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

Answer:

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

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

Answer:

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

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 = $\frac{\Delta H}{T}$ .............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 =  $\frac{40.5*10^3}{285}$

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