Question

When `0.15` mol of CO taken in a `2.5 L` flask is maintained at `750 K` along with a catalyst, the following reaction takes place
`CO(g)+2H_(2)(g) hArr CH_(3)OH(g)`
Hydrogen is introduced until the total pressure of the system is `8.5` atm at equilibrium and `0.08` mol of methanol is formed.
Calculate
a. `K_(p)` and `K_(c)`
b. 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 1

Correct Answer - A::B::C::D
`{:(CO,+,2H_(2),hArr,CH_(3)OH),(0.15 "mol",,,,0.08):}`
`0.15-0.08=0.017 "mol"`
Total volume `=2.5 L`
`P=8.5 "atm"`
`T=750 K`
`PV=nRT`
`8.5xx2.5=nxx0.0821xx750`
`n=0.345 "mol"`
Number of moles of `[H_(2)]` at equilibrium
`=0.345-(0.017+0.08)`
`=0.248 "mol"`
`P_(CO)=0.017/0.345xx8.5 "atm" =0.42 "atm"`
`P_(H_(2))=0.248/0.345xx8.5 "atm" =6.11 "atm"`
`P_(CH_(3)OH)=0.08/0.345xx8.5 "atm"=1.97 "atm"`
`K_(p)=(P_(CH_(3)OH))/(P_(CO)xxP_(H_(2)))=1.97/(0.42xx(6.11)^(2))=0.1256`
`K_(c)=(0.0812//2.5)/((0.017/2.5)(0.248/2.5)^(2))=478.2`
b. Number of mole of `H_(2)` taken initially `=0.248+2xx0.08`
`=0.308`
Number of mole of `CO` taken initially `=0.15`
Total Number of mole `=0.308+0.15=0.458`
`PV=nRT`
`Pxx25=0.458xx0.0821xx750`
`P=11.28 "atm"`