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

`{:(,CO_2(g)+C(s)hArr , 2CO(g)),("Initial moles",1,0),("Moles of equi.",1-alpha, 2alpha):}` (`alpha` is the degree of dissociation )
Total moles =`1-alpha+2alpha=1+alpha`
Partial pressure of `CO_2, p(CO_2)=((1-alpha)/(1+alpha))xx5`
Partial pressure of CO, `p(CO)=((2alpha)/(1+alpha))xx5`
`K_p=(p(CO)^2)/(p(CO_2))`
`10=((2alpha)/(1+alpha)xx5)^2/(((1-alpha)/(1+alpha)xx5))`
`=(20alpha^2)/(1-alpha^2)`
or `20alpha^2=10-10alpha^2`
`30alpha^2 =10`
`therefore alpha=sqrt(1/3)` =0.577
(a) At equilibrium
`therefore` Moles of `CO_2 = 1-alpha=1-0.577`=0.423
Moles of CO=`2alpha=2xx0.577` =1.154
(b)% of `CO_2` =5% by volume
% of CO =95% by volume
Then `p(CO_2)=5/100xxp` =0.05 p
`p(CO)=95/100xxp`=0.95 p
`K_p=(0.95p)^2/(0.05p)` =18.05 p
or 10=18.05 p
`therefore p=10/18.05` = 0.554 atm.