Chemical Equilibrium | General Chemistry 2

Chemical equilibrium:

Chemical equilibrium is a state in which the concentrations of reactants and products remain constant over time because the rates of the forward and reverse reactions are equal. In this dynamic state, reactants continuously convert into products and vice versa, but these processes occur at the same rate, resulting in no net change in concentrations. This balance is reached when the system stabilizes, with both reactions ongoing but perfectly balanced.

⇒ [reactant]_eq = constant; [product]_eq = constant

Dynamic nature of equilibrium:

Chemical equilibrium is a dynamic equilibrium: At equilibrium, the forward and reverse reactions occur at the same rate, leading to no net change in the concentrations of the involved species.

⇒ rate of forward reaction = rate of reverse reaction
 

N₂​O₄ ​(g) $\rightleftharpoons$ 2 NO₂ ​(g)

• Initially, when N₂O₄ is introduced into a container, it begins to decompose into NO₂, increasing the concentration of NO₂ and changing the color of the gas mixture from colorless to brown.
• Over time, the system reaches equilibrium where the rate of N₂O₄ decomposition equals the rate of NO₂ recombination. At this point, the concentrations of N₂O₄ and NO₂ remain constant, reflecting the established equilibrium.

Reversibility and equilibrium:

Chemical equilibrium is not only dynamic but also reversible. A chemical equilibrium can be attained from any direction:

• Starting with all reactants: The system will shift toward producing products until equilibrium is reached.
• Starting with all products: The system will shift toward producing reactants until equilibrium is reached.
• Starting with a mixture of reactants and products: The system will adjust until the equilibrium state is achieved.

Regardless of the initial concentrations, the system will always reach the same equilibrium state at a given temperature, where the ratio of the concentration of products to reactants remains constant. This ratio is expressed as the equilibrium constant, K, which is unique for each reaction at a specific temperature.

Equilibrium equation:

The equilibrium constant K quantifies the ratio of the concentrations of products to reactants at equilibrium for a chemical reaction. This ratio is derived from the law of mass action, which states that, at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. The equilibrium constant expression for a general reaction a A + b B $\rightleftharpoons$ c C + d D is:
 

The concentrations in the equilibrium-constant expression are considered to be concentration ratios, where the molarity of each substance is divided by its molarity (1 M) in the thermodynamic standard state. Because the units cancel out, the concentration ratios and the values of K_c are dimensionless.

Important points:

• Temperature dependence: The value of K remains constant only at a specific temperature. Any change in temperature will affect the value of the equilibrium constant.

• Stoichiometric coefficients: In the equilibrium expression, the concentration of each species is raised to the power corresponding to its coefficient in the balanced chemical equation.

• Pure solids and liquids: Pure solids and pure liquids do not appear in the equilibrium-constant expression as their activity is defined as 1 by convention.
   

For the reaction: 2 H₂ (g) + CO (g) $\rightleftharpoons$ CH₃OH (g)
The equilibrium constant expression is: K_C = $\frac{\left[{\mathrm{CH}}_3\mathrm{OH}\right]}{{\left[{\mathrm{H}}_2\right]}^2 \left[\mathrm{CO}\right]}$

For the reaction: PCl₃ (l) + Cl₂ (g) $\rightleftharpoons$ PCl₅ (s)
The equilibrium constant expression is: K_C = $\frac{1}{\left[{\mathrm{Cl}}_2\right]}$   
(By convention [PCl₃] = 1 and [PCl₅] = 1; pure liquid and pure solid)

Relationship with chemical kinetics:

Let's consider the general, reversible reaction: A + B $\rightleftharpoons$ C + D and assume that the forward and reverse reactions are elementary reactions. We can write the following rate laws:
 

Equilibrium constant K_p:

In gaseous equilibria, the equilibrium constant can be expressed in terms of either the concentrations of the gases (using K_c) or the partial pressures of the gases (using K_p​). For a general gas-phase reaction: a A + b B ⇌ c C + d D, K_p is given by:
 

For the reaction: N₂O₄ (g) $\rightleftharpoons$ 2 NO₂ (g)
The equilibrium constant in terms of partial pressures is: K_p = $\frac{{\left({\mathrm{P}}_{{\mathrm{NO}}_2}\right)}^2}{{\mathrm{P}}_{{\mathrm{N}}_2{\mathrm{O}}_4}}$

Relationship between K_c and K_p:

The relationship between K_c and K_p is derived from the ideal gas law, PV = nRT. For component A, for example:
 

Therefore, the relationship between K_p and K_c is: 
 

For the reaction: C (s) + CO₂ (g) $\rightleftharpoons$ 2 CO (g)

