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Gibbs Free Energy & Spontaneity

The equilibri...

Question

The equilibrium constant for a reaction is 10. What will be the value of $Δ G^{θ}$ ? $R = 8.314 J K^{- 1} m o l^{- 1} T = 300 K$ .

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Solution

From the expression,
$Δ G^{θ} = - 2.303 R T l o g K_{e q}$
$Δ G^{θ}$ for the reaction,
$= (2.303) (8.314 J K^{- 1} m o l^{- 1}) (300 K) l o g_{10} = - 5744.14 J m o l^{- 1} = - 5.744 k J m o l^{- 1}$

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{Δ G}^{o}

(R = 8.314 J K^{- 1} m o l^{- 1}

T = 300 K)

Δ G

R = 8.314 J K^{- 1}^{- 1}, T = 300 K

10

△ G^{o}

(R = 8 J K^{- 1} m o l^{- 1}, T = 300 K)

Δ G^{\circ}

(R = 8 J K^{- 1} m o l^{- 1}, T = 300 K)

△ G^{⊖}

J K^{- 1} m o l^{- 1}

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