• University of Ottawa

  CHG 3326

12.33 A 20 mol% LiCl/H2O solution is made by six mixing processes:

a) mix LiCl(s) with H2O(l)

b) mix H2O with 25 mol% LiCl/H2O solution

c) Mix LiCl.H2O(s) with H2O (l)

d) Mix LiCl(s) with a 10 mol% LiCl/H2) solution

e) Mix a 25 mol% LiCl/H2O solution with a 10 mol% LiCl/H2O solution

f) Mix LiCl.H2O(s) with a 10 mol% LiCl/H2O solution

Mixing in all cases is isothermal at 25 oC. For each process determine the heat

effect in J mol-1 of solution.

Solution :

Final solution ; 20 mol% LiCl/H2O, T = 25 + 273.15 = 298.15 K

a) LiCl + 4 H2O → LiCl (4H2O)

~

ΔH = -25.2 kJ mol-1 of LiCl from Fig.12.14

ΔH20% = (0.2 mol)(-25.2) = -5.1kJ mol-1 of solution

b) LiCl(3H2O) → LiCl + 3H2O (1)

LiCl + 4H2O → LiCl(4H2O) (2)

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LiCl(3H2O) + H2O → LiCl(4H2O)

ΔH1= ΔH(LiCl(s)) – ΔH (LiCl in 3 mol H2O)

=- 408.61 – (-429.366) = 20.756 kJ mol-1 from page 457

~

ΔH2 = -25.5 KJ mol-1of LiCl from Fig. 12.14

ΔH = (0.2 mol)(ΔH1 + ΔH2) = - 0.949 kJ mol-1 of 20 mol% solution

c) LiCl.H2O → Li +1/2 Cl2 + H2 + 1/2 O2 (1)

H2 + 1/2 O2 → H2O (2)

Li + 1/2 Cl2 → LiCl (3)

LiCl + 4 H2O → LiCl(4 H2O) (4)

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LiCl.H2O + 3 H2O → LiCl(4 H2O)

ΔH1= 712.58 kJ/mol from p.457 for LiCl.H2O(s)

ΔH2= -258.83 kJ/mol from table C.4 ΔHf H2O(l)

ΔH3= - 408.61 kJ/mol from p.457 for LiCl(s)

ΔH4= -25.5 kJ/mol from Fig.12.14

ΔH= (0.2 mol)(ΔH1+ ΔH2+ ΔH3+ ΔH4)

= 712.58 -258.83 – 408.61-25.5

= 3,928 kJ mol-1 of solution

1

d) LiCl + 4H2O → LiCl(4H2O) (1)

4/9 (LiCl(9H2O) → LiCl + 9H2O) (2)

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5/9 LiCl + 4/9 LiCl(9H2O) → LiCl(4H2O)

ΔH1= -25.5 kJ/mol from Fig. 12.14

ΔH2= (4/9)(32.4) = 14.4 kJ/mol from Fig. 12.14

ΔH= (0.2 mol)( ΔH1 + ΔH2 )= -2.22 kJ

e) 5/6 (LiCl(3H2O) → LiCl + 3H2O) (1)

1/6 (LiCl(9H2O) → LiCl + 9H2O) (2)

LiCl + 4 H2O → LiCl(4 H2O) (3)

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5/6 LiCl(3H2O) + 1/6 LiCl(9H2O) → LiCl(4 H2O)

ΔH1= (5/6)(20.756) = 17.3 kJ/mol from p.457

ΔH2= (1/6)(32.4) = 5.4 kJ/mol from Fig. 12.14

ΔH3= -25.5 kJ/mol from Fig. 12.14

ΔH= (0.2 mol)( ΔH1+ ΔH2+ ΔH3)= -1.561 kJ

f) 5/8(LiCl(H2O) → Li +1/2 Cl2 + H2 + 1/2 O2) (1)

5/8(H2 + 1/2 O2 → H2O) (2)

3/8 (LiCl(9H2O) → LiCl + 9H2O) (3)

5/8 (Li +1/2 Cl2 → LiCl) (4)

LiCl + 4 H2O → LiCl(4 H2O) (5)

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5/8 LiCl(H2O) + 3/8 LiCl(9H2O) → LiCl(4 H2O)

ΔH1= (5/8)(712.58) = 445.4 kJ/mol from p.457 for LiCl(H2O)

ΔH2= (5/8)(-258.83) = -161.8 kJ/mol from table C.4 ΔHf H2O(l)

ΔH3= (3/8)(32.4) = 12.2 kJ/mol from Fig. 12.14

ΔH4= (5/8)(- 408.61) = -255.4 kJ/mol from p.457 for LiCl

ΔH5= -25.5 kJ/mol from Fig. 12.14

ΔH= (0.2 mol)( ΔH1+ ΔH2+ ΔH3+ ΔH4+ ΔH5)= - 0.403 kJ