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  • 4.0 moles of argon and 5.0 moles of  are introduced into an evacuated flask

4.0 moles of argon and 5.0 moles of \mathrm{PCl}_{5} are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The \mathrm{K}_{\mathrm{p}} for the reaction is 

\left[\text { Given: } \mathrm{R}=0.082 \mathrm{~L} \operatorname{atm~} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right] 

Option: 1

2.25


Option: 2

6.24


Option: 3

12.13


Option: 4

15.24


Answers (1)

best_answer

Total pressure inside the flask

\mathrm{P_{\text {Total }}=\frac{9 \times 0.082 \times 610}{100}=4.5 \mathrm{~atm}} \\

\mathrm{P_{\mathrm{PCl_5 }}=\frac{5}{9} \times 4.5=2.5 \mathrm{~atm}} \\

\mathrm{P_{\text {Ar }}=\frac{4}{9} \times 4.5=2 \mathrm{~atm} }

Now,

\mathrm{P_{\text {Total }} \text { at\: equilibrim } =2.5-p+p+p+p_{Ar}=6} \\

\mathrm{p =1.5} \\

\mathrm{k_{p}=\frac{1.5 \times 1.5}{2.5-1.5} =2.25 }

Hence the correct answer is option 1

Posted by

Rishi

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