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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 24
Total: 73
Calculate (a) ΔG° and (b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K
NO(g)+
1
2
O2(g) ⇌ NO2(g)
where ΔfG°(NO2) = 52.0 kJ/mol, ΔfG°(NO) = 87.0 kJ/mol, ΔfG°(O2) = 0 kJ/mol
Solution:  
(a) ΔG° = ΣΔfG° (Products) − ΣΔfG° (Reactants)
= ΔfG°(NO2) – [ΔfG°(NO)+
1
2
ΔfG°(O2)] = 52.0 - (87.0+
1
2
×0) = - 35.0 kJmol−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 × 106
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