Correct Answer - 75 atm
`CH_3OCH_3(g)toCH_4(g)+H_2(g)+CO(g)`
Pressure at t=0 0.40
Pressure at t=12 min (0.40-P) P P P
For ideal gas behaviour moles `prop` pressure
`:.aprop0.40, " " (a-X)prop(0.40-P)`
`because K=2.303/t "log" a/(a-X) or " " 0.693/14.5=2.303/12 "log" 0.40/(0.40-P)`
`:. P=0.175` atm
Thus, total pressure after 12 minutes=0.40-P+P+P+P
=0.40+2P=0.40+2x0.175=0.75 atm