Ncert Exercises & Solutions

Q.20. The equilibrium constant for a reaction is 10. What will be the value of $∆ G^{θ}$ ? $R = 8.314 J K^{- 1} m o l^{- 1}, T = 300 K .$

NEETprep Answer:

From the expression,

∆ G^{θ} = - 2.303 R T \log K_{e q} ∆ G^{θ} f o r t h e r e a c t i o n, = (2.303) (8.314 J K^{- 1} m o l^{- 1}) (300 K) \log 10 = - 5744.14 J m o l^{- 1} = - 5.744 k J m o l^{- 1}