Final answer:
The entropy change when 0.200 mol of potassium freezes at 63.7°C is -1.418 J/K. This is calculated using the enthalpy of freezing and the temperature in Kelvin.
Step-by-step explanation:
The change in entropy when 0.200 mol of potassium freezes at 63.7°C can be found using the formula ΔS = ΔH/T, where ΔS is the entropy change, ΔH is the enthalpy change of the phase transition, and T is the temperature in kelvins. Given that the enthalpy of fusion (ΔHfus) for potassium is 2.39 kJ/mol, the entropy change can be calculated as follows:
- First, convert the given temperature from °C to K: T(K) = 63.7 + 273.15 = 336.85 K
- Use the enthalpy of fusion for the calculation: ΔHfus = -2.39 kJ/mol (negative because freezing is exothermic)
- Calculate the entropy change: ΔS = ΔH/T = (-2.39 kJ/mol) / (336.85 K) = -0.0071 kJ/(mol·K) = -7.09 J/(mol·K)
- Since we have 0.200 mol, the total entropy change is: -7.09 J/(mol·K) × 0.200 mol = -1.418 J/K
The entropy change for the system of 0.200 mol of potassium freezing at 63.7°C is -1.418 J/K.