{"id":866386,"date":"2025-08-08T18:08:26","date_gmt":"2025-08-08T12:38:26","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=866386"},"modified":"2025-08-08T18:08:26","modified_gmt":"2025-08-08T12:38:26","slug":"ncert-notes-class-11-chemistry-part-i-chapter-6-equilibrium","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-6-equilibrium\/","title":{"rendered":"NCERT Notes Class 11 Chemistry (Part-I) Chapter 6: Equilibrium (Free PDF)"},"content":{"rendered":"\n<p>NCERT Notes Class 11 Chemistry (Part-1) Chapter 6: Equilibrium introduces the concept of equilibrium in both physical and chemical processes in a more comprehensive way. You will learn how reversible reactions reach a dynamic balance and how to express this using equilibrium constants. The chapter also covers factors affecting equilibrium, such as temperature and pressure, along with Le Chatelier\u2019s Principle, all of which are essential for understanding reactions in chemistry and real-life applications.<\/p>\n\n\n\n\n\n\n<p><strong>Explore Notes of Class 11 Chemistry<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#f1c9d3\"><tbody><tr><td><strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-1-some-basic-concepts-of-chemistry\/\">Chapter 1<\/a><\/strong><\/td><td><strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-2-structure-of-atom\/\">Chapter 2<\/a><\/strong><\/td><td><strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-3-classification-of-elements-and-periodicity-in-properties\/\">Chapter 3<\/a><\/strong><\/td><td><strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-class-11-chemistry-part-i-chapter-iv-chemical-bonding-and-molecular-structure\/\">Chapter 4<\/a><\/strong><\/td><td><strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-5-thermodynamics-free-pdf\/\">Chapter 5<\/a><\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#b5f6de\"><tbody><tr><td><a href=\"https:\/\/drive.google.com\/file\/d\/1YQyr5Lw60YOu6_CphsZuP2DzjgcHhKNp\/view?usp=drive_link\"><strong>Download PDF of NCERT Notes Class 11 Chemistry (Part-I) Chapter-6: Equilibrium<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-importance-of-chemical-equilibrium\">Importance of Chemical Equilibrium<\/h2>\n\n\n\n<p>Chemical equilibria are significant in various biological and environmental processes. A key example is the equilibrium between O\u2082 molecules and hemoglobin, which plays a vital role in the transport and delivery of oxygen from the lungs to the muscles. Similarly, an equilibrium involving carbon monoxide (CO) and hemoglobin is responsible for CO toxicity, as CO interferes with oxygen transport in the body.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-equilibrium-in-physical-processes\">Equilibrium in Physical Processes<\/h2>\n\n\n\n<p>When a liquid evaporates in a closed container, molecules with higher kinetic energy escape from the liquid surface into the vapour phase. At the same time, vapour molecules collide with the surface and return to the liquid phase. Eventually, a stage is reached where the rate of molecules leaving the liquid equals the rate of return from the vapour; this is known as the equilibrium state. At equilibrium: Rate of evaporation = Rate of condensation<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-classification-of-reactions-based-on-the-extent-of-equilibrium\">Classification of Reactions Based on the Extent of Equilibrium<\/h2>\n\n\n\n<p>Reactions can be grouped into three categories based on how far they proceed before equilibrium is established:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Reactions that proceed nearly to completion\n<ul class=\"wp-block-list\">\n<li>Only negligible amounts of reactants remain.<\/li>\n\n\n\n<li>Sometimes, the remaining reactants are undetectable experimentally.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Reactions that form very small amounts of products\n<ul class=\"wp-block-list\">\n<li>Most reactants remain unchanged even at equilibrium.