For the equilibrium in gaseous phase in 2 lit flask we start with 2 moles of SO_(2) and 1 mole of O_(2) at 3 atm, 2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)). When equilibrium is attained, pressure changes to 2.5 atm. Hence, equilibrium constant K_(c)is :

For the equilibrium in gaseous phase in 2 lit flask we start with 2 mo

1.	For the equilibrium in gaseous phase in 2 lit flask we start with 2 moles of SO_(2) and 1 mole of O_(2) at 3 atm, 2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)). When equilibrium is attained, pressure changes to 2.5 atm. Hence, equilibrium constant K_(c)is :
Answer» Solution :`UNDERSET(2-2x)underset(2)(2SO_(2(g)))+underset(1-x)underset(1)(O_(2(g))) harr underset(2x)underset(0)(2SO_(3(g)))` Moles at equilibrium = 3-x At equilibrium pressure becomes 2.5 atm since pv= NRT Hence `p alpha n` Thus `(3-x)/(3)=(2-5)/(3)`, x=0.5 `[SO_(2)]=(2-2x)/(2)=0.5M` `[O_(2)]=(1-x)/(2)=0.25M, (SO_(3))=(2x)/(2)=0.5M` `K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=((0.5)^(2))/((0.5)^(2) xx 0.25)=4`