Calculate the standard entropy, delta S degree_rxn, of the following reaction at

Question

Calculate the standard entropy, delta S degree_rxn, of the following reaction at 25.0 degree C using the data in this table. The standard enthalpy of the reaction, delta H degree_rxn, is -44.2 kJ middot mol^-1 C_2H_4 (g) + H_2O (l) rightarrow C_2H_5OH(l) delta S degree_rxn = J middot K^-1 middot mol^-1 Then, calculate the standard Gibbs free energy of the reaction, delta G degree_rxn. delta G degree_rxn = kJ middot mol^-1 Finally, determine which direction the reaction is spontaneous as written at 25.0 degree C and standard pressure forward reverse both neither

Explanation / Answer

1)

From data table:

Sof(C2H4(g)) = 219.56 J/mol.K

Sof(H2O(l)) = 69.91 J/mol.K

Sof(C2H5OH(l)) = 160.7 J/mol.K

Balanced chemical equation is:

1C2H4(g) + 1H2O(l) ---> 1C2H5OH(l)

delta So rxn = 1*Sof(C2H5OH(l)) - 1*Sof( C2H4(g)) - 1*Sof(H2O(l))

delta So rxn = 1*160.7 - 1*219.56 - 1*69.91

delta So rxn = -128.77 J/mol.K

Answer: -128.8 J/mol.K

2)

Given:

delta Ho rxn = -44.2 KJ/mol

Now we have:

delta H = -44.2KJ/mol

delta S = -128.8J/mol.k = -0.1288 KJ/mol.K

T = 25.0 oC =(25.0+ 273) K = 298.0 K

use:

delta G = delta H - T*delta S

delta G = -44.2 - 298.0 *( -0.1288)

delta G = -5.82 KJ/mol

Answer: -5.82 KJ/mol

3)

Since Gorxn is positive, the reaction in forward direction will be spontaneous

Answer: forward

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