Lec 27 March 13 Fri

Lec 27 March 13 Fri - 3.17 Eg Is it feasible to make...

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3.17 Δ H º f ( C 2 H 5 OH (g) ) = -235.1 kJ/mol Δ H º f ( H 2 O (g) ) = -241.8 kJ/mol Δ H º f ( C 2 H 4 (g) ) = 52.26 kJ/mol Δ H º rxn = -45.56 kJ Δ S º ( C 2 H 5 OH (g) ) = 282.7 J/Kmol Δ S º ( H 2 O (g) ) = 188.8 J/Kmol Δ S º ( C 2 H 4 (g) ) = 219.6 J/Kmol Δ S º rxn = -125.7 J/K Eg. Is it feasible to make ethanol (see above) at 110 º C, but under standard pressure conditions ? At this temp. rxn is: C 2 H 4 (g) + H 2 O (g) C 2 H 5 OH (g) Now, use Δ G º = Δ H º - T Δ S º (convert everything to either kJ or J) Δ G º = -4.556 x 10 4 J - (383 K)(-125.7 J/K) = 2.58 kJ Now Δ G º is positive . -- It is no longer feasible to do this reaction at 110 º C. -- So increasing T will not make it speed up, but makes it stop -- If you want to speed up this reaction, you have to use a catalyst, and keep the temperature low Summary : (see Table 19.1) Very important table, but don’t memorize Remember: Δ G º = Δ H º - T Δ S º Δ H Δ S Result 1. positive negative Never spontaneous
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Lec 27 March 13 Fri - 3.17 Eg Is it feasible to make...

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