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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 24 of 73
Marks: +1, -0
Calculate (a) ΔG° and (b) the equilibrium constant for the formation of NO2\mathrm{NO}_2 from NO and O2\mathrm{O}_2 at 298 K
NO(g)+12O2(g)\mathrm{NO}_{(g)} + \frac{1}{2} \mathrm{O}_{2(g)} ⇌ NO2(g)\mathrm{NO}_{2(g)}
where ΔfG∘(NO2)\Delta_f G^\circ(\mathrm{NO}_2) = 52.0 kJ/mol, ΔfG∘(NO)\Delta_f G^\circ(\mathrm{NO}) = 87.0 kJ/mol, ΔfG∘(O2)\Delta_f G^\circ(\mathrm{O}_2) = 0 kJ/mol
Solution:  
(a) ΔG° = ΣΔfG∘\Sigma \Delta_f G^\circ (Products) − ΣΔfG∘\Sigma \Delta_f G^\circ (Reactants)
= ΔfG∘(NO2)\Delta_f G^\circ(\mathrm{NO}_2) – [ΔfG∘(NO)+12ΔfG∘(O2)]\left[ \Delta_f G^\circ(\mathrm{NO}) + \frac{1}{2} \Delta_f G^\circ(\mathrm{O}_2) \right] = 52.0 - (87.0+12×0)\left(87.0 + \frac{1}{2} \times 0\right) = - 35.0 kJ mol−1\mathrm{kJ\,mol^{-1}}
(b) –ΔG° = 2.303 RT log K
Hence, –(–35000) = 2.303 × 8.314 × 298 × log K
or log K = 6.1341 or K = antilog (6.1341) or K = 1.361 × 10610^6
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