For the reaction
`CO(g)+2H_(2)(g)hArrCH_(3)OH(g)`
Hydrogen gas is introduced into a five-litre flask at `327^(@)C`, containing `0.2` mol of `CO(g)` and a catalyst, untill the pressure is `4.92 atm`. At this point, `0.1` mol of `CH_(3)OH(g)` is formed. Calculate the equilibrium constants `K_(p)` and `K_(c )`.

Explanation

`{:(,,CO_((g)),+,2H_(2(g)),hArr,CH_(3)OH_((g))),(,"Initial mole",0.2,,a,,0),(,"Mole at equilibrium",(0.2-0.1),,(a-0.2),,0.1):}`
Now total mole at equilibrium `=0.1+a-0.2+0.1=a`
Also Mole `(n)=(PV)/(RT)=(4.92xx5)/(0.0821xx600)=0.499`
`a=0.499`
`:. [CH_(3)OH]=(0.1)/(5)`, `[CO]=(0.2-0.1)/(5)=(0.1)/(5)`,
`[H_(2)]=(0.499-0.20)/(5)=(0.299)/(5)`
`:. K_(c)=([CH_(3)OH])/([CO][H_(2)]^(2))=(0.1//5)/(0.1//5xx(0.299//5)^(2))`
`=279.64 litre^(2)mol^(-2)`
`K_(p)=K_(c)(RT)^(Deltan)`
`=279.64xx(0.0821xx600)^(-2)`
`=0.115 atm^(-2)`