`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

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» Correct Answer - A `{:(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"`

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`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.

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