Question

Calculate the standard entropy of vaporization of ethanol at its boiling point 285 K. The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ/mol

Answer 1

The standard entropy of vaporization is 142 J/mol*K

Step-by-step explanation:

Step 1: Data given

The boiling point = 285 Kelvin

The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ/mol

Step 2: Calculate the standard entropy of vaporization

ΔS = ΔH /T

⇒ with ΔS = the change in entropy

⇒ with ΔH = enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 *10³ J/mol

⇒ with T = the temperature =285 Kelvin

ΔS = (40.5 * 10³ J/mol)/ 285 Kelvin

ΔS = 142.1 ≈ 142 J/mol*K

The standard entropy of vaporization is 142 J/mol*K

(Note: The boiling point of ethanol is not 285K but 352 K)

Answer 2

standard entropy of vaporization of ethanol = 142.105 J/K-mol

Step-by-step explanation:

given data

enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 ×
`10^(3)` J/mol

entropy of vaporization of ethanol boiling point = 285 K

to find out

standard entropy of vaporization of ethanol

solution

we get here standard entropy of vaporization of ethanol that is expess as

standard entropy of vaporization of ethanol ΔS =
`(\Delta H)/(T)` .............1

here ΔH is enthalpy of vaporization of ethanol and T is temperature

put value in equation 1

standard entropy of vaporization of ethanol ΔS =
`(40.5*10^3)/(285)`

standard entropy of vaporization of ethanol = 142.105 J/K-mol