The thermodynamic equilibrium constant at 25°C for the chemical equation
$CH_{3}COOH(aq) + H_{2}O(l) ⇋ H_{3}O^+(aq) + CH_{3}COO^−(aq)$
is $K = 1.8 × 10^{−5}$ . Calculate the value of $ΔG^{\circ}_{rxn}$ at 25°C.

Question

The thermodynamic equilibrium constant at 25°C for the chemical equation

[latex]CH_{3}COOH(aq) + H_{2}O(l) ⇋ H_{3}O^+(aq) + CH_{3}COO^−(aq)[/latex]

is [latex]K = 1.8 × 10^{−5}[/latex]. Calculate the value of [latex]ΔG^{\circ}_{rxn}[/latex] at 25°C.

Accepted Answer

The value of [latex]ΔG^{\circ}_{rxn}[/latex] can be calculated from the value of K by using Equation 23.19:

[latex]ΔG^{\circ}_{rxn}= –RT\ ln\ K = –(8.3145\ J·mol^{–1}·K^{–1})(298\ K)ln(1.8 × 10^{−5})\= +2.71 × 10^{4}\ J·mol^{–1} = +27.1\ kJ·mol^{–1}[/latex]

The positive value of [latex]ΔG^{\circ}_{rxn}[/latex] means that the reaction described by

[latex]CH_{3}COOH(aq,\ 1\ M) + H_{2}O(l) ⇋ H_{3}O^+(aq,\ 1\ M) + CH_{3}COO^−(aq,\ 1\ M)[/latex]

will proceed spontaneously from right to left under standard conditions. We say that the reaction as written is not spontaneous from left to right under standard conditions. The units of [latex]kJ·mol^{–1}[/latex] mean that this is the value of [latex]ΔG^{\circ}_{rxn}[/latex] for one mole of the reaction as written.

Question 23.7: The thermodynamic equilibrium constant at 25°C for the chemi......

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