Question

Calculate ΔS if 2.10 mol of liquid water is heated from 0.00 ∘C to 13.5 ∘C under constant pressure if CP,m=75.3J⋅K−1⋅mol−1.

The answer the the problem above is 7.62 J/K

The melting point of water at the pressure of interest is 0.00 ∘C, and the enthalpy of fusion is 6.010 kJ⋅mol−1. The boiling point is 100. ∘C, and the enthalpy of vaporization is 40.65 kJ⋅mol−1. Calculate ΔS for the transformation of the same amount of water

H2O(s,0.00∘C) →H2O(g,100.∘C)

Can you help me solve the second question?

Answer 1

There are 3 processes in H₂O (s, 0.00 ^oC) H₂O (g, 100 ^oC)

Process 1 : Conversion of solid water to liquid water at 0.00 ^oC

S₁ = (n * H_fus) / T

where

n = moles of water = 2.10 mol

H_fus = enthalpy of fusion at melting point = 6.010 kJ/mol = 6.010 x 10³ J/mol

T = melting point temperature = 0.00 ^oC = 273.15 K

Substituting the values,

S₁ = (2.10 mol * 6.010 x 10³ J/mol) / (273.15 K)

S₁ = 46.21 J/K

Process 2 : Heating of liquid water from 0.00^oC to 100 ^oC

S₂ = (n) * (C_p) * ln(T₂ / T₁)

where

n = moles of water = 2.10 mol

C_p = heat capacity of liquid water = 75.3 J/mol-K

T₂ = final temperature = 100 ^oC = 373.15 K

T₁ = initial temperature = 0.00 ^oC = 273.15 K

Substituting the values,

S₂ = (2.10 mol) * (75.3 J/mol-K) * ln(373.15 K / 273.15 K)

S₂ = 49.33 J/K

Process 3 : Conversion of liquid water to gas water at 100 ^oC

S₃ = (n * H_vap) / T

where

n = moles of water = 2.10 mol

H_vap = enthalpy of vaporization at boiling point = 40.65 kJ/mol = 40.65 x 10³ J/mol

T = boiling point temperature = 100. ^oC = 373.15 K

Substituting the values,

S₃ = (2.10 mol * 40.65 x 10³ J/mol) / (373.15 K)

S₃ = 228.77 J/K

S = S₁ + S₂ + S₃

S = (46.21 J/K) + (49.33 J/K) + (228.77 J/K)

S = 324.30 J/K