For , A + B hArr C. the equilibrium concentration of A and B at a temperäture are 15 mol lit^(-1) When volume is doubled the reaction has equilibrium concentration of A as 10 mol lit^(-1) then

For , A + B hArr C. the equilibrium concentration of A and B at a temp

1.	For , A + B hArr C. the equilibrium concentration of A and B at a temperäture are 15 mol lit^(-1) When volume is doubled the reaction has equilibrium concentration of A as 10 mol lit^(-1) then
Answer» `K_(c) = 2` mole `lit^(-)` concentration of C in original equilibrium is 45 M `K_(c) = 0.2 mol^(-1)` litre as increasing volume the reaction PROCEEDS in the backward direction Solution : Since, on increasing volume, pressure decreases and the reaction proceeds in the direction where it shows an increase in mole i.e backward reaction. con. II EQULIBRIUM `[(15)/(2)+x][(15)/(2)+x][(a)/(2)-x]` `:. (15)/(2) +x=10 :. x=(5)/(2)` Now `Kc=([C])/([A][B])=([(a)/(2)-(5)/(2)])/([(15)/(2)+(15)/(2)][(15)/(2)+(15)/(2)])` ...(1) `Kc=([C])/([A][B])=(a)/(15 xx 15)` ...(2) Kc are same `:. (((a-5))/(2))/(10 xx 10)=(a)/(15 xx 15) implies a=45M` Now, `Kc=(45)/(15 xx 15)=0.2 "mole"^(-1)` lit