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

Given: If total mass of the mixture of `CO "and" CO_(2) "is" 100g,` then
`CO=90.55 "and" CO_(2)=100-90.55=9.45g`
Asked: `K_(c)=?`
Formulae: `K_(P)=(2_(CO)^(2))/P_(CO_(2))`
Substitution & Calculation:
Number of moles of `CO=90.55//28=3.234`
Number of moles of `CO_(2)=9.45//44=0.215`
`therefore P_(CO)=(3.234)/(3.234+0.215)xx1"atm"=0.938`atm
`P_(CO_(2))=(0.215)/(3.234+0.215)xx1"atm"=0.062` atm
`K_(P)=P_(CO)^(2)/(P_(CO_2))=(0.938)^(2)/(0.062)=14.19`
`Deltan_(g)=2-1=1 thereforeK_(P)=K_(c)(RT)"or" K_(c)=(K_(P))/(RT)=(14.19)/(0.0821xx1127)=0.153`