Introduction
Equilibrium is a fundamental concept in chemistry that describes the state at which the concentrations of reactants and products remain constant over time. This topic is crucial for JEE Advanced Chemistry, as it helps in understanding various chemical processes, reactions, and the principles governing them. This study note will break down the concept of equilibrium into smaller, digestible sections and elaborate on each part with detailed explanations, examples, and tips.
Chemical Equilibrium
Dynamic Nature of Equilibrium
Chemical equilibrium is a dynamic state where the rate of the forward reaction equals the rate of the backward reaction. This implies that even though the concentrations of reactants and products remain constant, the reactions continue to occur at the molecular level.
Consider the reversible reaction: $$ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) $$
At equilibrium, the rate of formation of $\text{NH}_3$ is equal to the rate of its decomposition back into $\text{N}_2$ and $\text{H}_2$.
Law of Mass Action
The Law of Mass Action states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants, each raised to a power equal to the stoichiometric coefficient in the balanced chemical equation.
For a general reaction: $$ aA + bB \rightleftharpoons cC + dD $$
The equilibrium constant expression ($K_c$) is given by: $$ K_c = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} $$
Equilibrium constants are dimensionless and depend only on temperature.
Types of Equilibrium Constants
- $K_c$ (Concentration-based equilibrium constant): Used when concentrations of reactants and products are given in mol/L.
- $K_p$ (Pressure-based equilibrium constant): Used for gaseous reactions, where partial pressures are considered.
- $K_w$ (Ion-product of water): Specific to the self-ionization of water.
The relationship between $K_c$ and $K_p$ for a gaseous reaction is given by: $$ K_p = K_c (RT)^{\Delta n} $$ where $\Delta n$ is the difference in the number of moles of gaseous products and reactants, $R$ is the universal gas constant, and $T$ is the temperature in Kelvin.
Le Chatelier's Principle
Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
Changes in Concentration
- Addition of Reactants/Products: Shifts the equilibrium to consume the added species.
- Removal of Reactants/Products: Shifts the equilibrium to produce more of the removed species.
For the reaction: $$ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) $$
Increasing $\text{N}_2$ or $\text{H}_2$ shifts the equilibrium to the right, producing more $\text{NH}_3$.
Changes in Pressure
- Increase in Pressure: Shifts the equilibrium towards the side with fewer moles of gas.
- Decrease in Pressure: Shifts the equilibrium towards the side with more moles of gas.
For the reaction: $$ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) $$
Increasing pressure shifts the equilibrium to the right, as there are fewer moles of gas on the product side.
Changes in Temperature
- Endothermic Reactions: Increase in temperature shifts the equilibrium to the right (favoring products).
- Exothermic Reactions: Increase in temperature shifts the equilibrium to the left (favoring reactants).
For an exothermic reaction: $$ \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D} + \text{heat} $$
Increasing temperature shifts the equilibrium to the left.
Remember that catalysts do not affect the position of equilibrium; they only speed up the rate at which equilibrium is achieved.
Solubility Equilibrium
Solubility Product Constant ($K_{sp}$)
The solubility product constant ($K_{sp}$) is used to describe the equilibrium between a solid and its ions in a saturated solution.
For a salt $AB$ that dissociates as: $$ AB(s) \rightleftharpoons A^+(aq) + B^-(aq) $$
The $K_{sp}$ is given by: $$ K_{sp} = [A^+][B^-] $$
For $\text{AgCl}$: $$ \text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) $$
$$ K_{sp} = [\text{Ag}^+][\text{Cl}^-] $$
Common Ion Effect
The common ion effect states that the solubility of a salt decreases in the presence of a common ion.
Adding $\text{NaCl}$ to a saturated solution of $\text{AgCl}$ will decrease the solubility of $\text{AgCl}$ due to the common $\text{Cl}^-$ ion.
Assuming that the solubility of a salt is unaffected by the presence of a common ion.
Acid-Base Equilibrium
Definitions
- Acids: Proton donors.
- Bases: Proton acceptors.
$K_a$ and $K_b$
- $K_a$ (Acid dissociation constant): Measures the strength of an acid in solution.
- $K_b$ (Base dissociation constant): Measures the strength of a base in solution.
For a weak acid $HA$: $$ HA \rightleftharpoons H^+ + A^- $$
$$ K_a = \frac{[H^+][A^-]}{[HA]} $$
For a weak base $BOH$: $$ BOH \rightleftharpoons B^+ + OH^- $$
$$ K_b = \frac{[B^+][OH^-]}{[BOH]} $$
Relationship Between $K_a$ and $K_b$
For a conjugate acid-base pair: $$ K_a \times K_b = K_w $$
where $K_w$ is the ion-product of water ($1.0 \times 10^{-14}$ at 25°C).
Use the relationship between $K_a$ and $K_b$ to find the dissociation constants of conjugate pairs.
Conclusion
Understanding equilibrium is essential for mastering JEE Advanced Chemistry. By breaking down the concepts into smaller sections and using examples, tips, and notes, one can grasp the nuances of equilibrium and apply them effectively in problem-solving. Remember to practice various problems to solidify your understanding and prepare thoroughly for the exam.
