At `1127 K` and `1 atm` pressure, a gaseous mixture of `CO` and `CO_(2)` in equilibrium with solid carbon has `90.55% CO` by mass:
`C_((s))+CO_(2(g))hArr2CO_((g))`
Calculate `K_(c)` for the reaction at the above temperature.

Explanation

If total mass of the mixture of `CO and CO_(2)` is 100 g , then
` :. "At eqm., " [BrCl] = ( 3*30 xx 10^(-3) - 3*0 xx 10^(-3))= 0*30 xx 10^(-3) = 3*0 xx 10^(-4) "mol"L^(-1)`
Number of moles of ` :. "At eqm., " [BrCl] = ( 3*30 xx 10^(-3) - 3*0 xx 10^(-3))= 0*30 xx 10^(-3) = 3*0 xx 10^(-4) "mol"L^(-1)`
` :. p_(CO) = (3*234)/(3*234 + 0*215 ) xx 1" atm " = 0*938 " atm" , p_(CO_(2)) = (0*215)/(3*234 + 0*215 ) xx 1 "atm " = 0*062 "atm "`
` k_(p) = (p_(CO)^(2))/p_(CO_(2)) = (0*938 )^(2)/(0*062 )= 14 *19` .
` Delta n_(g) = 2-1 =1 :. K_(p) = K_(c) (RT) or K_(c) = (K_(p))/(RT) = (14*19)/(0*0821 xx 1127)=0*153 ` .