Question

Calculate the standard molar entropy change for the formation of gaseous propane `(C_(3)H_(8))` at `293K`.
`3C("graphite") +4H_(2)(g) rarr C_(3)H_(8)(g)`
Standard molar entropies `S_(m)^(Theta) (JK^(-1)mol^(-1))`are:
`C("graphite") = 5.7, H_(2)(g) = 130.7,C_(3)H_(5)(g) = 270.2`

Answer 1

Standard molar entropy change,
`Delta_(r)S_(m)^(Theta) = S_(m)^(Theta) (C_(3)H_(8))-[3S_(m)^(Theta) (C ) +4S_(m)^(Theta) (H_(2))]`
`= 270.2 -[3 xx 5.7 +4 xx 130.7]`
`= 270.2 - 539.9 =- 269.7 J K^(-1)`