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Alenkinab [10]
3 years ago
11

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.
Chemistry
1 answer:
Rudiy273 years ago
4 0

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

\Delta H_{fus} = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{freezing}}{T_m}

\Delta S=\frac{-3170J/mol}{175.4K}

\Delta S=-18.07J/mol.K

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

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