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  

5 Views|Posted 8 months ago
Asked by Shiksha User
1 Answer
V
8 months ago

(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

Chemistry Ncert Solutions Class 11th 2023

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