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Chapter 7: Problem 106

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) \(K_{1},+K_{2}+K_{3}\)

Short Answer

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

Step by step solution

01

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 \( \mathrm{K}_{3} \).
02

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. \( \mathrm{HPO}_{4}^{2-} \rightarrow \mathrm{H}^{+} + \mathrm{PO}_{4}^{3-} \).
03

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: \[ \mathrm{K} = \mathrm{K}_{1} \times \mathrm{K}_{2} \times \mathrm{K}_{3} \].
04

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).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The relation between \(\mathrm{K}_{\mathrm{p}}\) and \(\mathrm{K}_{c}\) for the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g})\) is (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-1}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}\) (c) \(K_{P}-K_{e} /(R T)^{2}\) (d) \(K_{p}=K_{c} / R T\)

Match the following Column-I (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) (b) \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}\) (s) \(+\mathrm{CO}_{2}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g})\) (d) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) Column-II (p) Unaffected by inert gas addition at constant volume (q) Forward shift by rise in pressure (r) Unaffected by increase in pressure (s) Backward shift by rise in pressure (t) reaction has \(\Delta n_{g}>0\)

If for the reaction given below \(2 \mathrm{PQ} \rightleftharpoons{\mathrm{B}} \rightleftharpoons{\mathrm{P}_{2}}+\underset{\mathrm{g}}{\mathrm{Q}_{2}} \mathrm{~K}_{1}=2.5 \times 10^{5}\) \(\mathrm{PQ}+\frac{1}{2} \mathrm{R}_{2} \rightleftharpoons{\mathrm{PQR}} \mathrm{K}_{2}=5 \times 10^{-3}\) find \(\mathrm{K}_{3}\) for the reaction \(\frac{1}{2} \mathrm{P}_{2}+\frac{1}{2} \mathrm{Q}_{2}+\frac{1}{2} \mathrm{R}_{2} \rightleftharpoons \mathrm{PQR}\) (a) \(2.5 \times 10^{-3}\) (b) \(2.5 \times 10^{3}\) (c) \(1 \times 10^{-3}\) (d) \(5 \times 10^{-3}\)

A reversible reaction is said to have attained equilibrium, when (a) backward reaction stops (b) both backward and forward reactions take place at equal speed (c) both backward and forward reactions stop (d) concentration of each of the reactants and products becomes equal

\(\mathrm{K}_{\mathrm{p}}\) has the value of \(10^{-6} \mathrm{~atm}^{3}\) and \(10^{-4} \mathrm{~atm}^{3}\) at \(298 \mathrm{~K}\) and \(323 \mathrm{~K}\) respectively for the reaction \(\mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_{4}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(\Delta_{r} \mathrm{H}^{\circ}\) for the reaction is: (a) \(7.7 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-147.41 \mathrm{~kJ} / \mathrm{mol}\) (c) \(147.41 \mathrm{~kJ} / \mathrm{mol}\) (d) None of these
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