Initially $$0.8$$ mole of $$PCl5$$ and $$0.2$$ mole of $$PCl3$$ are mixed in one litre vessel. At equilibrium $$0.4$$ mole of $$PCl3$$ is present. The value of Kc for the $$reactionPCl5(g)$$⇔$$PCl3(g)+Cl2(g)$$

Question

Initially $$0.8$$ mole of $$PCl5$$      and $$0.2$$ mole of $$PCl3$$      are mixed in  one litre vessel.  At equilibrium $$0.4$$  mole of $$PCl3$$      is present.  The value of Kc      for the $$reactionPCl5(g)$$⇔$$PCl3(g)+Cl2(g)$$

Infinity Learn · Accepted Answer

$$PCl5$$⇌$$PCl3+Cl2$$                                           Given (0.2+x)=0.4$$       $$0.8$$                $$0.2$$         $$0$$                                                                 ⇒$$x=0.2$$         x                  x            $$x_$$     $$(0.8$$−$$x)     $$0.2+x$$   x             ∴$$Kc=0.4$$×$$0.20.6$$                       $$=0.133$$ M     $$=0.133$$ mole.lit−$$1$$

Initially 0.8 mole of PCl5 and 0.2 mole of PCl3 are mixed in one litre vessel. At equilibrium 0.4 mole of PCl3 is present. The value of Kc for the reactionPCl5(g)⇔PCl3(g)+Cl2(g)

Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!