Thermodynamic Equilibrium: The Ultimate State of Balance
When you drop an ice cube into hot water, the system changes rapidly. Ice melts, water cools down, and heat flows. But eventually, everything stops changing. The water reaches a uniform temperature, and no further macroscopic changes occur. This final, tranquil state is called Thermodynamic Equilibrium.
Table of Contents
1. What is Thermodynamic Equilibrium?
A system is said to be in a state of strict thermodynamic equilibrium if its macroscopic properties (like temperature, pressure, density, and chemical composition) do not change with time, and if it is isolated from its surroundings, no spontaneous change occurs.
It is an idealized macroscopic concept. On a microscopic level, molecules are still colliding, and forward/reverse reactions might still be occurring, but they happen at identical rates. Thus, there are no net macroscopic fluxes of matter or energy.
2. The Three Pillars of Equilibrium
For a system to be in complete thermodynamic equilibrium, it must simultaneously satisfy three independent conditions. If any one of these is violated, the system is in a state of non-equilibrium.
The Three Pillars of Thermodynamic Equilibrium
Thermal Equilibrium
No temperature gradients. Heat transfer has ceased.
Mechanical Equilibrium
No unbalanced forces. Pressure is uniform.
Chemical Equilibrium
No net reactions. Chemical composition is constant.
1. Thermal Equilibrium: Dictated by the Zeroth Law of Thermodynamics. Two parts of a system are in thermal equilibrium if they have the same temperature ($T$). If there is a temperature gradient, heat will spontaneously flow from hot to cold until equilibrium is achieved.
2. Mechanical Equilibrium: There must be no unbalanced macroscopic forces acting on or within the system. For a fluid system, this generally means the pressure ($P$) is uniform throughout (ignoring gravitational effects on tall columns of fluid).
3. Chemical Equilibrium: This is the most complex. The chemical composition of the system must not change with time. This implies two things: (a) No net chemical reactions are occurring, and (b) No net transfer of matter between different phases (e.g., liquid evaporating to gas) is taking place. This is governed by the equality of Chemical Potential ($\mu$).
3. Mathematical Criteria for Equilibrium
How do we mathematically know if a system has reached equilibrium? It depends entirely on the boundary constraints of the system (e.g., is it isolated? is it kept at constant temperature?). Let's derive the criteria.
Criterion 1: Isolated System (Constant $U, V$)
According to the Second Law of Thermodynamics, the entropy ($S$) of an isolated system (the universe) always increases during a spontaneous process ($dS_{iso} > 0$).
Because entropy cannot increase forever, it will eventually reach a maximum value. Once it hits this maximum peak, it can no longer change. Thus, the mathematical condition for an isolated system at equilibrium is:
Criterion 2: Closed System at Constant Temperature & Volume ($T, V$)
Chemists rarely work with perfectly isolated systems. If we have a rigid, sealed flask (constant $V$) submerged in a large water bath (constant $T$), we use the Helmholtz Free Energy ($A = U - TS$).
The differential is $dA = dU - TdS - SdT$. At constant $T$, $dT=0$, so $dA = dU - TdS$. From the Clausius inequality, $TdS \ge dq$. Also, from the First Law at constant volume, $dU = dq$. Therefore, $TdS \ge dU$, which rearranges to $dU - TdS \le 0$.
Criterion 3: Closed System at Constant Temperature & Pressure ($T, P$)
This is the most crucial criterion for chemistry, as most reactions happen in open beakers (constant atmospheric $P$) and at room temperature (constant $T$). Here, we use Gibbs Free Energy ($G = H - TS$).
We derived in a previous article that the fundamental equation is $dG = VdP - SdT$. If we hold both Temperature and Pressure constant, then $dT = 0$ and $dP = 0$.
For a spontaneous process, the system's Gibbs energy decreases ($dG < 0$) until it reaches the lowest possible energy valley. At the absolute bottom of this thermodynamic valley, the derivative is zero.
4. Chemical Potential ($\mu$) and Phase Equilibrium
When dealing with mixtures or multiple phases (like ice melting into water), the simple $dG = VdP - SdT$ equation is incomplete because the number of moles ($n$) of the substances are changing.
Willard Gibbs introduced the concept of Chemical Potential ($\mu$). It is defined as the partial molar Gibbs free energy. For a substance $i$:
The fundamental equation for Gibbs Free Energy expands to include chemical work:
Deriving Phase Equilibrium
Imagine a closed system with two phases, Phase $\alpha$ (e.g., liquid water) and Phase $\beta$ (e.g., water vapor), at constant $T$ and $P$.
