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Calculate the change in entropy that occurs in the system when 48.6 g of water (h2o) vaporizes from a liquid to a gas at its boiling point (100.0 ∘c). the heat of vaporization is 40.7 kj/mol.

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ChemistryHelper2024
The water is vaporizing at 100°C. The vaporizing water is the system to be studied. Given the mass of water is 48.6 g. The chemical formula of water is H₂O. The molar mass of water is 18 g/ mol
moles of water are:
moles = mass / molar mass = 48.6 / 18 = 2.7 mol
The water is vaporizing, Thus, it gains the heat energy to vaporize
The amount of heat gained by the water is:
Q = n ΔH⁰ vap
   = 2.7 mol x 40.7 kJ/mol x 1000J / 1kJ = 109890 J
Q is the amount of heat lost or gained, n is the number of moles of water and ΔH⁰vap is heat of vaporization.
The expression for entropy change is:
ΔS = Q / T = 109890 J / 373 K = 294.6 J/K
The change in the entropy of the system to three significant digits is 295 J/K

dsdrajlin

When 48.6 g of water vaporizes at its boiling point (100.0 °C), the change in the entropy is 0.292 kJ/K.

First, we will calculate the change in the enthalpy (ΔH) when 48.6 g of water vaporizes considering the following relationships.

  • The heat of vaporization of water is 40.7 kJ/mol.
  • The molar mass of water is 18.02 g/mol.

[tex]\Delta H = 48.06 g \times \frac{1mol}{18.02g} \times \frac{40.7kJ}{mol} = 109kJ[/tex]

Then, we will convert 100.0 °C (T) to Kelvin using the following expression.

[tex]T = K = \° C + 273.15 = 100.0\° C + 273.15 = 373.2 K[/tex]

Finally, we will calculate the change in the entropy (ΔS) for this process using the following expression.

[tex]\Delta S = \frac{\Delta H }{T} = \frac{109kJ}{373.2K} = 0.292kJ/K[/tex]

When 48.6 g of water vaporizes at its boiling point (100.0 °C), the change in the entropy is 0.292 kJ/K.

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