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? Please report your answer two points past the decimal with the unit J/molK. ∆H˚fus = 3.17 kJ/mol.

Entropy is defined as the measure of possible arrangement of the atoms or molecules. Entropy change for the the freezing process mole of 1 liquid methanol at -97.6 degree celcius and 1 atm will be -18.07 J/mol.K

Explanation:

The entropy change can be calculated by the given formula:

[tex]\begin{aligned} \Delta \text S &= \frac{\Delta \text H_{\text {freezing}}}{\text T_{f}} \end{aligned}[/tex]

in which,

[tex]\Delta \text S[/tex] = change in entropy

[tex]\Delta \text H_{\text {fusion}} &=[/tex] change in enthalpy of fusion that is -3.17 kJ/mol (as given).

By the formula of enthalpy change,

[tex]\begin{aligned} \Delta \text H_{\text {fusion}} &= - \Delta \text H_{\text {freezing}} &= -3.17 {\text {kJ/mol}} &= -3170 \text {J/mol}\end{aligned}[/tex]

Now,

[tex]{\text T_{f}}[/tex] = Freezing Point Temperature = [tex]-97.6^{o} \text C[/tex] = 273 + (-97.6) = -175.4 K

Substituting the values in above equation of entropy change, we get,

[tex]\begin{aligned} \Delta \text S &= \frac{\Delta \text H_{\text {freezing}}}{\text T_{f}} \end{aligned}[/tex]

[tex]\begin{aligned} \Delta \text S &= \frac{-3170 \text {J/mol}}{175.4 \text {K}} \\\\Delta \text S& = -18.07 \text {J/mol.K}\end{aligned}[/tex]

Thus, the value of change in entropy for freezing process of 1 mole of liquid methanol is -18.07 J/mol.k

For Further Reference:

https://brainly.com/question/14591611?referrer=searchResults