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.

Explanation

`{:(,CO(g),+,2H_(2)(g),hArr,CH_(3)OH(g)),("moles before reaction",0.15,,a,,0),("moles after",(0.15-x),,(a-2x),,x),("reaction and",,,,,x=0.08):}`
Total moles at eqilibrium
`=0.15-x+a-2x+ x=0.15+a-2x`
`=0.15+a-0.16=a-0.01`
Also total moles at equilibrium are obtained by `n=(PV)/(RT)`
`{:(n=(8.5xx2.5)/(0.0821xx750),:',P=8.5 "atm"),(,,V=2.5 L),( :. n=0.345,,T=750 K):}} {:(,,),("at equilibrium",,),(,,):}`
`:. a-0.01=0.345`
`:. a=0.355`
At equilibrium:
"moles" of `CO=0.15-0.08=0.07`
"moles" of `H_(2)=0.355-0.16=0.195`
"moles" of `CH_(3)OH=0.08`
`:.K_(c )=([CH_(3)OH])/([H_(2)]^(2)[CO])=(0.08.//2.5)/((0.07//2.5)(0.195//2.5)^(2))`
`=187.85 "mol"^(-2) L^(2)`
Also
`K_(p)=K_(c )(RT)^(Deltan)=187.85xx(0.0821xx750)^(-2)`
`=0.05 "atm"^(-2)`
b. If reaction does not take place, then
moles of `CO=0.15 "mol"`
moles of `H_(2)=0.355 "mol"`
`:.` Total moles `=0.505`
Using relation `PVxxnRT`
`:. Pxx2.5=0.505xx0.0821xx750`
`P=12.438 "atm"`