Given that ΔHo(fusion,0°C) of water is 6025 J/mol, Cp,m(ice) = 37.7 J K–1 mol–1, and Cp,m(water,liquid) = 75.3 J K–1 mol–1, calculate ΔG (in J) for the freezing of 8 moles of water at 1 atm pressure and the constant temperature T = 8.6 °C.
To calculate ΔG (the change in Gibbs free energy) for the freezing of water at a constant temperature, we can use the equation:
ΔG = ΔH - TΔS
Where:
- ΔH is the enthalpy change
- T is the temperature in Kelvin
- ΔS is the entropy change
First, let's calculate the enthalpy change of the freezing process, ΔH:
ΔH = n * ΔH(fusion, 0°C)
Where:
- n is the number of moles of water
- ΔH(fusion, 0°C) is the standard enthalpy of fusion of water at 0°C, which is given as 6025 J/mol
So, ΔH = 8 mol * 6025 J/mol = 48200 J
Next, let's calculate the entropy change of the freezing process, ΔS:
ΔS = n * Cp,m(ice) * ln(T2/T1)
Where:
- Cp,m(ice) is the molar heat capacity of ice, given as 37.7 J K–1 mol–1
- T2 and T1 are the final and initial temperatures, respectively, both in Kelvin
To find T1 in Kelvin, we need to convert 8.6°C to Kelvin:
T1 = 8.6 + 273.15 = 281.75 K
Since water freezes to ice at 0°C (273.15 K), the final temperature T2 is 273.15 K.
ΔS = 8 mol * 37.7 J K–1 mol–1 * ln(273.15 K / 281.75 K) = -49.22 J/K
Now, let's calculate ΔG using the equation mentioned earlier:
ΔG = ΔH - TΔS = 48200 J - 8.6 °C * 4.184 J/g⋅°C * 8.314 J/mol⋅K * 273.15 K
We need to convert grams to moles, so we use the molar mass of water (H2O), which is approximately 18.015 g/mol.
ΔG = 48200 J - 8.6 °C * 4.184 J/g⋅°C * (8.314 J/mol⋅K / 18.015 g/mol) * 273.15 K
Simplifying the expression gives us the final answer for ΔG.
Note: Make sure to double-check the values and units given in the question and ensure consistency in the units used throughout the calculations.