Question #147852
Calculate the standard entropy of formation of liquid ethanol at 25°C. [S0(Cgraphite) = 160.70J/K•mol, S0(O2) = 205.14 J/K•mol, S0(H2) = 130.68 J/K•mol, S0(C2H5OH) = 130.68 J/K•mol]
1
Expert's answer
2020-12-08T05:29:52-0500

The formal reaction of formation of liquid ethanol is:

2Cgraphite + 1/2O2 + 3H2 \rightarrow C2H5OH.

Using Hess law, the standard entropy change of this reaction is:

S=Sethanol03SH201/2SO202Sgraphite0∆S = S^0_{ethanol} - 3S^0_{H_2} - 1/2S^0_{O_2} - 2S^0_{graphite}

S=130.683130.680.5205.142160.70∆S = 130.68 - 3·130.68-0.5·205.14-2·160.70

S=685.33∆S = -685.33 J/K·mol.

The standard entropy of formation of liquid ethanol is -685.33 J/K·mol.


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