Comprehensive Study Notes on Dynamic Chemical Equilibrium

Fundamental Definition of Chemical Equilibrium

Chemical equilibrium is defined as a specific type of dynamic equilibrium. This state occurs within a chemical system when the speed of the forward reaction, denoted as $R_f$, becomes exactly equal to the speed of the backward reaction, denoted as $R_b$. At this point of equilibrium, the macro-properties of the system remain constant over time; specifically, the concentration of the products and the concentration of the reactants do not change. For a system containing reactants $A + B$ and products $C + D$, this is expressed as maintaining a constant value for $[C+D]$ and $[A+B]$. This constancy applies to partial pressures as well in gaseous systems.

The Law of Mass Action and Active Mass

The Law of Mass Action dictates the rate at which a chemical reaction proceeds. It is fundamentally based on the concept of Active Mass. In the context of these reactions, active mass is defined as the molarity (concentration measured in moles per unit volume) of the substances that participate in the reaction. For a general reversible reaction $A + B \rightleftharpoons C + D$, the Rate of Forward Reaction ($R_f$ or $ROF$) is directly proportional to the product of the active masses of the reactants: $R_f \propto [A][B]$. Similarly, the Rate of Backward Reaction ($R_b$ or $ROB$) is proportional to the active masses of the products: $R_b \propto [C][D]$.

Characteristics of Chemical Equilibrium

There are several critical characteristics that define the state of chemical equilibrium. First, at equilibrium, the Rate of Forward reaction ($ROF$) must equal the Rate of Backward reaction ($ROB$). Second, the equilibrium state can be achieved from either direction, whether one starts with purely reactants or purely products. Third, the initial concentrations of the substances involved have no effect on the eventual equilibrium constant value, although they determine the actual amounts present at equilibrium. Fourth, chemical equilibrium can only be achieved and maintained within a closed vessel, particularly for reactions involving gases, to prevent the escape of matter.

The Role of a Catalyst in Equilibrium

A catalyst serves a specific function in a reversible system: it enhances the speeds of both the forward and backward reactions equally. While a catalyst does not change the position of equilibrium or the concentrations of the species at equilibrium, it significantly helps the system reach the equilibrium state much earlier than it would otherwise.

Mathematical Expressions of Equilibrium Constants

The equilibrium constant is a ratio that describes the extent of a reaction at equilibrium. When writing these expressions, the stoichiometric coefficients of the reactants and products from the balanced chemical equation become the powers (exponents) to which their concentrations or pressures are raised.

$K_c$ is the equilibrium constant expressed in terms of molarity, with units typically given as $mol\,dm^{-3}$. For example, in the dissociation of dinitrogen tetroxide $N_2O_4(g) \rightleftharpoons 2NO_2(g)$, the expression is written as $K_c = \frac{[NO_2]^2}{[N_2O_4]}$.

$K_p$ is the equilibrium constant expressed in terms of partial pressures, designated for systems involving gases. For the same reaction, it is expressed as $K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}$.

A crucial convention in these calculations is that the active mass of pure solids ($s$) and pure liquids ($l$) is taken as unity ($1$) when compared to gaseous components in the same system.

Relationship Between $K_p$ and $K_c$

There is a universal relationship that links the equilibrium constant in terms of pressure to the equilibrium constant in terms of concentration. This is expressed by the formula $K_p = K_c(RT)^{\Delta n_g}$. In this equation, $R$ represents the universal gas constant and $T$ represents the temperature in Kelvin. The term $\Delta n_g$ represents the change in the number of moles of gaseous substances, calculated as the total moles of gaseous products minus the total moles of gaseous reactants. For a theoretical reaction $2A(g) + 3B(g) \rightleftharpoons C(g)$, the value would be $\Delta n_g = 1 - (2 + 3) = -4$.

Numerical Application: Moles Remaining at Equilibrium

Consider a scenario where $3\,moles$ of $H_2$ and $3\,moles$ of $I_2$ are placed in a $10\,L$ container at $25^{\circ} C$. The equilibrium constant $K_c$ for the reaction $H_2 + I_2 \rightleftharpoons 2HI$ is $49$. To find the moles of $I_2$ remaining at equilibrium, we set up the following logic:

Initially, the moles are $3$ for $H_2$ and $3$ for $I_2$. Let $x$ be the amount of moles reacted at equilibrium. The final moles will be $(3-x)$ for $H_2$, $(3-x)$ for $I_2$, and $2x$ for $HI$. The concentrations are these mole values divided by the volume ($10\,L$).

$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(2x/10)^2}{(3-x/10)(3-x/10)} = \frac{4x^2}{(3-x)^2}$

Setting the equation to the known constant:

$49 = \frac{(2x)^2}{(3-x)^2}$

Taking the square root of both sides:

$7 = \frac{2x}{3-x}$

Solving for $x$:

$21 - 7x = 2x$

$9x = 21$

$x = 2.33$

The remaining moles of $I_2$ are calculated as $3 - x = 3 - 2.33 = 0.6667\,moles$.

Numerical Application: Calculation of $K_c$ with Percent Dissociation

In a separate example involving the synthesis of ammonia, $1\,mole$ of $N_2$ and $3\,moles$ of $H_2$ are placed in a solution. If $25\%$ of the initial $N_2$ is consumed to produce $NH_3$, we can determine the equilibrium composition. The balanced equation is $N_2 + 3H_2 \rightleftharpoons 2NH_3$.

Initial moles: $N_2 = 1$, $H_2 = 3$, $NH_3 = 0$.

If $25\%$ of $N_2$ is used, the change in moles for $N_2$ is $0.25$. Following the stoichiometry, the change for $H_2$ is $3 \times 0.25 = 0.75$, and the change for $NH_3$ is $2 \times 0.25 = 0.5$.

Equilibrium moles:

$N_2 = 1 - 0.25 = 0.75$

$H_2 = 3 - 0.75 = 2.25$

$NH_3 = 0.5$

The formula for $K_c$ for this reaction is $K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$. Based on these values and the volume $V$, the equilibrium constant can be computed.

Numerical Application: Calculation of $K_p$ from Partial Pressures

For the gaseous reaction $A_2(g) + B_2(g) \rightleftharpoons 2AB(g)$, the equilibrium state contains $0.2\,moles$ of $A_2$, $0.3\,moles$ of $B_2$, and $0.1\,moles$ of $AB$. The total pressure of the system is given as $2\,atm$. To find $K_p$, we first determine the total number of moles:

$n_{total} = 0.2 + 0.3 + 0.1 = 0.6\,moles$

The partial pressure ($P$) of each component is calculated using the mole fraction ($X$) multiplied by the total pressure ($P_T$):

$P_{A_2} = X_{A_2} \times P_T = \frac{0.2}{0.6} \times 2 = \frac{2}{3}\,atm$

$P_{B_2} = X_{B_2} \times P_T = \frac{0.3}{0.6} \times 2 = 1\,atm$

$P_{AB} = X_{AB} \times P_T = \frac{0.1}{0.6} \times 2 = \frac{1}{3}\,atm$

The expression for $K_p$ is:

$K_p = \frac{(P_{AB})^2}{(P_{A_2})(P_{B_2})} = \frac{(1/3)^2}{(2/3)(1)} = \frac{1/9}{2/3} = \frac{1}{6}$

This yields a value of approximately $0.1667$ for the equilibrium constant $K_p$.