Question

What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (-97.6˚C) and 1 atm?

a) ΔS = (-3.17 kJ/mol) / (-97.6 °C)
b) ΔS = (3.17 kJ/mol) / (-97.6 °C)
c) ΔS = (-3.17 kJ/mol) / (-97.6 K)
d) ΔS = (3.17 kJ/mol) / (-97.6 K)

Answer 1

The entropy change for the freezing of 1 mole of liquid methanol at its freezing temperature of -97.6°C is calculated using the formula ΔS = ΔH/T, with T in Kelvin. The correct answer is that the entropy change is ΔS = (-3.17 kJ/mol) / 175.6 K.

Step-by-step explanation:

To determine the entropy change (ΔS) for the freezing process of 1 mole of liquid methanol at its freezing temperature of -97.6°C (175.6 K) and 1 atm, you must consider that the freezing process is the reverse of melting. Therefore, the entropy change of freezing is the negative of the entropy change of melting. Given that the enthalpy change (ΔH) for freezing methanol is -3.17 kJ/mol, you can calculate the entropy change using the formula:

ΔS = ΔH/T

However, the temperature must be in Kelvin. To convert -97.6°C to Kelvin, add 273.15:

-97.6°C + 273.15 = 175.6 K

Substituting the values into the equation, we get:

ΔS = (-3.17 kJ/mol) / 175.6 K

ΔS = -0.01805 kJ/K ≈ -18.05 J/K (per mole)

Thus, the correct answer is:

c) ΔS = (-3.17 kJ/mol) / 175.6 K