11.1 Nature of Dynamic Equilibrium & Equilibrium Expressions (Kc, Kp)

11.1 Nature of Dynamic Equilibrium & Equilibrium Expressions (Kc, Kp)

Quick Summary: Dynamic chemical equilibrium is established in a closed system when the rates of the forward and reverse reactions become equal, maintaining constant macroscopic concentrations while microscopic molecular transformations continue. The Law of Mass Action defines the equilibrium constant, formulated as $K_c$ for molarities or $K_p$ for gas partial pressures, linked by $K_p = K_c(RT)^{\Delta n_{\text{gas}}}$. Pure solids and liquids exhibit constant thermodynamic activities of unity and are excluded from equilibrium expressions. Modifying chemical equations algebraically inverts, exponentiates, or multiplies $K$.

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1. The Nature of Dynamic Equilibrium

In a closed system, reversible chemical reactions proceed in both directions. As reactants form products, product accumulation accelerates the reverse reaction. Eventually, the system achieves dynamic chemical equilibrium, defined by the equality of opposing rates: $\text{Rate}_{\text{forward}} = \text{Rate}_{\text{reverse}} > 0$

Dynamic equilibrium exhibits four fundamental characteristics:

1. Macroscopic Constancy: Measurable bulk properties—concentrations, partial pressures, color, and temperature—remain constant over time.
2. Microscopic Dynamism: Molecular collisions and chemical interconversions continue incessantly at identical forward and reverse rates.
3. Equal Rates vs. Unequal Concentrations: Equilibrium requires equal reaction rates, not equal reactant and product concentrations. The equilibrium composition depends entirely on thermodynamic stability.
4. Reversibility of Approach: Equilibrium can be reached starting from pure reactants, pure products, or any mixture of both.

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2. The Law of Mass Action & Formulation of $K_c$

Formulated by Cato Guldberg and Peter Waage (1864), the Law of Mass Action states that for a reversible reaction at constant temperature, the ratio of product concentrations to reactant concentrations raised to their stoichiometric coefficients equals a constant $K$. For a general reaction: $a\,\text{A} + b\,\text{B} \rightleftharpoons c\,\text{C} + d\,\text{D}$

The concentration equilibrium constant ($K_c$) is: $K_c = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b}$

Thermodynamically, brackets represent dimensionless activities ($a_i = [i] / c^\circ$, where standard concentration $c^\circ = 1.0\,\text{M}$). Thus, equilibrium constants are formally dimensionless, though subscript $c$ indicates molarity values.

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3. Gas-Phase Equilibria: $K_p$ and Its Relation to $K_c$

Gas-phase equilibria are frequently expressed using partial pressures in atmospheres: $K_p = \frac{(P_{\text{C}})^c (P_{\text{D}})^d}{(P_{\text{A}})^a (P_{\text{B}})^b}$

By the Ideal Gas Law, partial pressure relates to molarity: $P_i = (n_i / V)RT = [i]RT$. Substituting into $K_p$ derives the fundamental conversion: $K_p = K_c (RT)^{\Delta n_{\text{gas}}}$

• $R = 0.08206\,\text{L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$
• $T$ is temperature in Kelvin ($\text{K}$)
• $\Delta n_{\text{gas}} = \sum n_{\text{gas, products}} - \sum n_{\text{gas, reactants}} = (c + d) - (a + b)$

When $\Delta n_{\text{gas}} = 0$ (equal moles of gas on both sides), $(RT)^0 = 1$, and $K_p = K_c$.

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4. Heterogeneous Equilibria: Pure Solids and Liquids

A heterogeneous equilibrium involves species in two or more physical phases. Pure solids ($s$) and pure liquids ($l$) have fixed densities and molar masses. Consequently, their molar concentrations ($n/V = \text{density}/\text{molar mass}$) are invariant, giving them constant thermodynamic activities of unity ($a = 1.0$). Therefore, pure solids and pure liquids are strictly excluded from equilibrium expressions.

For the thermal decomposition of limestone: $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ $K_c = [\text{CO}_2], \quad K_p = P_{\text{CO}_2}$

Similarly, liquid solvent water ($\text{H}_2\text{O}(l)$) in dilute aqueous equilibria is omitted from $K_c$.

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5. Mathematical Rules for Manipulating $K$

Modifying balanced equations changes $K$ according to specific algebraic rules:

1. Reversing an Equation: Inverting direction inverts the equilibrium constant: $K_{\text{reverse}} = \frac{1}{K_{\text{forward}}} = (K_{\text{forward}})^{-1}$
2. Multiplying Coefficients by $n$: Raising coefficients to a factor $n$ raises $K$ to the $n$-th power: $K' = (K)^n \quad (\text{e.g., dividing by 2 yields } K' = \sqrt{K})$
3. Adding Chemical Equations: When sequential reactions are summed, their equilibrium constants multiply: $K_{\text{overall}} = K_1 \times K_2 \times K_3 \times \dots$

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6. Magnitude and Significance of $K$

The magnitude of $K$ indicates the extent of reaction at equilibrium:

• $K \gg 1$ ($K > 10^3$): Product-favored; forward reaction goes virtually to completion.
• $K \ll 1$ ($K < 10^{-3}$): Reactant-favored; very little product forms.
• $K \approx 1$ ($10^{-3} \le K \le 10^3$): Comparable amounts of reactants and products coexist. Note: $K$ reflects thermodynamic stability, revealing nothing about reaction rate.

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7. Equilibrium Expression Rules & Manipulation Summary

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8. Worked Problem: $K_p$ to $K_c$ Conversion

Problem: At $500.0\,\text{K}$, $K_p = 1.45 \times 10^{-5}$ for $\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)$. Calculate $K_c$ ($R = 0.08206\,\text{L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$).

1. Calculate $\Delta n_{\text{gas}}$: $\Delta n_{\text{gas}} = 2 - (1 + 3) = -2$
2. Rearrange for $K_c$: $K_p = K_c (RT)^{\Delta n_{\text{gas}}} \implies K_c = K_p (RT)^{-\Delta n_{\text{gas}}} = K_p (RT)^2$
3. Compute: $RT = 0.08206 \times 500.0 = 41.03$ $K_c = (1.45 \times 10^{-5}) \times (41.03)^2 = 2.44 \times 10^{-2}$