The equilibrium constant for a reaction is 10. What will be the value of ΔG^θ? R = 8.314 JK^–1 mol^–1, T = 300 K.

Open in App

Solution

From the expression,

ΔG^θ = –2.303 RT logK_eq

ΔG^θ for the reaction,

= (2.303) (8.314 JK^–1 mol^–1) (300 K) log10

= –5744.14 Jmol^–1

= –5.744 kJ mol^–1

Suggest Corrections

Similar questions

{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}

Join BYJU'S Learning Program