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Reactions where concentrations of reactants and products are comparable\n<ul class=\"wp-block-list\">\n<li>Both are present in significant amounts at equilibrium.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p class=\"has-pale-ocean-gradient-background has-background\"><strong>Also Read:<\/strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-geography-fundamentals-of-physical-geography-chapter-11-world-climate-and-climate-change\/\"><strong>NCERT Solutions Class 11 Geography Fundamentals of Physical Geography Chapter 11: World Climate and Climate Change (Free PDF)<\/strong><\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-factors-affecting-the-extent-of-equilibrium\">Factors Affecting the Extent of Equilibrium<\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The extent to which a reaction proceeds depends on various experimental conditions, such as:\n<ul class=\"wp-block-list\">\n<li>Concentration of reactants<\/li>\n\n\n\n<li>Temperature<\/li>\n\n\n\n<li>Other system-specific factors<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>In industrial and laboratory settings, optimising these operational conditions is essential to:\n<ul class=\"wp-block-list\">\n<li>Shift the equilibrium toward the desired product<\/li>\n\n\n\n<li>Maximise efficiency and yield<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-equilibrium-in-physical-processes-0\">Equilibrium in Physical Processes<\/h2>\n\n\n\n<p>Physical equilibria occur in phase transformations where two states of matter coexist in a closed system under specific conditions of temperature and pressure. These include:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Solid \u21cc Liquid<\/li>\n\n\n\n<li>Liquid \u21cc Gas<\/li>\n\n\n\n<li>Solid \u21cc Gas<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-solid-liquid-equilibrium\">Solid\u2013Liquid Equilibrium<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Example<\/strong>: Ice and water at 273 K in a perfectly insulated thermos flask (no heat exchange with surroundings) are in equilibrium.<\/li>\n\n\n\n<li>Observations:\n<ul class=\"wp-block-list\">\n<li>The mass of ice and water does not change with time.<\/li>\n\n\n\n<li>Temperature remains constant.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>However, the equilibrium is dynamic, not static:\n<ul class=\"wp-block-list\">\n<li>Liquid water molecules collide with ice and adhere to it.<\/li>\n\n\n\n<li>Simultaneously, ice molecules escape into the liquid phase.<\/li>\n\n\n\n<li>Since the rate of melting = rate of freezing, there is no net change in mass.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>This equilibrium occurs only at a specific temperature and pressure.<\/li>\n\n\n\n<li>The temperature at which solid and liquid coexist at atmospheric pressure is called the normal melting point or freezing point.<\/li>\n<\/ul>\n\n\n\n<p><strong>Inferences:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Both opposing processes (melting and freezing) occur simultaneously.<\/li>\n\n\n\n<li>Both occur at the same rate, keeping the mass of each phase constant.<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-liquid-vapour-equilibrium\">Liquid\u2013Vapour Equilibrium<\/h3>\n\n\n\n<p><strong>Experimental setup<\/strong>: A transparent box with a U-tube manometer and a drying agent like anhydrous CaCl\u2082 is used. After placing a watch glass containing water inside the box and removing the drying agent:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The mercury level in the manometer\u2019s right limb increases and becomes constant, indicating a rise in pressure.<\/li>\n\n\n\n<li>Water volume decreases as it evaporates.<\/li>\n\n\n\n<li>Initially, the box contains very little or no water vapour. As water evaporates:\n<ul class=\"wp-block-list\">\n<li>Pressure rises due to the formation of water vapour.<\/li>\n\n\n\n<li>With time, condensation begins, balancing evaporation.