Show that the reaction
`CO(g) +(1//2)O_(2)(g) rarr CO_(2)(g)`
at `300K` is spontaneous and exothermic, when the standard entropy change is `-0.094 k J mol^(-1) K^(-1)`. The standard Gibbs free energies of formation for `CO_(2)`and `CO` are `-394.4`and `-137.2kJ mol^(-1)`, respectively.

Explanation

For the given reaction, we have : `CO(g)+(1)/(2)O_(2)(g)rarr CO_(2)(g) Delta_(r )S^(Θ)=-0.094 kJ K^(-1)mol^(-1)`
The free - energy change of the reaction is :
`Delta_(r )G^(Θ)=Delta_(f)G_(CO_(2))^(Θ)-Delta_(f)G_(CO)^(Θ)-(1)/(2)Delta_(f)G_(O_(2))^(Θ)`
`=(-394.4+137.2)kJ mol^(-1) = -257.2 kJ mol^(-1) " " [because Delta_(f)G_(O_(2))^(Θ)=0]`
Since `Delta_(r )G^(Θ)` is negative, the reaction is spontaneous. The enthalpy change of the reaction is :
`Delta_(r )H^(Θ)=Delta_(r )G^(Θ)+T Delta_(r )S^(Θ)`
`=[-257.2+(300)(-0.094)]kJ mol^(-1)=-285.4 kJ mol^(-1)`
Since `Delta_(r )H^(Θ)` is negative, the reaction is exothermic.