Calculate the standard Gibbs energy change for the formation of propane at 298 K:
`3C("graphite") + 4H_(2)(g) to C_(3)H_(8)(g)`
`Delta_(f)H^(@)` for propane, `C_(3)H_(8)(g) = -103.8 kJ mol^(-1)`.
Given : `S_(m)^(0)[C_(3)H_(8)(g)] = 270.2 J K^(-1) "mol"^(-1)`
`S_(m)^(@)("graphite") = 5.70 J K^(-1) "mol"^(-1)`
and `S_(m)^(0)[H_(2)(g)] = 130.7 J K^(-1) "mol"^(-1)`.

Answer

`-23.43kJ mol^(-1)`

Explanation

`Delta_( r)S^(@) = S^(@) ( C_(3)H_(8)) -[3S^(@) ` ( graphite) ` + 4S^(@) ( H_(2))]= 270.2- ( 3 xx 5.7 xx 4xx 130.7 ) = - 269.7 JK^(-1) mol^(-1)`
`Delta_(r)G^(@) = Delta_(r)H^(@) - T Delta_(r)S^(@) = - 103.8 kJ mol^(-1) - 298 K xx ((-269.7)/( 1000) kJ K^(-1) mol^(-1))`
`= - 103.8 + 80.37 kJ mol^(-1) = - 23.43 kJ mol^(-1)` .