<\/li>\n\n\n\n<li>Finally, the rate of evaporation = the rate of condensation:<br>H\u2082O (l) \u21cc H\u2082O (vap)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The constant pressure at equilibrium is called the equilibrium vapour pressure.<\/li>\n\n\n\n<li>Vapour pressure is temperature-dependent and is higher for more volatile liquids.<\/li>\n\n\n\n<li>Comparison:\n<ul class=\"wp-block-list\">\n<li>Different liquids like acetone, methyl alcohol, and ether have different vapor pressures at the same temperature.<\/li>\n\n\n\n<li>A higher vapor pressure implies: Greater volatility and a Lower boiling point.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Observation in open systems:\n<ul class=\"wp-block-list\">\n<li>When liquids are exposed to the atmosphere, evaporation occurs, but condensation is negligible. Equilibrium is not attained due to continuous dispersion of vapour.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Boiling point:\n<ul class=\"wp-block-list\">\n<li>For water at 1.013 bar (atmospheric pressure), the boiling point = 100\u00b0C.<\/li>\n\n\n\n<li>The normal boiling point is defined as the temperature at which liquid and vapour are at equilibrium at 1.013 bar.<\/li>\n\n\n\n<li>Boiling point varies with altitude due to atmospheric pressure differences.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-solid-vapour-equilibrium\">Solid\u2013Vapour Equilibrium<\/h3>\n\n\n\n<p><strong>Example<\/strong>: When solid iodine is placed in a closed container:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Violet vapours fill the vessel.<\/li>\n\n\n\n<li>Intensity increases over time, then becomes constant, indicating equilibrium.<\/li>\n\n\n\n<li>The equilibrium can be represented as:<br>I\u2082 (solid) \u21cc I\u2082 (vapour)<\/li>\n\n\n\n<li>Other examples:\n<ul class=\"wp-block-list\">\n<li>Camphor (solid) \u21cc Camphor (vapour)<\/li>\n\n\n\n<li>NH\u2084Cl (solid) \u21cc NH\u2084Cl (vapour)<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-equilibrium-involving-dissolution-of-solids-or-gases-in-liquids\">Equilibrium Involving Dissolution of Solids or Gases in Liquids<\/h3>\n\n\n\n<p><strong>Solids in Liquids: <\/strong>Dissolving sugar in water.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Only a limited amount of sugar dissolves at room temperature.<\/li>\n\n\n\n<li>On cooling a hot saturated solution, sugar crystallises.<\/li>\n\n\n\n<li>A saturated solution is one where no more solute dissolves at a given temperature.<\/li>\n\n\n\n<li>At this stage, a dynamic equilibrium exists between the solid sugar and sugar in solution.<\/li>\n\n\n\n<li>Sugar (solution) \u21cc Sugar (solid)<\/li>\n\n\n\n<li>Rate of dissolution = Rate of crystallisation<\/li>\n\n\n\n<li>Confirmed using radioactive sugar: When added to a saturated solution of non-radioactive sugar.<\/li>\n<\/ul>\n\n\n\n<p><strong>Gases in Liquids: Soda water bottle<\/strong>.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>When opened, CO\u2082 escapes due to lower pressure outside.<\/li>\n\n\n\n<li>There is an equilibrium between dissolved CO\u2082 and gaseous CO\u2082:<br>CO\u2082 (gas) \u21cc CO\u2082 (in solution)<\/li>\n\n\n\n<li>This is governed by Henry\u2019s Law:\n<ul class=\"wp-block-list\">\n<li>The mass of gas dissolved is proportional to the pressure of gas above the liquid.<\/li>\n\n\n\n<li>Solubility decreases with temperature.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Soda bottles are sealed under pressure to maintain high CO\u2082 solubility.<\/li>\n\n\n\n<li>On opening:\n<ul class=\"wp-block-list\">\n<li>Pressure drops<\/li>\n\n\n\n<li>CO\u2082 escapes<\/li>\n\n\n\n<li>A new equilibrium is established<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Eventually, soda goes flat as gas escapes continuously into the open air.