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) rarr CH_(3)OH(g)`
`implies x = 4.92xx5/(0.082 xx 600) = 0.5 implies `moles of `H_(2)` at equilibrium =` x - 0.2 = 0.3`
Partial pressures : C`O = 0.1/0.5p , H_(2) = 0.3/0.5 p`,
`CH_(3)OH = 0.1/0.5 p`
`Kp = p/5/((p/5)(3/5 p )^(2)) = 25/9p^(2) = 25/9(4.92)^(2) = 0.10 atm^(-2)`
Concentrations : `[CO] = 0.1/5 M, [H_(2)] = 0.3/5 M`,
`[CH_(3)OH] = 0.1/5 M implies Kc = (0.1//5)/((0.1//5)(0.3//5)^(2)) = 277.77 M^(-2)`