K_c = $\frac{{\left[\mathrm{CO}\right]}^2}{\left[{\mathrm{CO}}_2\right]}$

K_p = $\frac{{{\mathrm{P}}_{\mathrm{CO}}}^2}{{\mathrm{P}}_{{\mathrm{CO}}_2}}$ = $\frac{{\left[\mathrm{CO}\right]}^2}{\left[{\mathrm{CO}}_2\right]}$ RT = K_c RT

Heterogeneous equilibria:

In a chemical equilibrium involving more than one phase, such as solids, liquids, and gases, the equilibrium is described as heterogeneous. Unlike homogeneous equilibria, where all reactants and products are in the same phase, heterogeneous equilibria involve substances in different states.

Key characteristics of the equilibrium expression:

• Pure solids and liquids exclusion: In heterogeneous equilibria, the concentrations of pure solids and pure liquids do not appear in the equilibrium expression because their concentrations remain constant throughout the reaction. These substances are assigned an activity of 1, which simplifies the equilibrium expression.
   

Consider the decomposition of calcium carbonate: CaCO₃ (s) $\rightleftharpoons$ CaO (s) + CO₂ (g)
The equilibrium expression for this reaction is: K_c = [CO₂]
Here, both CaCO₃ and CaO, being pure solids, are omitted from the expression.

• Multiple phases: Heterogeneous equilibria can involve solids, liquids, and gases. The key is to include only the concentrations of the species that are in the gas or aqueous phase in the equilibrium expression.
   

For the reaction: PbCl₂ (s) $\rightleftharpoons$ Pb²⁺ (aq) + 2 Cl⁻ (aq)
The equilibrium expression is: K_c = [Pb²⁺][Cl⁻]^2​

Manipulation of equilibrium constants:

• If a reaction is reversed, the equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant:
   

• If the entire reaction is multiplied by a factor n, the equilibrium constant for the new reaction is the original equilibrium constant raised to the power of n:
   

• If two reactions are added together, the equilibrium constant for the overall reaction is the product of the equilibrium constants for the individual reactions:
   

Judging the extent of a reaction:

The magnitude of the equilibrium constant K (whether K_c or K_p​) provides insight into the extent of a reaction at equilibrium:

• Large K (> 10³): Indicates that the equilibrium position is far to the right, favoring the formation of products. Most of the reactants are converted into products at equilibrium.
• Small K (< 10^-3): Indicates that the equilibrium position is far to the left, favoring the reactants. Very little product is formed at equilibrium.
• Intermediate K (10^-3 < K < 10³): Suggests that both reactants and products are present in significant amounts at equilibrium, indicating a balanced reaction.

Predicting the direction of a reaction:

To predict the direction in which a reaction will proceed to reach equilibrium, the reaction quotient Q is used and compared to the equilibrium constant K.

The reaction quotient Q represents the ratio of the concentrations (or partial pressures) of products to reactants at any point in time during the reaction. It is calculated using the same formula as K, but with non-equilibrium concentrations. For the general reaction: a A + b B  $\rightleftharpoons$ c C + d D, the reaction quotient is given by:
 

For the reaction: CO₂ (g) + H₂ (g) $\rightleftharpoons$ CO (g) + H₂O (g)

The reaction quotient at time t is: Q_C = $\frac{{\left[\mathrm{CO}\right]}_{\mathrm{t}} {\left[{\mathrm{H}}_2\mathrm{O}\right]}_{\mathrm{t}}}{{\left[{\mathrm{CO}}_2\right]}_{\mathrm{t}} {\left[{\mathrm{H}}_2\right]}_{\mathrm{t}}}$

Comparing Q and K:

• If Q < K: The reaction will proceed in the forward direction, converting reactants into products until equilibrium is reached.
• If Q > K: The reaction will proceed in the reverse direction, converting products back into reactants until equilibrium is reached.
• If Q = K: The system is at equilibrium, and no shift will occur.

How to calculate equilibrium concentrations:

To find the equilibrium concentrations of reactants and products, a systematic approach is used:

1. Write the balanced equation.
2. Set up an ICE (initial, change, equilibrium) table:
   - Initial: Record the initial concentrations (or partial pressures) of the reactants and products.
   - Change: Represent the change in concentrations as the system moves toward equilibrium using a variable x.
   - Equilibrium: Express the equilibrium concentrations in terms of the initial values and the change.
3. Write the equilibrium expression.
4. Substitute known values and solve for x.
5. Calculate the equilibrium concentrations from the calculated value of x.
6. Check the results by substituting them into the equilibrium equation..

