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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` |
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Answer» 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)` |
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