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Question
Calculate a) ΔG°and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K
\[\ce{NO(g) + 1/2 O_2 (g) <=> NO_2(g)}\]
where ΔfG⊝ (NO2) = 52.0 kJ/mol
ΔfG⊝ (NO) = 87.0 kJ/mol
ΔfG⊝ (O2) = 0 kJ/mol
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Solution
a) For the given reaction,
ΔG⊝ = ΔG⊝ ( Products) – ΔG°( Reactants)
ΔG⊝ = 52.0 – {87.0 + 0}
= - 35.0 kJ mol–1
b) We know that,
ΔG⊝ = RT log Kc
ΔG⊝ = 2.303 RT log Kc
`"K"_"c" = (-35.5xx10^(-3))/(-2.303xx8.314xx298)`
= 6.134
`therefore "K"_"c" = "antilog" (6.134)`
`= 1.36 xx 10^6`
Hence, the equilibrium constant for the given reaction Kc is 1.36 × 106
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