At `817^(@)C`, `K_(p)` for the reaction between `CO_(2(g))` and excess hot graphite `(s)` is `10 atm`.
(`a`) What are the equilibrium concentration of the gases at `817^(@)C` and a total pressure of `5 atm?`
(`b`) At what total pressure, the gas contains `5% CO_(2)` by volume?

Explanation

(`a`) The given equilibrium is ,
`{:(,,CO_(2(g)),+,C_((s)),hArr,2CO_((g))),(,"Initial mole",1,,,,0),(,"Final mole",(1-alpha),,,,2alpha):}`
`K_(p)=((n_(CO))^(2))/(n_(CO_(2)))xx[(P)/(sumn)]^(1)=((2alpha)^(2))/((1-alpha))xx[(5)/(1+alpha)]`
`10=(20alpha^(2))/(1-alpha^(2))`
or `10-10alpha^(2)=20alpha^(2)`
`:. alpha^(2)=(10)/(30)`
`:. alpha=sqrt((1)/(3))=0.577`
Thus, mole of `CO_(2)` at equilibrium `=1-alpha`
`=1-0.577=0.423`
and mole of `CO` at equilibrium
`=2alpha=2xx0.577=1.154`
Total moles present at equilibrium
`=0.423+1.154=1.577`
At equilibrium, `PV=nRT`
`P=5 atm`, `n=1.577`, `T=817+273=1090 K`
`5xxV=1.577xx0.0821xx1090`
`:. V=28.22 litre`
`:. [CO]` at equilibrium `=(1.154)/(28.22)=0.041 mol litre^(-1)`
`[CO_(2)]` at equilibrium `=(0.423)/(28.22)=0.015 mol litre ^(-1)`
(`b`) `{:(For,,CO_(2(g)),+,C_((s)),hArr,2CO_((g))),(,"Initial mole",1,,0,,0),(,"Final mole",(1-a),,,,2a):}`
`:. ` Total mole at equilibrium `=1-a+2a=1+a`
Given `[(1-a)/(1+a)]=(5)/(100)`
`:. a=(95)/(100)`
`:' K_(p)=((n_(CO))^(2))/(n_(CO_(2)))xx[(P)/(sumn)]^(1)`
`:. 10=([(2xx95)/(100)]^(2))/(((5)/(100)))xx(P)/(((105)/(100)))`
`:. P=0.145 atm`