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

(`i`) Given,
`{:(,,CO_((g)),+,2H_(2(g)),hArr,CH_(3)OH_((g))),(,"Mole before reaction",0.15,,a,,0),(,"Mole after reaction",(0.15-x),,(a-2x),,x):}`
and `x=0.08`
Total mole at equilibrium
`=0.15-x+a-2x+x=0.15+a-2x`
`=0.15+a-0.16=a-0.01`
Also total mole at equilibrium are obtained by,
`n=(PV)/(RT)`
`n=(8.5xx2.5)/(0.0821xx750) :. P=8.5 atm, V=2.5 litre, T=750 K` at eqm.
`:. n=0.345`
`:. a-0.01=0.345`
`:. a=0.355`
At equilibrium,
Mole of `CO=0.15-0.08=0.07`
Mole of `H_(2)=0.355-0.16=0.195`
Mole 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)litre^(2)`
Also `K_(p)=K_(c)(RT)^(Deltan)=187.85xx(0.0821xx750)^(2)`
`=0.05 atm^(-2)`
(`ii`) If reaction does not take place, then
Mole of `CO=0.15`
Mole of `H_(2)=0.355`
`:. `Total mole `=0.505`
`:. Pxx2.5=0.505xx0.0821xx750`
`:. P=12.438 atm`