7.24. Calculate (a) ∆Gϴ and (b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K
NO (g) + ½ O2 (g)⇌ NO2 (g)
where
ΔfGϴ (NO2) = 52.0 kJ/mol
ΔfGϴ (NO) = 87.0 kJ/mol
ΔfGϴ (O2) = 0 kJ/mol
7.24. Calculate (a) ∆Gϴ and (b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K
NO (g) + ½ O2 (g)⇌ NO2 (g)
where
ΔfGϴ (NO2) = 52.0 kJ/mol
ΔfGϴ (NO) = 87.0 kJ/mol
ΔfGϴ (O2) = 0 kJ/mol
(a) ΔG? = Δf?G?(NO2?) − [Δf?G?(NO) + ½ ?Δf?G?(O2?)]
= 52.0 − 87.0 −1/2 * 0
= −35 kJ/mol
(b) logK = − ΔG? / 2.303RT = − 35*103 / (2.303 * 8.314 * 298) ?
= 6.314
K = antilog (6.314)
= 1.362 * 106
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Chemistry Ncert Solutions Class 11th 2023
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