Chemical Equilibrium – Chemistry Study Notes

Equilibrium Constants: $K_c$, $K_p$, and $K_x$

When a reversible chemical reaction reaches equilibrium at a constant temperature, the ratio of product concentrations to reactant concentrations remains constant. Each concentration is raised to the power of its respective stoichiometric coefficient. This constant is known as the equilibrium constant.

For a general gaseous reaction expressed as $aA + bB \rightleftharpoons cC + dD$, we can define equilibrium constants based on molar concentration ($K_c$), partial pressures ($K_p$), and mole fractions ($K_x$).

The concentration equilibrium constant, $K_c$, is expressed using molar concentrations in mol/L:

$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

For systems involving gases, expressing equilibrium in terms of partial pressures is often more convenient. The pressure equilibrium constant, $K_p$, replaces molar concentrations with the partial pressure of each gaseous species:

$K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}$

A fundamental thermodynamic derivation bridges these two constants by utilizing the ideal gas law equation, $PV = nRT$. This yields the partial pressure of a gas $P_i = \left(\frac{n_i}{V}\right)RT = C_iRT$.

Substituting this into the $K_p$ expression generates the classic interconversion formula:

$K_p = K_c (RT)^{\Delta n_g}$

Here, $\Delta n_g$ represents the change in the number of moles of gas. It is calculated strictly as $\Delta n_g = (\text{moles of gaseous products}) – (\text{moles of gaseous reactants})$. If $\Delta n_g = 0$, then $K_p = K_c$.

Additionally, the mole fraction equilibrium constant $K_x$ is related to $K_p$ through the total pressure ($P_t$) of the system via the relation $K_p = K_x \cdot (P_t)^{\Delta n_g}$.

Reaction Quotient and Direction of Reaction

While the equilibrium constant $K$ strictly applies when a system has attained dynamic balance, the reaction quotient ($Q$) evaluates the state of a reaction at any arbitrary point in time. It is formulated identically to the equilibrium constant expression using current non-equilibrium concentrations or pressures ($Q_c$ or $Q_p$).

Thus, $Q$ serves as a mathematical compass predicting the spontaneous direction in which the reaction will proceed to achieve equilibrium. By comparing the magnitude of the reaction quotient against the established equilibrium constant, competitive exam aspirants can immediately deduce system behavior.

The criteria for reaction direction are:

• $Q < K$: The forward reaction is favored. The system contains an excess of reactants relative to products, and the reaction shifts spontaneously from left to right until $Q = K$.
• $Q > K$: The backward reaction is favored. The system contains an excess of products relative to reactants, and the reaction shifts spontaneously from right to left to consume products.
• $Q = K$: The system is at dynamic chemical equilibrium. The rates of the forward and reverse processes are precisely equal, and no net concentration changes occur.

Understanding $Q$ is crucial for industrial applications where maintaining optimal yield requires constantly monitoring reactor conditions. Operators must inject or venting specific reagents to ensure $Q$ remains offset from $K$ in the desired product-forming direction.

Degree of Dissociation and Vapour Density Relations

The degree of dissociation ($\alpha$) is defined as the fraction of one mole of a reactant that dissociates into product molecules under specific conditions of temperature and pressure. It ranges from $0$ (no dissociation) to $1$ (complete dissociation, often expressed as a percentage between $0\%$ and $100\%$).

Determining $\alpha$ accurately is a staple of numerical problems in competitive examinations. Consider a general dissociation equilibrium for a binary gas: $A \rightleftharpoons nB$.

If we start with $1$ mole of $A$ and let $\alpha$ denote the degree of dissociation, the equilibrium moles of each species can be mapped using an ICE (Initial, Change, Equilibrium) table:

• Initial moles: $A = 1$, $B = 0$
• Change in moles: $A = -\alpha$, $B = +n\alpha$
• Equilibrium moles: $A = 1 – \alpha$, $B = n\alpha$
• Total moles at equilibrium ($N_{tot}$) = $1 – \alpha + n\alpha = 1 + (n – 1)\alpha$

A classic exam application involves relating the degree of dissociation to vapour density ($V.D.$) or molecular weight. Vapour density is directly proportional to the molar mass of a gas mixture via $M_{mix} = 2 \times V.D.$.

