For the autoionization of water at 25°C, H2O(l) <---> H+ (aq) + OH-(aq) Kw is 1.0 X 10^-14. What is (Delta Gº) for the process? ​

ΔG° = 80 Kj/mole

Explanation:

ΔG° = -R·T·ln(Kw)

R = 0.008314 Kj/mol·K

T = 25°C = 298K

Kw = 1 x 10⁻¹⁴

ΔG° = -R·T·ln(Kw) -(0.08314Kj/mol·K)(298K)ln(1 x 10⁻¹⁴)

= -(0.008314)(298)(-32.2)Kj/mole

= 79.78Kj/mole ≈ 80Kj/mole (1 sig.-fig.)

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