For the reaction
`CO(g)+2H_(2)(g) hArr CH_(3)OH(g)`
hydrogen gas is introduced into a `5-L` flask at `327^(@)C` containing `0.2` mol of `CO(g)` and a catalyst, until the pressure is `4.92` atm. At this point, `0.1` mol of `CH_(3)OH(g)` is formed. Calculate the equilibrium constant `K_(p)` and `K_(c)`.

Explanation

Suppose x moles of `H_2` are introduced into the flask
`"0.2 x 0 Initial moles"`
`CO+2H_(2) Leftrightarrow CH_(3)OH`
`"(0.2 -0.1) (x-0.2) 0.1 Moles at eqb".`
`((0.1)/(5)=0.02)" "((x-0.2)/5)" "((0.1)/(5)=0.02)" Concentration at eqb."`
Total moles `=0.1x-0.2+0.1=x`
`K_(p) =(p_(CH_(3)OH))/(p_(CO) xx p_(H_(2))^(2))=((0.1)/(x) xx 4.92)/((0.1)/(x) xx 4.92) ((x-0.2)/(x) xx 4.9)`
`or K_(p)=[(x)/((x-0.2)4.9)]^(2) .....(1)`
Further, `K_(c)=([CH_(3)OH])/([CO] [H_2]^2)=(0.02)/((0.02 xx ((x-0.2))/(5))^(2))=((5)/(x-0.2))^(2) …………(2)`
Apply `K_p=K_c (RT)^(triangle n)`
`K_(p)/K_(c)=(RT)^(trianglen)=(0.0821 xx 600)^(-2) ……..(3)`
From eqns (1) and (2), we have
`K_(p)/K_(c)=((x)/(x-0.2) 4.92)^(2) ((x-0.2)/5)^(2)=((x)/(4.92 xx 5))^(2).......(4)`
From eqns (3) and (4) we have
`((x)/(4.92 xx 5))^(2)=(0.0821 xx 600)^(-2), x=0.5`
Substituting x in eqns (1) and (2) we get
`K_(p)=0.1147 atm^(-2)`
`K_(c)=277.78 ("moles/litre")^(-2)`