Problem 106 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant for the reaction \(\mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)is \(\mathrm{K}_{1}\) for reaction \(\mathrm{H}_{2} \mathrm{PO}_{4} \longrightarrow \mathrm{H}^{+}+\mathrm{HPO}_{4}^{2-}\) is \(\mathrm{K}_{2}\) and for reaction \(\mathrm{HPO}_{4}^{2-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-}\) is \(\mathrm{K}_{3}\). The equilibrium constant (K) for \(\mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-}\) will be (a) \(\mathrm{K}_{1}, \times \mathrm{K}_{2} \times \mathrm{K}_{3}\) (b) \(\mathrm{K}_{1} / \mathrm{K}_{2} \mathrm{~K}_{3}\) (c) \(\mathrm{K}_{2} / \mathrm{K}_{1} \mathrm{~K}_{3}\) (d) \(K_{1},+K_{2}+K_{3}\)

Step-by-Step Solution

Answer

The equilibrium constant is \( \mathrm{K}_{1} \times \mathrm{K}_{2} \times \mathrm{K}_{3} \) (option a).

Step 1: Identify the Reactions

List the given reactions and their equilibrium constants: 1. \( \mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) with \( \mathrm{K}_{1} \).2. \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HPO}_{4}^{2-} \) with \( \mathrm{K}_{2} \).3. \( \mathrm{HPO}_{4}^{2-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-} \) with \( \mathrm{K}_{3} \).

Step 2: Combine the Reactions

To find the equilibrium constant for the reaction \( \mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-} \),you need to combine the given reactions: - The overall reaction is the sum of all steps: 1. \( \mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow \mathrm{H}^{+} + \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) 2. \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightarrow \mathrm{H}^{+} + \mathrm{HPO}_{4}^{2-} \) 3. \( \mathrm{HPO}_{4}^{2-} \rightarrow \mathrm{H}^{+} + \mathrm{PO}_{4}^{3-} \).

Step 3: Find the Equilibrium Constant of the Combined Reaction

The equilibrium constant for a reaction that is the sum of several sequential reactions is the product of the equilibrium constants of those reactions.- Multiply the equilibrium constants: \[ \mathrm{K} = \mathrm{K}_{1} \times \mathrm{K}_{2} \times \mathrm{K}_{3} \].

Step 4: Select the Correct Option

The corresponding option for the equilibrium constant \( \mathrm{K} = \mathrm{K}_{1} \times \mathrm{K}_{2} \times \mathrm{K}_{3} \) is (a).

Key Concepts

Acid-Base Equilibrium

Acid-base equilibrium is a fascinating topic that involves the balance between acids and bases in a chemical system. These systems often involve weak acids and bases that partly dissociate into ions in solution. Let's take phosphoric acid (\(\text{H}_3\text{PO}_4\)) as an example. It dissociates in stages: initially releasing one hydrogen ion (\(\text{H}^+\)) to form dihydrogen phosphate (\(\text{H}_2\text{PO}_4^-\)), and finally forming phosphate (\(\text{PO}_4^{3-}\)) through subsequent dissociations.
This learnable process follows specific equilibrium constants (\(K_1, K_2, \text{ and } K_3\)). These constants measure the extent of dissociation and change with different conditions, such as temperature. Understanding acid-base equilibrium is pivotal since it explains why certain solutions are acidic or alkaline and helps in calculating pH levels. It's the heart of many chemical reactions, from industrial processes to biological systems.

Chemical Equilibrium

Chemical equilibrium occurs when the forward and reverse reactions in a chemical process happen at the same rate, leading to a constant concentration of products and reactants. This state doesn't mean the quantities of substances are equal, but they remain steady.

Thermodynamics

Thermodynamics delves into the energy dynamics of reactions, particularly in cases like chemical equilibria. The first law of thermodynamics, also known as the law of energy conservation, tells us that the energy in the universe remains constant. However, it can transition between forms, like from heat to work in a chemical reaction.
In equilibrium, thermodynamics helps understand how energy exchanges affect equilibrium positions. A reaction such as the dissociation of phosphoric acid involves changes in enthalpy and entropy, critical thermodynamic properties that influence whether reactions are endothermic (absorbing energy) or exothermic (releasing energy).

• Enthalpy change (\(\Delta H\)) refers to the heat change at constant pressure.
• Entropy change (\(\Delta S\)) measures disorder or randomness.
• The Gibbs free energy change (\(\Delta G = \Delta H - T\Delta S\)) determines the spontaneity of a reaction.

Thermodynamics aids in predicting how factors such as temperature play into equilibrium constants. Higher temperatures might increase endothermic reactions but decrease exothermic ones, thus shifting the equilibrium position.