Questions: The heat of fusion ΔHf of acetone (OCC3(CH3)2) is 5.7 kJ / mol. Calculate the change in entropy ΔS when 20 g of acetone freezes at -94.7 C. Be sure your answer contains a unit symbol and the correct number of significant digits.

Solution Steps

First, we need to convert the mass of acetone to moles. The molar mass of acetone (C$_3$H$_6$O) is calculated as follows: \[ \text{Molar mass of acetone} = 3 \times 12.01 + 6 \times 1.008 + 16.00 = 58.08 \, \text{g/mol} \] Given mass of acetone is 20 g: \[ \text{Moles of acetone} = \frac{20 \, \text{g}}{58.08 \, \text{g/mol}} = 0.3443 \, \text{mol} \]

The entropy change (\(\Delta S\)) for a phase transition can be calculated using the formula: \[ \Delta S = \frac{\Delta H_f}{T} \] where \(\Delta H_f\) is the heat of fusion and \(T\) is the temperature in Kelvin.

First, convert the temperature from Celsius to Kelvin: \[ T = -94.7^\circ \text{C} + 273.15 = 178.45 \, \text{K} \]

Given \(\Delta H_f = 5.7 \, \text{kJ/mol}\), we need to convert this to J/mol: \[ \Delta H_f = 5.7 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 5700 \, \text{J/mol} \]

Now, calculate the entropy change per mole: \[ \Delta S_{\text{per mole}} = \frac{5700 \, \text{J/mol}}{178.45 \, \text{K}} = 31.94 \, \text{J/(mol·K)} \]

Finally, multiply the entropy change per mole by the number of moles to get the total entropy change: \[ \Delta S = 31.94 \, \text{J/(mol·K)} \times 0.3443 \, \text{mol} = 11.00 \, \text{J/K} \]

Final Answer