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_3OH(g)),("Initial mole",0.2,a,"",0),("mole at eqm",(0.2-0.1),(1-0.2),"",0.1):}`
Now total mole at eqm. =0.1 +a-0.2 +0.1=a
Also, `n_("total")=(PV)/(RT)=(4.92xx5)/(0.0821xx600=0.499`
`therefore a=0.499`
`therefore [CH_3OH]=(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_3OH])/([CO][H_2]H_2)=((0.1)/5)/((0.1)/5xx((0.299)/5)^2)`
`=279.64 L^2mol^(-2)`
`K_c=279.64 L^2mol^(-2)`
`K_p=K_c(RT)^(Deltang)=279.64 xx(0.0821xx600)^(-2)`
`=0.115 atm^(-2)`