<\/li>\n<\/ul>\n\n\n\n<p class=\"has-pale-ocean-gradient-background has-background\"><strong>Also Read:&nbsp; <\/strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-english-snapshots-chapter-1-the-summer-of-the-beautiful-white-horse\/\"><strong>NCERT Notes Class 11 English Snapshots Chapter 1: The Summer of the Beautiful White Horse (Free PDF)&nbsp;<\/strong><\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-general-characteristics-of-equilibria-involving-physical-processes\">General Characteristics of Equilibria Involving Physical Processes<\/h2>\n\n\n\n<p>All physical equilibria discussed above share these common features:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Equilibrium is possible only in a closed system at a fixed temperature.<\/li>\n\n\n\n<li>Opposing processes occur at equal rates, leading to a dynamic but stable state.<\/li>\n\n\n\n<li>Measurable properties remain constant during equilibrium.<\/li>\n\n\n\n<li>Equilibrium is characterised by constant values of measurable parameters like:\n<ul class=\"wp-block-list\">\n<li>Vapour pressure<\/li>\n\n\n\n<li>Boiling\/melting point<\/li>\n\n\n\n<li>Solubility, etc.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The magnitude of these parameters reflects the extent to which the physical process has occurred before reaching equilibrium.<\/li>\n<\/ol>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-reversible-and-irreversible-reactions\">Reversible and Irreversible Reactions<\/h2>\n\n\n\n<p>A chemical reaction is said to be reversible if the products formed can react again to give back the original reactants. In a closed system, reversible reactions reach chemical equilibrium, where: Rate of forward reaction = Rate of backward reaction. At this point, concentrations of reactants and products remain constant over time. Representation of a general reversible reaction:<br>A + B \u21cc C + D<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-law-of-chemical-equilibrium-and-equilibrium-constant\">Law of Chemical Equilibrium and Equilibrium Constant<\/h2>\n\n\n\n<p>For the general reversible reaction: aA + bB \u21cc cC + dD<br>The equilibrium constant in terms of concentration (Kc) is given by: Kc = [C]\u1d9c [D]\u1d48 \/ [A]\u1d43 [B]\u1d47. Square brackets [ ] represent the molar concentration of the substances at equilibrium. Kc remains constant at a fixed temperature for a particular reaction.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-homogeneous-equilibria\">Homogeneous Equilibria<\/h3>\n\n\n\n<p>In homogeneous equilibria, all the reactants and products are in the same physical phase, typically gaseous or aqueous.<\/p>\n\n\n\n<p><strong>Examples:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>N\u2082(g) + 3H\u2082(g) \u21cc 2NH\u2083(g)<br>Equilibrium constant:<br>Kc = [NH\u2083]\u00b2 \/ [N\u2082][H\u2082]\u00b3<\/li>\n\n\n\n<li>H\u2082(g) + I\u2082(g) \u21cc 2HI(g)<br>Equilibrium constant:<br>Kc = [HI]\u00b2 \/ [H\u2082][I\u2082]<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-heterogeneous-equilibria\">Heterogeneous Equilibria<\/h3>\n\n\n\n<p>In heterogeneous equilibria, reactants and products exist in different phases. The concentrations of pure solids and pure liquids are not included in the expression for the equilibrium constant.<\/p>\n\n\n\n<p>Example: CaCO\u2083(s) \u21cc CaO(s) + CO\u2082(g)<br>Equilibrium constant:<br>Kc = [CO\u2082]<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-applications-of-equilibrium-constants\">Applications of Equilibrium Constants<\/h2>\n\n\n\n<p>Let Qc be the reaction quotient (same form as Kc, but using current concentrations):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If Qc &lt; Kc, the reaction proceeds in the forward direction.<\/li>\n\n\n\n<li>If Qc &gt; Kc, the reaction proceeds in the reverse direction.<\/li>\n\n\n\n<li>If Qc = Kc, the system is at equilibrium.<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-gibbs-free-energy-and-chemical-reactions\">Gibbs Free Energy and Chemical Reactions<\/h2>\n\n\n\n<p>The spontaneity of a chemical reaction is determined by the Gibbs free energy change (\u0394G).