Suppose an infinitesimal amount of matter, $dn$, transfers from Phase $\alpha$ to Phase $\beta$.
- Change in moles in Phase $\alpha$: $dn_\alpha = -dn$
- Change in moles in Phase $\beta$: $dn_\beta = +dn$
The total change in Gibbs Free Energy ($dG$) at constant $T$ and $P$ is the sum of the changes in both phases:
For the system to be in equilibrium, $dG_{total}$ must be zero ($dG = 0$):
Since $dn$ is not zero, the term in parentheses must be zero. This yields the profound universal law of phase equilibrium:
Meaning: Matter will spontaneously flow from a phase with higher chemical potential to a phase with lower chemical potential until the potentials equalize.
5. Connecting $\Delta G^\circ$ to the Equilibrium Constant ($K_{eq}$)
For a chemical reaction taking place in a solution or gas phase: $aA + bB \rightleftharpoons cC + dD$
The Gibbs free energy of the reaction mixture at any given moment is given by the reaction isotherm equation:
Where:
- $\Delta G$ is the free energy change at given conditions.
- $\Delta G^\circ$ is the Standard free energy change (all reactants/products at $1 \text{ bar}$ or $1 \text{ M}$).
- $R$ is the universal gas constant ($8.314 \text{ J K}^{-1} \text{ mol}^{-1}$).
- $T$ is absolute temperature in Kelvin.
- $Q$ is the Reaction Quotient (ratio of product activities to reactant activities at that moment).
The Magic of Equilibrium:
When the reaction finally reaches chemical equilibrium, two critical things happen:
- The system is at a minimum energy state, so macroscopic driving force ceases: $\Delta G = 0$.
- The reaction quotient $Q$ becomes exactly equal to the thermodynamic Equilibrium Constant: $Q = K_{eq}$.
Substituting these two facts into the isotherm equation yields the most important equation in chemical thermodynamics:
(Or in Base 10 log: $\Delta G^\circ = -2.303 RT \log_{10} K_{eq}$)
This equation proves that the standard free energy ($\Delta G^\circ$) is not the criterion for spontaneity (that's $\Delta G$). Instead, $\Delta G^\circ$ dictates how far a reaction will proceed before it reaches equilibrium (the magnitude of $K_{eq}$).
6. Extensive Solved Problems (JEE / NEET Level)
Calculating Equilibrium Constant from Free Energy
Question:
Calculate the equilibrium constant ($K_p$) for the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ at $298 \text{ K}$. Given that the standard free energies of formation ($\Delta G_f^\circ$) for $NH_3(g)$, $N_2(g)$, and $H_2(g)$ are $-16.4 \text{ kJ/mol}$, $0$, and $0$ respectively. (Use $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$).
Solution:
Step 1: Calculate the Standard Gibbs Free Energy of Reaction ($\Delta G_{rxn}^\circ$).
Step 2: Relate $\Delta G^\circ$ to the Equilibrium Constant.
Step 3: Substitute the values and solve for $\log_{10} K_p$.
Step 4: Take the antilog to find $K_p$.
Answer: The equilibrium constant $K_p$ is $5.62 \times 10^5$. Because $K_p$ is massive, the formation of ammonia is highly favored at equilibrium at 298 K.
Predicting Spontaneity using the Reaction Quotient
Question:
For a reaction $A + B \rightleftharpoons C$, the standard free energy change $\Delta G^\circ$ is $+5.0 \text{ kJ/mol}$ at $300 \text{ K}$. In a particular mixture, the partial pressures are $P_A = 2 \text{ atm}$, $P_B = 2 \text{ atm}$, and $P_C = 0.1 \text{ atm}$. Calculate $\Delta G$ for this mixture. Is the reaction spontaneous in the forward direction under these specific conditions?
Solution:
Step 1: Calculate the Reaction Quotient ($Q_p$).
Step 2: Use the reaction isotherm equation.
Step 3: Substitute values (ensure units are consistent in Joules).
- $\Delta G^\circ = +5000 \text{ J/mol}$
- $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$
- $T = 300 \text{ K}$
- $\ln(0.025) \approx -3.689$
Answer: $\Delta G = -4.2 \text{ kJ/mol}$.
Analysis: Even though the standard state reaction is non-spontaneous ($\Delta G^\circ > 0$), the actual reaction mixture has so few products and so many reactants that the actual free energy change $\Delta G$ is negative. Therefore, the reaction IS spontaneous in the forward direction under these specific conditions until equilibrium is reached.
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