Ncert Exercises & Solutions

7.24 Calculate a) $∆ G °$ and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K

NO (g) + ½ O2 (g) NO2 (g)

where

∆_fG° (NO₂) = 52.0 kJ/mol

∆_fG° (NO) = 87.0 kJ/mol

∆_fG° (O₂) = 0 kJ/mol

(a) For the given reaction,

∆ G °

∆ G °

(Products)-

∆ G °

(Reactants)

∆ G °

= 52.0 - {87.0 + 0)

= -35.0 kJ

{mol}^{- 1}

(b) We know that,

∆ G ° = RT \log K_{c} ∆ G ° = 2.303 RT \log K_{c} \log K_{c} = \frac{- 35.0 \times 10^{- 3}}{- 2.303 \times 8.314 \times 298} = 6.134 ∴ K_{c} = antilog (6.134) = 1.36 \times 10^{6}

Hence, the equilibrium constant for the given reaction

K_{c} is 1.36 \times 10^{6}