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 \(\Delta G^\circ\).
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.
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\(\ln(1.0\times10^{-14}) = -14(2.303) = -32.24\)
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:\(\\Delta G^\circ = +79.9\ \mathrm{kJmol^{-1}}\)
A balanced state where opposing processes occur at equal rates, maintaining constant concentrations of reactants and products.
Interpretation
In chemistry, equilibrium is not just a definition; it is a natural principle that governs how systems behave. Whether it is a chemical reaction, a physical change, or even a biological regulation, systems tend to move toward a state where opposing processes balance each other.
At its core, equilibrium means no observable change with time, but this does not mean that everything has stopped. Instead, it reflects a perfect balance of opposing tendencies. This distinction is important because many students mistakenly think equilibrium is a static condition, when in reality it is often highly dynamic at the microscopic level.
Concept Application
Let us connect this idea to chemical systems.
In most chemical reactions, the process does not simply stop after forming products. Instead, as products accumulate, they begin to react backward to reform reactants. This creates two competing processes:
Forward reaction: Reactants → Products
Reverse reaction: Products → Reactants
Equilibrium is reached when these two processes occur at equal rates.
At this point:
Concentrations of reactants and products become constant
The reaction appears to have stopped
But molecular activity continues continuously
This understanding allows us to classify equilibrium meaningfully.
What is Equilibrium?
Equilibrium is the state of a system in which no net change occurs with time, because opposing processes exactly balance each other. Even if the system is disturbed, natural tendencies act to restore this balance.
This idea extends beyond chemistry. For instance, a book resting on a table remains stationary because the downward gravitational force is exactly balanced by the upward normal force. Such a balance ensures stability.
Thus, equilibrium represents a condition of perfect balance between opposing influences, leading to a stable system.
Types of Equilibrium
Equilibrium can broadly be understood in two forms:
1. Dynamic Equilibrium
Dynamic equilibrium applies to both physical and chemical processes. It occurs when two opposite processes happen at the same rate.
Consider a closed container containing water:\(\text{H}_2\text{O (l)} \rightleftharpoons \text{H}_2\text{O (g)}\)
Water molecules continuously evaporate into vapor
Simultaneously, vapor molecules condense back into liquid
When both rates become equal, equilibrium is established
At this stage:
The amount of liquid and vapor remains constant
But molecules are continuously exchanging between phases
This is why it is called dynamic equilibrium; there is continuous activity, yet no visible change.
2. Chemical Equilibrium
Chemical equilibrium specifically refers to reversible chemical reactions: \(\text{Reactants} \rightleftharpoons \text{Products}\)
Initially, the forward reaction dominates. As products form, the reverse reaction begins. Eventually, a state is reached where:\(\text{Rate of forward reaction} = \text{Rate of reverse reaction}\)
At equilibrium:
Concentrations remain constant
Reactions continue at the molecular level
The system is stable but not static
Types Based on Phases
Homogeneous Equilibrium All species are in the same phase:\(\text{N}_2 (g) + 3\text{H}_2 (g) \rightleftharpoons 2\text{NH}_3 (g)\)
Heterogeneous Equilibrium Different phases are involved:\(\text{CaCO}_3 (s) \rightleftharpoons \text{CaO} (s) + \text{CO}_2 (g)\)
Equilibrium Constant (K)
To describe equilibrium quantitatively, we use the equilibrium constant.
For a general reaction:\(aA + bB \rightleftharpoons cC + dD\)
The equilibrium constant is:\(K = \frac{[C]^c [D]^d}{[A]^a [B]^b}\)
This expression tells us the relative amounts of products and reactants at equilibrium.
For gaseous systems:
In terms of concentration → \(K_c\)
In terms of pressure → \(K_p\)
Their relationship is:\(K_p = K_c (RT)^{\Delta n}\)
Where
If \(\Delta n = 0\), then:\(K_p = K_c\)
Also, pure solids and liquids are not included in equilibrium expressions because their effective concentration remains constant.
Significance of Equilibrium Constant
The value of \(K\) tells us how far a reaction proceeds:
\(K \gg 1\) → Products are favored; reaction proceeds nearly to completion
\(K \ll 1\) → Reactants are favored; very little product forms
\(K \approx 1\) → Both reactants and products are present in comparable amounts
Thus, \(K\) acts as a measure of the position of equilibrium.
Insight
A powerful way to remember equilibrium is this:
Equilibrium is not about stopping; it is about balancing.
Also keep in mind:
Equal rates do not mean equal concentrations
A large \(K\) does not mean fast reaction—it only indicates extent, not speed
If you internalize these two ideas, equilibrium will start to feel intuitive rather than abstract.
At equilibrium, we express the equilibrium constant in terms of concentrations (\(K_c\)) or partial pressures (\(K_p\)) using the law of mass action.
However, there is a crucial refinement students must internalize:
Pure solids and pure liquids do not appear in equilibrium expressions because their concentrations (or activities) remain constant and are taken as unity.
This is not a shortcut—it reflects physical reality. A solid’s “effective concentration” does not change during the reaction, so including it would not affect the equilibrium ratio.
For gases:
Use \(K_c\) in terms of molar concentration Use \(K_p\) in terms of partial pressures They are related by:
\(K_p=K_c (RT)^{\Delta n}\)
where \(\Delta n = \text{moles of gaseous products} – \text{moles of gaseous reactants}\)
Concept Application
For each reaction, we will:
Identify phases carefully Exclude solids and liquids
Include only: gases (for both \(K_c\) and \(K_p\)) aqueous species (only in \(K_c\)) Compute \(\Delta n\) only using gaseous species (for \(K_p\))
Solution
(1) 3Fe (s) + 4H₂O (g) ⇌ Fe₃O₄ (s) + 4H₂ (g)
Only gases are included: H₂O(g) and H₂(g)
\(K_c = \frac{[H_2]^4}{[H_2O]^4}\)
For \(K_p\):
\(K_p = \frac{(P_{H_2})^4}{(P_{H_2O})^4}\)
Now check \(\Delta n\):
\(\Delta n = 4 – 4 = 0\)
So,
\(K_p = K_c\)
(2) HF (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + F⁻ (aq)
Here:
H₂O is a pure liquid → excluded All others are aqueous species
\(K_c = \frac{[H_3O^+][F^-]}{[HF]}\)
No \(K_p\) expression exists because there are no gaseous species.
P₄ is a solid → excluded PCl₃ is a liquid → excluded Only Cl₂(g) remains
\(K_c = \frac{1}{[Cl_2]^6}\)
\(K_p = \frac{1}{(P_{Cl_2})^6}\)
Now,
\(\Delta n = 0 – 6 = -6\)
So,
\(K_p = K_c (RT)^{-6}\)
Insight
The fastest way to get these right in exams is to build a mental filter:
“If it’s a solid or liquid → ignore it. If it’s gas or aqueous → include it.”
Also remember:
If only one gaseous species appears, it will sit alone in the denominator or numerator—this often surprises students. When \(\Delta n = 0\), \(K_p = K_c\), which is a powerful shortcut worth spotting instantly.
If you consistently apply these filters, equilibrium expressions stop being memorization—and become almost automatic.
