Question

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1.00x10-3 mol PCls is introduced into a 250 ml flask and equilibrium is established at 284 °C: PC1s(g) - PC13(g) + Cl2(g) The quantity of Cl2 present at equilibrium is found to be 9.65x10-4 mol. What is the value of Kc for the dissociation reaction at 284 °C?

Answer 1

"sluoe* ) Given Reaction is, at 204c Роза? Pesc) + Cleg) Given that 1.00x 103 mol Plus in 250ml flask and at equim the mole of 9.65x15 mol Ke for Dissonation of Pelsis Ke [Puig] [cly] [Pug] Let initially puls is 'a'rol and Plz & Cly are o mol. and at equilibrium time the pets of Clo let semole then puls Remaine (4-2) mol as given below, also to convert therse mol into concentration which is mol volume - PC3 PU5 a x Initial to equil" ta teq d-x Conch V <lx xo yn o x X1> arx x 2² V Ke = [Pels] [Cl] [Pus Caru) (a-x)v flere given a = 1.00* 10?mol za mol of 2 = mol of Puz 9-65x105 Va 250 ml = 250 LE 0.250L 1000 Kea (9665x104) 0.93x10-6 T100x10^3-9-65x104)0-250) 80 75x10-6 Kc = 0.106 2 Ans: