Question

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.

Answer 1

(I).Calculation of `K_(p)` for the reaction
Let the total mass of the gasesous mixture =100 g

Mass of CO in the mixture =90.55 g
Mass of`CO_(2)` in the mixture =(100 -90 .55 ) =9.45 g
No. of moles of `CO =(90.55 g)/((28 g "mol"^(-1)) =3.234 " mol"`
No. of moles of `CO_(2) =(90 .55 g)/((44 g "mol"^(-1)) =0.215 " mol "`
`pCO ` in the mixture `=((3.234"mol"))/((3.449 "mol")) xx 1 " atm " = ((3.234 "mol"))/((3.449 "mol")) xx 1 " atm " =0. 938 " atm "`
`pCO_(2)` in the mixture `=((3.215 "mol"))/((3.449 "mol")) xx 1 " atm " =0. 062 " atm "`
`C(s) + CO_(2) (g) hArr 2CO(g)`
`"Eqm. pressure" " "0.062 "atm" " "0.938 "atm"`
`K_(p) =(P^(2)CO)/(pCO_(2)) =((0.938 "at")^(2))/((0.062 "atm")) =14.19 " atm "`
(II) . Calculation of `K_(c)` for the rection.
` K_(c) =(K_(p))/((RT)^(Deltang))`
`K_(p) = 14.19 " atm " ,R =0.0821 L " atm" K^(-1) "mol"^(-1) , T =1127 K, Delta^(ng) =2 -1 =1`
`:. K_(c) =((14.19 " atm "))/((0.821 L " atm " K^(-1) " mol "^(-1)) xx (1127 K^(1))) =0.153 " mol " L^(-1).`