0.15 mole of `CO` taken in a `2.5L` flask is maintained at `750K` along with a catalyst so that the following reaction can take place:
`CO(g)+2H_(2)(g)hArrCH_(3)OH(g)`
Hydrogen is introduced until the total pressure of the system is 8.5 atm at equilibrium and 0.08 mole of methanol is formed. Calculare (i) `K_(p)` and `K_(c)` and (ii) the final pressure if the same amount of `CO` and `H_(2)` as before are used, but with no catalyst so that the reaction does not take place.

Answer

(i) `K_(c)=(20000)/(343)=58.3 mol^(-2)L^(2)K_(P)=(58.3)/(41xx41)=0.035atm^(-2)` (ii) `P=8.2` atm

Explanation

(i) `CO(g)+2H_(2)(g)hArrCH_(3)OH(g)`
`0.15 a`
`0.15-x a-2x x rArr x=0.08`
`0.15-x+a-2x+x=0.5 PV=nRT`
`a-2x=0.35 n+(8.2xx2.5)/(0.082xx500)=0.5`
`K_(C)=((0.08)/(2.5))/(0.07/(2.5)x((0.35)/(2.5))^(2))=(20000)/(343)=58.3`
`K_(P)=58.3xx(RT)^(-2)=(58.3)/((0.082xx500)^(2))=0.035`
(ii) Total pressure will remain `8.2` atm as catalyst only time taken to achieve equilibrium, does not affect equilibrium condition//concentration.