Le Châtelier’s principle:

Le Châtelier’s principle states that if a stess is applied to a system at equilibrium, the system will respond by shifting in the direction that minimizes the effect of the stress.

This principle can be applied to understand how the following changes affect an equilibrium system:

• Addition of a reactant or product
• Removal of a reactant or product
• Change in volume (resulting in a change in concentration or partial pressure)
• Change in temperature

Summary of shifts:

Consider the equilibrium: A (g) + B (g) $\rightleftharpoons$ C (g). According to Le Chatelier’ principle, the position of equilibrium will move in such a way as to counteract any disturbance:
 

Effect of a catalyst:

Adding a catalyst increases the rate at which equilibrium is achieved by speeding up both the forward and reverse reactions equally. However, it does not shift the position of the equilibrium; the equilibrium constant K remains unchanged.

Addition of a reactant or product:

• Adding a reactant: the equilibrium shifts to the right (toward the products) to consume the added reactant and produce more product.
• Adding a product: the the equilibrium shifts to the left (toward the reactants) to reduce the concentration of the added product.
   

For the equilibrium: N₂ (g) + 3 H₂ (g) $\rightleftharpoons$ 2 NH₃ (g)

• Adding N₂ or H₂ will shift the equilibrium to the right, producing more NH₃.
• Adding NH₃ will shift the equilibrium to the left, increasing the concentrations of N₂ and H₂.

Removal of a reactant or product:

• Removing a reactant: the equilibrium shifts to the left (toward the reactants) to produce more of the removed substance and reduce the concentration of the products.
• Removing a product: the equilibrium shifts to the right (toward the products) to produce more of the removed product.
   

For the equilibrium: N₂ (g) + 3 H₂ (g) $\rightleftharpoons$ 2 NH₃ (g)

• Removing N₂ or H₂ will shift the equilibrium to the left, reducing the concentration of NH₃.
• Removing NH₃ will shift the equilibrium to the right, increasing the concentration of NH₃.

Effect of pressure changes:

Pressure changes significantly affect equilibrium when gases are involved. The response of the system depends on the number of moles of gas on each side of the balanced equation.

• Increasing pressure: the equilibrium will shift toward the side with fewer moles of gas. This shift reduces the total number of gas molecules, thereby lowering the pressure.
• Decreasing pressure: the equilibrium will shift toward the side with more moles of gas. This increases the total number of gas molecules, which raises the pressure.
   

For the equilibrium: N₂ (g) + 3 H₂ (g) ⇌ 2 NH₃ (g)

Increasing the pressure will shift the equilibrium to the right, favoring the production of NH₃​, which has fewer moles of gas (2 moles) compared to the reactants (4 moles).

Effect of volume changes:

Volume changes are closely related to pressure changes. According to Boyle’s Law (P ∝ $\frac{1}{V}$), pressure and volume are inversely related, so a change in volume will have the opposite effect on pressure, and consequently, on the position of equilibrium.

• Decreasing volume: the pressure increases, causing the equilibrium to shift toward the side with fewer moles of gas.
• Increasing volume: the pressure decreases, causing the equilibrium to shift toward the side with more moles of gas.

Equal moles of gas on both sides:

If the number of moles of gas is the same on both sides of the equation, changes in pressure or volume will have no effect on the position of equilibrium.
 

In the reaction: H₂ (g) + I₂ (g) $\rightleftharpoons$ 2 HI (g)

Both sides have the same number of moles of gas (2 moles), so changes in pressure or volume will not shift the equilibrium.

Exothermic reactions:

In an exothermic reaction, heat is released as a product. The equilibrium can be viewed as: Reactants $\rightleftharpoons$ Products + Heat.

• Increasing temperature: the equilibrium will shift to the left (toward the reactants) to absorb the excess heat and reduce the temperature.
• Decreasing temperature: the equilibrium will shift to the right (toward the products) to release more heat and increase the temperature.
   

For the exothermic reaction: N₂ (g) + 3 H₂ (g) ⇌ 2 NH₃ (g) (ΔH < 0),

The equilibrium can be viewed as: N₂ (g) + 3 H₂ (g) ⇌ 2 NH₃ (g) + Heat. Increasing the temperature will shift the equilibrium to the left, reducing the amount of NH₃​ produced.

Endothermic reactions:

In an endothermic reaction, heat is absorbed as a reactant. The equilibrium can be viewed as: Reactants + Heat $\rightleftharpoons$ Products.

• Increasing temperature: the equilibrium will shift to the right (toward the products) to consume the added heat.
• Decreasing temperature: the equilibrium will shift to the left (toward the reactants) to reduce heat consumption.