Question

At room temperature, the equilibrium constant (Kw) for the self-ionization of water is 1.00 x 10-¹⁴. Calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water.

a) Positive free energy change.
b) Negative free energy change.

c) Zero free energy change.
d) Insufficient information to determine.

Answer 1

The standard free energy change for the reaction of hydrogen ion with hydroxide ion to produce water at room temperature is negative, indicating a spontaneous reaction under standard conditions.

Step-by-step explanation:

The reaction for the self-ionization of water is H2O(l) ⇌ H+(aq) + OH-(aq) and its equilibrium constant is Kw = [H+] [OH-]. At 25°C, Kw = 1.00 × 10-14.

To find the standard free energy change (ΔG°) for this reaction, we can use the thermodynamic relationship between the equilibrium constant and the free energy change:

ΔG° = -RT ln(Kw)

Where R is the universal gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298 K at 25°C). Plugging in the values gives us:

ΔG° = - (8.314 J/mol·K) (298 K) ln(1.00 × 10-14)

Calculating this provides a negative free energy change, indicating that the reaction is spontaneous under standard conditions. Therefore, the correct answer is b) Negative free energy change.