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1 mole of PCl_(5) is placed in a closed vessel at 523K. At equilibrium, if it dissociates to an extent of 35%, calculate K_(p)" for "PCl_(5)hArrPCl_(3)+Cl_(2). Equilibrium pressure is found to be 5xx10^(5) Pa.

Answer» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;</p>Solution :Initial no. of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a>: `underset(1)(PCl_(5))hArrunderset(0)(PCl_(3))+underset(0)(Cl_(2))` <br/> `{:("No. of moles present"),("at equilibrium"):}}1-x=0.65" "x""x""(x=0.35)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> number of moles present at equilibrium = 0.65 + 0.35 + 0.35 = 1.35 <br/> We know that partial <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> = Mole fraction = Mole fraction `xx` total pressure <br/> `thereforeP_(PCl_(5))=(0.65/1.35)xx5xx10^(5)<a href="https://interviewquestions.tuteehub.com/tag/pa-1145246" style="font-weight:bold;" target="_blank" title="Click to know more about PA">PA</a>,P_(PCl_(3))=(0.35/1.35)xx5xx10^(5)Pa,P_(Cl_(2))=(0.35/1.35)xx5xx10^(5)Pa` <br/> `K_(p)=(P_(PCl_(3))xxP_(Cl_(2)))/(P_(PCl_(5)))=((0.35/1.35)xx5xx10^(5)(0.35/1.35)xx5xx10^(5))/((0.65/1.35)xx5xx10^(5))=6.98xx10^(4)Pa`</body></html>


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