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

Explanation

`C("graphite")+O_(2)(g) to CO_(2)(g), Delta G_(f)^(0) [ CO_(2)(g)]= - 394.4 " kJ mol"^(-1)`
`C("graphite")+(1)/(2)O_(2)(g) to CO(g)`,
`Delta G_(f)^(0)[CO(g)] = - 137.2 " kJ mol"^(-1)`
Subtracting eq. (2) from eq. (1),
`CO(g)+(1)/(2)O_(2)(g) to CO_(2)(g)`
`Delta G^(0)= [ -394.4 -(-137.2)] kJ = - 257.2 kJ`
Since `Delta G^(0) lt 0`, the reaction is spontaneous at 300 K.
Given : `T= 300 K and Delta S^(0)= - 0.094 " kJ mol"^(-1) K^(-1)`
We know, `Delta G^(0)= Delta H^(0)- T Delta S^(0)`
`therefore - 257.2 = Delta H^(0)-300(-0.094)`
`therefore Delta H^(0)= - 285.4 kJ`
Since, `Delta H^(0) lt 0`, the reaction is exothermic.