`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

A

Explanation

`{:(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"`