Question

Τhe enthalpy of vaporization of ammonia is 23.35 kJ/mol at its boiling point (–33.4°C). Calculate the value of ∆S when 1.00 mole of ammonia is vaporized at –33.4°C and 1.00 atm

Answer 1

Answer:

ΔS = 0.0975 KJ/K

Explanation:

Trouton's rule relates the enthalpy and entropy of vaporization at the normal boiling point.

• ΔSvap = ΔHvap / Tbp

∴ Tbp = - 33.4°C ≅ 239.5 K

∴ ΔHvap = 23.35 KJ/mol

⇒ ΔS = (23.35 KJ/mol) / (239.5 K) = 0.0975 KJ/mol.K

for 1 mol of NH3:

⇒ ΔS = 0.0975 KJ/K

Answer 2

The heat or the enthalpy of vaporization is the amount of energy needed to change the substance from liquid to gas.

When a liquid substance is heated at a given temperature and pressure then, the liquid transforms into a gaseous phase.

Given,

• Enthalpy of vaporization of ammonia (ΔHvap) = 23.35 kJ/mol

• The boiling point of ammonia (Tbp)= - 33.4°C

To convert Celcius into Kelvin:

T (K) = T (°C) + 273.15

= - 33.4°C + 273.15

= 239.75 K

To calculate the enthalpy and entropy of vaporization at the normal boiling point the following formula will be used:

$\Delta\;\text{S}&=\;\Delta\;\dfrac{\text{Hvap}}{\text{Tbp}}$

$\Delta\;\text{S}&=\;\Delta\;\dfrac{\text{23.35 KJ/mol }}{\text{239.5 K}}\;=\text{0.0975 KJ/mol K}$

= 0.0975 KJ/K

Therefore, when 1.00 mole of ammonia is vaporized at –33.4°C and 1.00 atm then the change in the entropy will be 0.0975 KJ/K.

To learn more about enthalpy and entropy of vaporization follow the given link:

https://brainly.com/question/15094832