An equilibrium mixture contains 0.5, 0.12 and 5 moles of SO_(2),O_(2)andSO_(3) respectively, in a one litre vessel at a certain temperature. How many mole of O_(2) must be forced into the reaction mixture in order to increase the conc. Of SO_(3) to 5.3 mole at the same temperature? (Given K_(c) for the reaction, 2SO_(2)+O_(2)iff2SO_(3) is 800)

An equilibrium mixture contains 0.5, 0.12 and 5 moles of SO_(2),O_(2)a

1.	An equilibrium mixture contains 0.5, 0.12 and 5 moles of SO_(2),O_(2)andSO_(3) respectively, in a one litre vessel at a certain temperature. How many mole of O_(2) must be forced into the reaction mixture in order to increase the conc. Of SO_(3) to 5.3 mole at the same temperature? (Given K_(c) for the reaction, 2SO_(2)+O_(2)iff2SO_(3) is 800)
Answer» `0.506` `0.908` `0.74` `0.45` Solution :`{:(,2SO_(2(g)),+,O_(2(g)),iff,2SO_(3(g))),("Initial","0.5 moles",,"0.12 moles",,"5 moles"),("At equilibrium","(0.5 - 0.3) moles",,"(0.12 + X - 0.15) moles",,"5.3 moles"):}` Suppose .x. moles of `O_(2)` be introduced into the vessel in ORDER to increase the conc. of `SO_(3(g))` to 5.3 moles/litre. IMPLIES 0.3 mole `SO_(2)` and 0.15 mole `O_(2)` will be consumed ACCORDING to the above equation. Hence at equilibrium, `[SO_(2)]=0.2M` `[O_(2)]=(x-0.03)M,[SO_(3)]=5.3M` `implies K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)xx[O_(2)])=((5.3)^(2))/((0.2)^(2)xx(x-0.03))=800` (given) `implies x-0.03=0.878impliesx=0.908` moles