<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If \u0394G &lt; 0, the reaction is spontaneous.<\/li>\n\n\n\n<li>If \u0394G &gt; 0, the reaction is non-spontaneous.<\/li>\n\n\n\n<li>If \u0394G = 0, the system is at equilibrium.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-gibbs-energy-and-equilibrium-constant\">Gibbs Energy and Equilibrium Constant<\/h3>\n\n\n\n<p>For a general reaction:<\/p>\n\n\n\n<p>A + B \u21cc C + D, the relation between Gibbs energy and equilibrium constant is:<\/p>\n\n\n\n<p>\u0394G = \u0394G\u00b0 + RT ln Q<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394G = Gibbs energy change at non-equilibrium conditions<\/li>\n\n\n\n<li>\u0394G\u00b0 = Standard Gibbs energy change<\/li>\n\n\n\n<li>R = Gas constant = 8.314 J mol\u207b\u00b9 K\u207b\u00b9<\/li>\n\n\n\n<li>T = Temperature in Kelvin<\/li>\n\n\n\n<li>Q = Reaction quotient<\/li>\n<\/ul>\n\n\n\n<p>At equilibrium, \u0394G = 0 and Q = K, so:<\/p>\n\n\n\n<p>\u0394G\u00b0 = -RT ln K<\/p>\n\n\n\n<p>Thus,<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If K &gt; 1, then ln K is positive, so \u0394G\u00b0 &lt; 0 \u2192 reaction is spontaneous.<\/li>\n\n\n\n<li>If K &lt; 1, then ln K is negative, so \u0394G\u00b0 &gt; 0 \u2192 reaction is non-spontaneous.<\/li>\n\n\n\n<li>If K = 1, then \u0394G\u00b0 = 0 \u2192 system is at equilibrium.<\/li>\n<\/ul>\n\n\n\n<p>This relation connects thermodynamics (\u0394G\u00b0) and equilibrium (K).<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-factors-affecting-equilibria\">Factors Affecting Equilibria<\/h2>\n\n\n\n<p>Once equilibrium is achieved in a chemical system, changes in conditions (such as concentration, pressure, or temperature) can disturb the equilibrium. The system then adjusts itself to re-establish a new equilibrium state. These changes and the system&#8217;s response are governed by Le Chatelier\u2019s Principle.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-le-chatelier-s-principle\">Le Chatelier\u2019s Principle<\/h2>\n\n\n\n<p>If a system at equilibrium is disturbed by a change in concentration, pressure, or temperature, the system will shift its equilibrium position to counteract the disturbance. The principle helps in:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Predicting the direction of shift in equilibrium.<\/li>\n\n\n\n<li>Optimizing reaction conditions in industrial and laboratory processes.<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-effect-of-concentration-change\">Effect of Concentration Change<\/h2>\n\n\n\n<p>The following are the effects of a concentration change.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Adding a reactant increases its concentration:\n<ul class=\"wp-block-list\">\n<li>The equilibrium shifts in the forward direction to reduce the added concentration.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Removing a reactant or adding a product:\n<ul class=\"wp-block-list\">\n<li>Equilibrium shifts in the reverse direction.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Example:\n<ul class=\"wp-block-list\">\n<li>In the equilibrium: N\u2082(g) + 3H\u2082(g) \u21cc 2NH\u2083(g)<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p class=\"has-pale-ocean-gradient-background has-background\"><strong>Also Read: <\/strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-geography-fundamentals-of-physical-geography-chapter-14-biodiversity-and-conservation\/\"><strong>NCERT Solutions Class 11 Geography Fundamentals of Physical Geography Chapter 14: Biodiversity and Conservation (Free PDF)<\/strong><\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-effect-of-pressure-change\">Effect of Pressure Change<\/h2>\n\n\n\n<p>Pressure changes affect gaseous reactions involving a change in the number of moles.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-increase-in-pressure\">Increase in Pressure<\/h3>\n\n\n\n<p>If pressure is increased by decreasing volume, the equilibrium shifts in the direction with fewer gas molecules. Example: N\u2082(g) + 3H\u2082(g) \u21cc 2NH\u2083(g) (4 moles \u21cc 2 moles). Increased pressure shifts the equilibrium toward NH\u2083.