Calculate `K_p` at 298K for the gas-phase reaction `CO+H_2O Leftrightarrow CO_2+H_2` from the following data.

Explanation

For the given reaction,
`triangleH^(0)=[triangleH_(f)^(0) (CO_(2))+triangleH_(f)^(0) (H_(2))]-[triangleH_(f)^(0) (CO)+triangleH_(f)^(0) (H_(2)O)]`
`=(-393.5+0)-(-110.5-241.8)`
`=-41.2kJ =-41200 "J/mole"`
`triangleS^(0)=[S^(0) (CO_(2))+S^(0) (H_(2))]-[S^(0)(CO)+S^(0) (H_(2)O)]`
`=(213.6+130.6)-(197.6+188.7)`
`=-42.1"J/mole"`
Applying the thermodynamic equation,
`triangleG^(0)=triangleH^(0)-T triangleS^(0)`
`=-41200-298 xx(-42.1)`
=-28655 J
Now, we have
`triangleG^(0)=-2.303 RT log K_P`
`-28655=-2.303 xx 8.314 xx 298 xx log K_(p)`
`log K_(p)=5.022`
`K_(p)=1.052 xx 10^(5)`.