The value ofK_(p)" for the reaction ", CO_(2) (g) + C (s) hArr2 CO (g),is 3.0at 1000 K. Initially , p_(co_(2)) = 0.48" bar "and P_(CO)= 0 bar and opure graphite is present , calculate the equilibrium partial pressures of CO and CO_(2)

The value ofK_(p)" for the reaction ", CO_(2) (g) + C (s) hArr2 CO (g)

1.	The value ofK_(p)" for the reaction ", CO_(2) (g) + C (s) hArr2 CO (g),is 3.0at 1000 K. Initially , p_(co_(2)) = 0.48" bar "and P_(CO)= 0 bar and opure graphite is present , calculate the equilibrium partial pressures of CO and CO_(2)
Answer» Solution :`{:(,CO_(2)(g) + C(s),hArr,2 CO (g)),("INTIAL",0.48 " bar ",,0 "bar"),("AT.eqm.",(0.48-p)"bar",,2" p bar"):}` ` K_(p) = (2 p)^(2)/(0.48-p )=3.0 " (Given) ".` ` :. 4 p^(2) = 1.44 - 3 p or 4 p^(2) + 3 p - 1.44=0` ` :.p = ( -b pm sqrt(b^(2) - 4 ac))/(2a)=(-3 pm sqrt ( 9-4 xx 4 ( - 1. 44)))/8` ` = ( -3 pm sqrt(32*04))/8= (-3 pm 5.66)/8 = 2.66/8 ""` (neglecting - ve value) = 0.33 ATM `:. p_(CO_(2))= 0.48 - 0.33 = 0.15 " bar" , p_(CO_(2)) = 2 xx 0.33 = 0. 66 " bar " `