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-decrease-in-pressure\">Decrease in Pressure<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If pressure is decreased, the equilibrium shifts in the direction with more gas molecules.<\/li>\n\n\n\n<li>If the number of moles on both sides is equal, the pressure change has <strong>no effect<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-effect-of-inert-gas-addition\">Effect of Inert Gas Addition<\/h2>\n\n\n\n<p>Addition of inert gas at constant volume: No effect on equilibrium as partial pressures remain unchanged. Addition of inert gas at constant pressure: Volume increases, partial pressures decrease. Equilibrium shifts toward the side with more gas molecules.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-effect-of-temperature-change\">Effect of Temperature Change<\/h2>\n\n\n\n<p>Endothermic reaction (heat is absorbed): A + B + heat \u21cc C + D<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>An increase in temperature shifts the equilibrium forward.<\/li>\n\n\n\n<li>A decrease in temperature shifts it backward.<\/li>\n<\/ul>\n\n\n\n<p>Exothermic reaction (heat is released): A + B \u21cc C + D + heat<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>An increase in temperature shifts the equilibrium <strong>backward<\/strong>.<\/li>\n\n\n\n<li>A decrease in temperature shifts it <strong>forward<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ionic-equilibrium-in-solutions\">Ionic Equilibrium in Solutions<\/h2>\n\n\n\n<p>Chemical reactions in aqueous solutions often involve the formation of ions and reach equilibrium. This equilibrium is referred to as ionic equilibrium. It deals with the behavior of acids, bases, salts, and the extent of their ionisation in water.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-acids-bases-and-salts\">Acids, Bases, and Salts<\/h2>\n\n\n\n<p>The concepts of acids, bases, and salts are given below.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Arrhenius Concept:\n<ul class=\"wp-block-list\">\n<li>Acids: substances that increase the H\u207a concentration in water.<\/li>\n\n\n\n<li>Bases: substances that increase OH\u207b concentration in water.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Br\u00f8nsted-Lowry Concept:\n<ul class=\"wp-block-list\">\n<li>Acids: proton (H\u207a) donors.<\/li>\n\n\n\n<li>Bases: proton (H\u207a) acceptors.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Lewis Concept:\n<ul class=\"wp-block-list\">\n<li>Acids: electron-pair acceptors.<\/li>\n\n\n\n<li>Bases: electron-pair donors.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Salts are products formed from acid-base neutralization.<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ionisation-of-acids-and-bases\">Ionisation of Acids and Bases<\/h2>\n\n\n\n<p>The following are the effects of the ionization of acids and bases.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Strong acids and bases completely ionise in aqueous solutions.<\/li>\n\n\n\n<li>Weak acids and bases only partially ionise.<\/li>\n\n\n\n<li>Ionisation equilibrium is established between unionised molecules and ions.<\/li>\n\n\n\n<li>Example for a weak acid:<br>CH\u2083COOH \u21cc CH\u2083COO\u207b + H\u207a<br><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-ionisation-constants-of-weak-acids\">Ionisation Constants of Weak Acids<\/h3>\n\n\n\n<p>The acid dissociation constant (Ka) quantifies the ionisation of a weak acid:<br>Ka = [H\u207a][A\u207b] \/ [HA]. Larger the Ka, the stronger the acid. The degree of ionization increases with:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Decrease in the concentration of acid.<\/li>\n\n\n\n<li>Increase in dielectric constant of the solvent.