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Show that the reaction : CO(g) + 1/2O(...

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

Text Solution

Verified by Experts

`Delta_(r)G^(@) = Delta_(f)G^(@)("products") - Delta_(f)G^(@)("reactants")`
` = [Delta_(f)G^(@(CO_(2))] - [Delta_(f)G^(@)(CO) + 1/2Delta_(f)G^(@)(O_(2))]`
`= -394.4 - [-137.2 + 0]`
` = -257.2 kJ`
Since `Delta_(r)G^(@)` is negative, the reaction is spontaneous. Now
`DeltaG^(@) = DeltaH^(@) -TDeltaS`
` -257.2 = DeltaH^(@) -[300 xx(-0.094)]`
`therefore DeltaH^(@) = -257.2 -28.2 = -285.4 kJ`
Since `DeltaH^(@)` is negative, the reaction is exothermic.
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