Because total moles increase upon dissociation while total mass remains conserved, the observed molar mass decreases. The experimental or average molar mass ($M_{avg}$ or $M_t$) is related to the theoretical initial molar mass ($M_{initial}$ or $D$) and $\alpha$ through the relation:

$\alpha = \frac{D – d}{d(n – 1)} = \frac{M_{initial} – M_{avg}}{M_{avg}(n – 1)}$

Here, $D$ (or $M_{initial}$) represents the theoretical vapour density of the undissociated gas. Conversely, $d$ (or $M_{avg}$) represents the experimental vapour density of the dissociated equilibrium mixture.

Le Chatelier’s Principle and Temperature Dependence

Le Chatelier’s Principle is a qualitative guideline stating that if a dynamic equilibrium is disturbed by changing conditions, the position of equilibrium shifts to counteract the change and re-establish equilibrium. Disturbances can take the form of concentration adjustments, pressure or volume shifts, or thermal fluctuations.

The influence of these stressors breaks down into clear operational rules:

• Concentration: Adding a reactant shifts the equilibrium toward the products; removing a product pulls the reaction forward to replace it.
• Pressure and Volume: Increasing the pressure (by decreasing the volume) shifts the equilibrium toward the side with fewer moles of gas ($\Delta n_g < 0$). Conversely, expanding the volume favors the side with more moles of gas. Inert gases added at constant volume have no effect on equilibrium, whereas addition at constant pressure increases total volume and shifts equilibrium toward the side with greater $\Delta n_g$.
• Temperature: Temperature is unique because it is the only factor that actually alters the numerical value of the equilibrium constant $K$.

The quantitative temperature dependence of the equilibrium constant is governed by the Van ‘t Hoff equation, which bridges thermodynamics and kinetics:

$\log\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{2.303 R} \left(\frac{T_2 – T_1}{T_1 T_2}\right)$

For an endothermic reaction ($\Delta H^\circ > \mathbf{+ve}$), raising the temperature ($T_2 > T_1$) increases the value of the equilibrium constant ($K_2 > K_1$), favoring the forward reaction.

For an exothermic reaction ($\Delta H^\circ < \mathbf{-ve}$), increasing the temperature decreases $K$, shifting the equilibrium backward and reducing product yield.

Important Facts / Formulas

Previous Year Question Hints

• Hint 1 (Vapour Density & $\alpha$): When given the vapour density of $N_2O_4$ decreasing upon heating due to dissociation into $NO_2$, apply the $\alpha = \frac{D-d}{d(n-1)}$ formula directly. Remember that $n=2$ for $N_2O_4 \rightleftharpoons 2NO_2$, simplifying the denominator to $d(2-1) = d$.
• Hint 2 (Unit Analysis of $K$): Always verify the units of $K_c$ ($(\text{mol L}^{-1})^{\Delta n_g}$) and $K_p$ ($(\text{atm})^{\Delta n_g}$) before solving stoichiometry-heavy numericals. Mismatching pressure units in mixed expressions is a common trap.
• Hint 3 (Temperature Shifts): If a question asks how industrial ammonia synthesis via the Haber process ($N_2 + 3H_2 \rightleftharpoons 2NH_3, \Delta H = -\text{ve}$) responds to cooling, remember that lower temperatures increase $K$ for exothermic reactions. However, kinetic rates must still be balanced using catalysts.

Quick Revision Summary

Chemical equilibrium is dynamic; forward and reverse reaction rates are equal, and macroscopic properties remain constant.
The equilibrium constant ($K_c$ or $K_p$) depends strictly on temperature and remains independent of initial concentrations, catalysts, or the presence of inert species at constant volume.
The reaction quotient ($Q$) determines reaction direction: $Q < K$ (forward), $Q > K$ (backward), and $Q = K$ (equilibrium).
$\Delta n_g$ dictates whether $K_p$ equals $K_c$; when $\Delta n_g = 0$, $K_p = K_c$.
Le Chatelier’s Principle predicts qualitative shifts when stress is applied via concentration, pressure, volume, or temperature changes.
Temperature is the only parameter that alters the magnitude of the equilibrium constant $K$, governed quantitatively by the Van ‘t Hoff equation.
Catalysts speed up the attainment of equilibrium by lowering activation energy equally for both forward and reverse paths, leaving the equilibrium position and $K$ unchanged.
The degree of dissociation ($\alpha$) relates initial theoretical vapour density ($D$) to observed equilibrium vapour density ($d$) via $\alpha = \frac{D-d}{d(n-1)}$.