CO+2H_(2) rarr CH_(3)OH (all gases). An equilibrium mixture consists of 2.0 atm CH_(3)OH, 1 atm CO and 0.1 atm H_(2). The volume, at same T. Find new equilibrium pressures.

CO+2H_(2) rarr CH_(3)OH (all gases). An equilibrium mixture consists o

1.	CO+2H_(2) rarr CH_(3)OH (all gases). An equilibrium mixture consists of 2.0 atm CH_(3)OH, 1 atm CO and 0.1 atm H_(2). The volume, at same T. Find new equilibrium pressures.
Answer» SOLUTION :`{:(CO(g),+,2H_(2)(g),hArr,CH_(3)OH(g)),(0.2 "mol",,-,,-),((0.2-0.1),,x,,0.1 "mol" larr "at equilibrium"):}` Total moles of equilibrium `=0.1+x+0.1=0.2+x` Also total moles `=(PV)/(RT)=(4.92xx5)/(0.0821xx600)=0.5` `rArr 0.5=0.2+x` `rArr x=0.3`="mol" of `H_(2)` at equilibrium `K_(C)=([CH_(3)OH])/([CO][H_(2)]^(2))=(0.1//5)/((0.1//5)(0.3//5)^(2))=277.8` If there is no catalyst, no REACTION occurs. `n=n_(CO)+n_(H_(2))=0.2+0.5=0.7` `P=n(RT)/V=6.88 "atm"`