Given are the following standard free energies of formation at 293 K.
`Delta_(f)G^(@) //kJ mol^(-1)" "underset(-137.17)(CO(g))" "underset(-394.36)(CO_(2)(g))" "underset(-228.57)(H_(2)O(g))" "underset(-237.13)(H_(2)O(l))" "underset("zero")(H_(2)(g))`
(a) Find `Ddelta_(r)G^(@) ` and the standard equilibrium constant `K_(p)^(0)` at 298 K for the reaction
`" "CO(g)+ H_(2)O(g) iff CO_(2)(g) + H_(2)(g)`
(b) If `CO, CO_(2)` and `H_(2)` are mixed so that the partical pressure of each is 101.325 kPa and the mixture is
brought into contact with excess of liquid water, what will be the partial pressure of gas when
equilibrium is attained at 298 K. The volume available to the gases is contant .

Answer

`P_(CO_(2)) = 202.65 kPa; p_(H_(2)O)=3.16 kPa; p_(CO)= 0.124 kPa`

Explanation

(i) `DeltaG^(@) = - 394.36 - [ -137.17 -228.57]`
`= - 28.62`
`= - 2.303 RT log K_(p)`
`K_(p)= 1.047 xx 10^(5)`
`By " "H_(2)O(l) hArr H_(2)O(g)`
`DeltaG^(@) = - 228.57 + 237.13`
` = 8.56 = RT In K_(p)`
`K_(p) = 0.0316 "bar"`
`P_(H_(2)O) (g) = 3.16 pa`
`underset("at eq. pressure X") underset("Initial pressure (101.325)")(CO) + underset(3.16 kpa)underset((3.16 kpa))(H_(2)O(g))" "hArr" "underset(202.65 kpa)underset((101.325))(CO_(2)) + underset(202 .65 kpa)underset((101.325))(H_(2))`
`1.047 xx 10^(5) =(202.65)^(5)/(X xx3.16)`
`X = 0.124 kPa`