Answer to Question #147852 in Analytical Chemistry for crisalde

Question

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]

Answer 1

The formal reaction of formation of liquid ethanol is:

2C_graphite + 1/2O₂ + 3H₂ $\rightarrow$ C₂H₅OH.

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

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

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

$∆S = -685.33$ J/K·mol.

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