For the autoionization of water at 25°C, what is ΔG° for the process?

For the autoionization of water at 25°C,

H₂O (l) ⟺ H⁺(aq) + OH^– (aq)

k_w= 1.0 × 10^-14 . What is ΔG° for this process?

Interpretation

The autoionization of water is the process in which a small fraction of water molecules transfer a proton to one another, producing hydrogen and hydroxide ions:\(\mathrm{H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)}\)

The key idea in this problem is the connection between equilibrium and thermodynamics. The equilibrium constant tells us how far a reaction proceeds, while the standard Gibbs free energy change tells us whether the reaction is thermodynamically favorable under standard conditions.

These quantities are related by:\(\Delta G^\circ = -RT\ln K\)

where:\(R = 8.314\ \mathrm{J,mol^{-1},K^{-1}}\)

and

\(T\) is the absolute temperature.

Because the ionization constant of water is extremely small,\(K_w = 1.0\times10^{-14}\)

we already expect the forward reaction to be unfavorable under standard conditions. Therefore, the answer should come out as a positive value of \(\Delta G^\circ\).

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Concept Application

For the given equilibrium,\(\mathrm{H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)}\)

the equilibrium constant is:\(K = K_w = 1.0\times10^{-14}\)

The temperature is:\(T = 25^\circ\mathrm{C} = 298.15\ \mathrm{K}\)

Substituting these values into the Gibbs free energy equation will directly give the standard free energy change for the autoionization process.

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Solution

Using\(\Delta G^\circ = -RT\ln K_w\)

Substituting the numerical values,\(\Delta G^\circ = -(8.314)(298.15)\ln(1.0\times10^{-14})\)

Now evaluate the logarithmic term:\(\ln(1.0\times10^{-14}) = -14\ln 10\)

Since\(\ln 10 = 2.303\)

we obtain\(\ln(1.0\times10^{-14}) = -14(2.303) = -32.24\)

Substituting back,\(\Delta G^\circ = -(8.314)(298.15)(-32.24)\)\(\Delta G^\circ = 79.9\times10^4\ \mathrm{J,mol^{-1}}\)

Converting joules to kilojoules,\(\Delta G^\circ = 79.9\ \mathrm{kJmol^{-1}}\)

Therefore,\(\Delta G^\circ = +79.9\ \mathrm{kJmol^{-1}}\)

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Insight

A useful relationship to remember is:

• If \(K>1\), then \(\Delta G^\circ<0\)
• If \(K<1\), then \(\Delta G^\circ>0\)

Since water has an extremely small ionization constant, \(K_w=10^{-14}\), the equilibrium lies overwhelmingly toward molecular water. The large positive value of \(\Delta G^\circ\) is simply the thermodynamic expression of that fact.

Final Answer:\(\\Delta G^\circ = +79.9\ \mathrm{kJmol^{-1}}\)