Answer :
Final answer:
The entropy of vaporization for potassium is calculated using the given molar enthalpy of vaporization and temperature, yielding 109.1 J K-1 mol-1. However, this answer is not among the provided choices, which may indicate an error in the data or options.
Explanation:
To calculate the entropy of vaporization (
Vap S) for potassium, we can use the Clausius-Clapeyron equation, which describes the relationship between the vapor pressure and temperature of a substance.
Vap S can be found using the equation
Vap S =
Vap H/T, where
Vap H is the enthalpy of vaporization and T is the absolute temperature at which the substance evaporates.
We're given that the molar enthalpy of vaporization (
Vap H) for potassium is 40.67 kJ mol-1 at 372.76 K. To find
Vap S in units of J K-1 mol-1, we first convert
Vap H to joules by multiplying by 1000, which yields
Vap H = 40,670 J mol-1.
Now we can calculate
Vap S using the formula:
Vap S =
Vap H/T = 40,670 J mol-1 / 372.76 K
Vap S = 109.10 J K-1 mol-1
After rounding to one decimal place, the entropy of vaporization for potassium is 109.1 J K-1 mol-1. However, this option is not listed in the multiple choice answers provided. Therefore, either the provided data or options may have an error or additional context may be required to correctly determine the entropy of vaporization for potassium from the given choices.