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

Let the total mass of the mixture of CO and `CO_2` is 100g, then
CO=90.55 g and `CO_2` =100-90.55 =9.45 g
Moles of CO=`90.55/28`=3.234
Moles of `CO_2=9.45/44`=0.215
Mole of fraction of CO=`3.234/(3.234+0.215)=0.938`
Mole fraction of `CO_2`=1-0.938=0.062
`therefore p_(CO)` =mole fraction x total pressure
= 0.938 x 1 atm = 0.938 atm
`p_(CO_2)` = 0.062 x 1 atm = 0.062 atm
`K_p` for the reaction
`C(s)+CO_2(g) hArr 2CO(g)`
`K_p=(p_(CO_2))^2/(p_(CO_2))=(0.938)^2/(0.062)`=14.19
Now , `Deltan_g`=2-1=1
`K_p=K_c(RT)^(Deltan_g)`
or `K_c=K_p/(RT)=14.19/(0.0821xx1127)`=0.153