Question

Calculate the Standard Molar Entropy of methanol ( CH3OH) and ethanol ( C2H5OH) at 25 celcius or 298.15 K. Once calculated explain WHY one has a high standard molar entropy.

Given = melting point of methanol is 175.47 K; enthaply of fusion is 3.18 KJ/mol; boiling point is 337.7K; enthalpy of vaporization is 35.21 KJ/mol

melting point of ethanol is 159.0K; enthalpy of fusion is 5.02KJ/mol; boiling point is 351.44 K; and enthalpy of vaporization is 38.56 KJ/mol

. . . . so I dont know what equations I can use given these knowns but here are some potential equations to use:

Gm(standard) = Hm(standard) - TSm(standard) . . . . m is molar so Gm= G / mol

Delta S of Vaporization = delta H vaporization / T

Delta S fusion = delta H fusion / T

I am supposed to be able to do the problem with only the GIVEN information so hopefully someone can help me! Thanks and I will rate.

Answer 1

To calculate the S0 you need to calculate the entropy from 0 K to 298 K in the following order:

S from the heat capacity 0K----Temperature of fusion

S from the change of state (fusion) using the Enthaply of formation

S from the heat capacity Temperature of fusion --- 298 K

The sum of all the entropies is S⁰

From the data you have, there are missing the heat capacities:

the average Cp (for solid methanol)~12.97J/K-mol, liquid methanol~81.6J/K-mol

the average Cp (for solid ethanol)~11.67 J/K-mol, liquid methanol~111.46J/K-mol

applying the equations above:

For methanol

For ethanol:

Theentropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. You can see that the Ethanol has more carbons that Methanol, also molecules with strong hydrogen bonds (Methanol stronger than Ethanol) have lower values of S, which reflects more order structures.