Respuesta :
when the entropy change of vaporization
= enthalpy of vaporization/boiling point temperature
when we have the enthalpy of vaporization = 38560 J/mol
and the boiling point temperature in Kelvin = 78.3 + 273 = 351.3 K
by substitution:
∴the entropy change of vaporization = 38560J/mol/351.3K
= 109.76 J/K/mol
and when the liquid has lesser entropy than the gas and we here convert from
gas to liquid so, the change in entropy = -109.76 J/K/mol
now, we need the moles of C2H5OH = mass/molar mass when the molar
mass of C2H5OH = 46 g/mol and mass = 42.2 g
∴ moles of C2H5OH = 42.2 g / 46 g/mol = 0.92 moles
when 1 mol of C2H5OH turns in liquid entropy change →-109.76 J/K/mol
∴ 0.92 mol of C2H5OH → X
∴ X entropy change when 0.92 mol = -109.76 *0.92 mol / 1 mol
= 84.64 J/K
= enthalpy of vaporization/boiling point temperature
when we have the enthalpy of vaporization = 38560 J/mol
and the boiling point temperature in Kelvin = 78.3 + 273 = 351.3 K
by substitution:
∴the entropy change of vaporization = 38560J/mol/351.3K
= 109.76 J/K/mol
and when the liquid has lesser entropy than the gas and we here convert from
gas to liquid so, the change in entropy = -109.76 J/K/mol
now, we need the moles of C2H5OH = mass/molar mass when the molar
mass of C2H5OH = 46 g/mol and mass = 42.2 g
∴ moles of C2H5OH = 42.2 g / 46 g/mol = 0.92 moles
when 1 mol of C2H5OH turns in liquid entropy change →-109.76 J/K/mol
∴ 0.92 mol of C2H5OH → X
∴ X entropy change when 0.92 mol = -109.76 *0.92 mol / 1 mol
= 84.64 J/K
The change in the randomness of the system is the entropy change. The entropy change after condensation at the standard boiling point is 84.64 J/K.
What is the entropy change?
When a system undergoes the addition or deletion of the reactant and the products, then the disorder of the system is known as entropy change.
Given,
Enthalpy of vaporization = 38560 J/mol
Boiling point temperature = 351.3 K
[tex]\begin{aligned}\text{Entropy change of vaporization} &= \dfrac{\text{enthalpy of vaporization}}{\text{boiling point temperature}}\\\\&= \dfrac{38560}{351.3}\\\\&=109.76 \;\rm J/K/mol\end{aligned}[/tex]
Here, liquid has less entropy than gas hence the change in entropy is -109.76 J/K/mol.
Moles of ethanol is calculated as:
[tex]\begin{aligned}\rm moles &= \dfrac{\rm mass}{\rm molar\; mass}\\\\&= \dfrac{42.2}{46}\\\\&= 0.92 \;\rm moles\end{aligned}[/tex]
If 1 mole of ethanol has an entropy change of -109.76 J/K/mol. Then, 0.92 moles will have,
[tex]\dfrac{-109.76 \times 0.92}{1} = 84.64\;\rm J/K[/tex]
Therefore, 84.64 J/K is the entropy change.
Learn more about entropy change here:
https://brainly.com/question/9864321
