An arbitrary compound $$P2Q$$ decomposes according to reaction:$$2P2Q(g)$$⇌ $$2P2(g)+Q2(g)If$$ one starts the decomposition reaction with $$4$$ moles of $$P2Qand$$ value of equilibrium constant Kp is numerically equal to total pressure at equilibrium. Then which option $$is/are$$ correct at equilibrium,.

Question

An arbitrary compound $$P2Q$$ decomposes according to reaction:$$2P2Q(g)$$⇌          $$2P2(g)+Q2(g)If$$ one starts the decomposition reaction with $$4$$ moles of $$P2Qand$$ value of equilibrium constant  Kp is numerically equal to total pressure at equilibrium. Then which option $$is/are$$ correct at equilibrium,.

Infinity Learn · Accepted Answer

$$2P2Q$$       ⇌                                               $$2P2+Q2t=0$$                               $$4$$                             $$0$$                    $$0t=eq$$           $$4(1$$−α$$)$$                 $$4$$α              $$2$$α$$Kp=4$$α$$4+2$$α×$$P22$$α$$4+2$$α×$$P4(1$$−α$$)4+2$$α×$$P2=2$$α$$3(1$$−α$$)2(4+2$$α$$)$$×PBut, $$Kp=P($$ given $$)$$∴$$2$$α$$3=(1-$$α$$)2(4+2$$α$$)$$∴$$-6$$α$$+4=0$$∴α$$=(2/3)Total$$ moles at $$eq=$$  $$4+2$$α$$=(16/3)nP2Q=nQ2=(4/3)$$,  $$nP2=(8/3)$$\ Total number of moles of products $$=$$  $$83+43=123=4$$.

An arbitrary compound P2Q decomposes according to reaction:2P2Q(g)⇌ 2P2(g)+Q2(g)If one starts the decomposition reaction with 4 moles of P2Qand value of equilibrium constant Kp is numerically equal to total pressure at equilibrium. Then which option is/are correct at equilibrium,.

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