The standard free energy of formation at 300 K of the given compounds are shown in the following table.
Compounds CaO CO2 N2O5 SO3 aCO3 Ca(NO3)2 CaSO4
Standard free
energy of
formation, –606 –393 134 –368
–1129 –740 –1317
Δ f G 0 at
300 K in
kJ/mole.-
R = 8.314 J K–1 mol–1 
(i) Among CO2, N2O5 and SO3 highest acidic oxide
is-
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)
Sol. CaO + CO 2 ⎯→ CaCO 3
Δ r G 0 = – 1129 (– 606) – (– 393) = – 130 
CaO + SO 3 ⎯→ CaSO 4
Δ r G 0 = – 1317 – (– 606) – (– 368) = – 343 
CaO + N 2 O 5 ⎯→ Ca(NO 3 ) 2
Δ r G 0 = – 740 – (– 606) – 134 = – 268 
(ii)
Sol. CaO (s) + CO 2 (g) ⎯→ CaCO 3 (s)
Δ r G 0 = + RT ln 
= – 130 × 10 3
∴ + 8.314 × 650 × ln 
= – 130 × 10 3 or ln 
= 
= 
~ – 24
or 
= e
–24 bar
(iii)
Sol. H+ (aq) + OH– (aq) ⎯→ H2O ( λ ) Δ r S 0 or Δ r S
0 = Δ form S 0 (H 2 O, λ )
– Δ form S 0 (H + , aq) – Δ form S 0 (OH – , aq)
or Δ r S 0 = 70 – 0 – (– 10.7)
= 70 + 10.7 = 80.7 J K –1 mol –1 Δ r G
0 = Δ r H 0 – T Δ r S 0 = – 57 – 300 × 10 –3 × 80.7
or Δ r G 0 = – 57 – 24.21 = – 81.21 kJ/mol
Δ r G 0 = Δ form G 0 (H 2 O, λ ) –
Δ form G 0 (H + , aq) – Δ form G 0 (OH – , aq)
– 81.21 = – 280 – Δ form G 0 (OH – , aq)
or – Δ form G 0 (OH – , aq) = – 280 + 81.21
= – 198.79 kJ/mol
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