Question

Calculate:

1. ∆GΘ and
2. the equilibrium constant for the formation of NO2 from NO and O2 at 298 K

$NO (g) + \frac 12O_2 (g) ⇋ NO_2 (g)$
where,
∆fGΘ (NO2) = 52.0 kJ/mol
∆fGΘ (NO) = 87.0 kJ/mol
∆fGΘ (O2) = 0 kJ/mol.

Answer 1

(a) For the given reaction,
ΔG° = ΔG°( Products) - ΔG°( Reactants)
ΔG° = 52.0 - {87.0 + 0}
ΔG° = - 35.0 kJ mol-1

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(b) We know that, ΔG° = RT log Kc
ΔG° = 2.303 RT log Kc
$K_c = \frac {-35.0×10^{-3}}{-2.303×8.314×298}$

$K_c= 6.134$
∴ $K_c$= antilog $(6.134)$
$K_c = 1.36×10^6$

Hence, the equilibrium constant for the given reaction $K_c$ is $1.36 × 10^6.$