The equilibrium constant for a reaction is 20. What will be the value of $Δ G$ ?
$R = 8.314 J K^{- 1}^{- 1}, T = 300 K$

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Solution

Expressing representing the Gibbs free energy equilibrium constant at given temperature is:

$Δ G^{0} = - 2.303 R T l o g K e_{q}$

$b y p u t t i n g t h e v a l u e o f R, T, a n d K w e g e t$ ;

$Δ G^{0} = (2.303) \times (8.314 J K^{- 1}) \times (300 K) l o g 20$

= $- 7467 J m o l^{- 2}$

= $- 7467 k J m o l^{- 1}$

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