Question

HI has a normal boiling point of –35.4°C, and its ∆Hvap is 21.16 kJ/mol. Calculate the molar entropy of vaporization (∆Svap) in J/K.mol

a) ∆Svap = 74.22J/K⋅mol
b) ∆Svap = -74.22J/K⋅mol
c) ∆Svap = 214.38J/K⋅mol
d) ∆Svap = -214.38J/K⋅mol

Answer 1

The molar entropy of vaporization (∆Svap) is calculated by dividing the ∆Hvap value by the temperature in Kelvin.

Step-by-step explanation:

To calculate the molar entropy of vaporization (∆Svap), we need to convert the ∆Hvap value from kJ/mol to J/mol. Given that the ∆Hvap is 21.16 kJ/mol, we multiply this value by 1000 to convert it to J/mol, resulting in 21160 J/mol. Since ∆Svap is defined as ∆Hvap/T, where T is the temperature in Kelvin, we need to convert the normal boiling point of -35.4°C to Kelvin. Adding 273.15 to -35.4 gives us 237.75 Kelvin. Therefore, ∆Svap = 21160 J/mol divided by 237.75 K, which equals 88.91 J/K.mol.