Question

At room temperature, the equilibrium constant (Kw) for the self-ionization of water is 1.00 × 10^−14. Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)

a) -2.303 RT
b) 2.303 RT
c) -RT/2.303
d) RT/2.303

Answer 1

The standard free energy change (∆G°) for the aqueous reaction of hydrogen ion with hydroxide ion to produce water is a. -2.303 RT.

Step-by-step explanation:

The equilibrium constant for the autoionization of water is referred to as the ion-product for water and is given the symbol Kw. The equation for the reaction is H₂O(l) → H⁺(aq) + OH⁻(aq). The value of Kw at room temperature is 1.00 × 10-14.

To calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water, we need to use the reverse reaction of the self-ionization of water. Since the reaction is reversed, the equilibrium constant is the reciprocal of Kw:

Kw reverse = 1/Kw = 1/(1.00 × 10-14)

The standard free energy change (∆G°) is related to the equilibrium constant by the equation ∆G° = -RT ln(K). Substituting the value of Kw reverse into this equation, we can solve for ∆G°:

∆G° = -RT ln(1/(1.00 × 10-14)) = a. -2.303 RT