: 2
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Option: 3
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Option: 4
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Answers (1)

best_answer

$\\\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\\ \mathrm{P_{\circ }-x}\;\; \; \; \; \mathrm{x}\;\; \; \; \; \; \; \;\; \mathrm{x}$

$\mathrm{K_{p}=\frac{x^{2}}{P_{0}-x}$

Now, initially $\mathrm{PCl_5}$ and $\mathrm{N_2}$ were added to the container. $\mathrm{N_2}$ is inert to the system and will not react under the given conditions

$\mathrm{ P_{N_{2}}=\frac{n R T}{V}=\frac{2 \times R \times 600}{200}=6 R}$

$\mathrm{P_{0}=P _{P C l_{5}}=\frac{n R T}{V}=\frac{5 \times R \times 600}{200}=15 R}$

Given,

$\mathrm{P_{eq}=P_{0}-x+x+x+P_{N_{2}}}$

$\begin{aligned} \mathrm{2.46 =P_{0}+x+P_{N_{2}}=15 R+x+6 R} \\ \end{aligned}$

$\begin{aligned} \mathrm{2.46 =21 R+x} \\ \end{aligned}$

$\begin{aligned} &\mathrm{x=2.46-21 R=2.46-21 \times 0.082 }\end{aligned}$

$\mathrm{x=2.46-1.722=0.738}$

and,

$\begin{aligned} \mathrm{P_{0}-x =15 R-0.738=1.23=0.492} \end{aligned}$

Now, $\mathrm{K_{p}=\frac{x^{2}}{P_{0}-x}=\frac{(0.738)^{2}}{0.492}=1.107}$

$\mathrm{K_{p}=1107 \times 10^{-3}}$

Hence, the answer is 1107

Posted by

Ritika Kankaria

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