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-ionization-of-weak-bases\">Ionization of Weak Bases<\/h3>\n\n\n\n<p>The base dissociation constant (Kb) is used for weak bases:<br>Kb = [BH\u207a][OH\u207b] \/ [B]. A higher Kb value indicates a stronger base.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-relation-between-ka-and-kb\">Relation between Ka and Kb<\/h3>\n\n\n\n<p>For a conjugate acid-base pair: Ka \u00d7 Kb = Kw<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Where Kw is the ionic product of water:<br>Kw = [H\u207a][OH\u207b] = 1.0 \u00d7 10\u207b\u00b9\u2074 mol\u00b2 L\u207b\u00b2 at 298 K<\/li>\n\n\n\n<li>Also: pKa + pKb = 14<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-factors-affecting-acid-strength\">Factors Affecting Acid Strength<\/h2>\n\n\n\n<p>The following are the key factors that affect the acid strength<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>More electronegative atoms bonded to H make the acid stronger.<\/li>\n\n\n\n<li>Larger size of the conjugate base leads to a weaker H\u2013X bond \u2192 stronger acid.<\/li>\n\n\n\n<li>Inductive effect, resonance, and hydration also influence strength.<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-common-ion-effect-in-the-ionisation-of-acids-and-bases\">Common Ion Effect in the Ionisation of Acids and Bases<\/h2>\n\n\n\n<p>Suppression of ionisation of a weak electrolyte by adding a strong electrolyte having a common ion. Example: Addition of HCl suppresses the ionization of CH\u2083COOH due to the common ion H\u207a.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-hydrolysis-of-salts-and-the-ph-of-their-solutions\">Hydrolysis of Salts and the pH of their Solutions<\/h2>\n\n\n\n<p>Reaction of salt with water to produce acidic or basic solutions. Depends on the nature of the acid and base that form the salt:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Salt of strong acid + strong base \u2192 neutral solution.<\/li>\n\n\n\n<li>Salt of weak acid + strong base \u2192 basic solution.<\/li>\n\n\n\n<li>Salt of strong acid + weak base \u2192 acidic solution.<\/li>\n\n\n\n<li>Salt of weak acid + weak base \u2192 pH depends on Ka and Kb.<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-buffer-solutions\">Buffer Solutions<\/h2>\n\n\n\n<p>Buffer solution: Resists change in pH on addition of small amounts of acid or base. Types:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Acidic buffer: Weak acid + its salt (e.g., CH\u2083COOH + CH\u2083COONa)<\/li>\n\n\n\n<li>Basic buffer: Weak base + its salt (e.g., NH\u2084OH + NH\u2084Cl)<\/li>\n\n\n\n<li>Henderson-Hasselbalch equation (for acidic buffer): pH = pKa + log([salt]\/[acid])<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-solubility-equilibria-of-sparingly-soluble-salts\">Solubility Equilibria of Sparingly Soluble Salts<\/h2>\n\n\n\n<p>Sparingly soluble salts establish an equilibrium between solid and dissolved ions. Example: BaSO\u2084(s) \u21cc Ba\u00b2\u207a(aq) + SO\u2084\u00b2\u207b(aq). Solubility product constant (Ksp): Ksp = [Ba\u00b2\u207a][SO\u2084\u00b2\u207b]. Used to predict precipitation and calculate solubility.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-important-formulas-in-ncert-notes-class-11-chemistry-part-i-chapter-6-equilibrium\">Important Formulas in NCERT Notes Class 11 Chemistry (Part-I) Chapter 6: Equilibrium<\/h2>\n\n\n\n<p>Here are the important formulas and equations from NCERT Class 11 Chemistry Chapter 6: Equilibrium:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Equilibrium Constant (Concentration):<\/strong><strong><br><\/strong>Kc = [C]^c [D]^d \/ [A]^a [B]^b<\/li>\n\n\n\n<li><strong>Equilibrium Constant (Partial Pressure):<\/strong><strong><br><\/strong>Kp = (PC)^c (PD)^d \/ (PA)^a (PB)^b<\/li>\n\n\n\n<li><strong>Relation Between Kp and Kc:<\/strong><strong><br><\/strong>Kp = Kc \u00d7 (RT)^\u0394n<br>where \u0394n = moles of gaseous products \u2212 moles of gaseous reactants<\/li>\n\n\n\n<li><strong>Gibbs Free Energy and Equilibrium Constant:<\/strong><strong><br><\/strong>\u0394G\u00b0 = \u2212RT ln K<\/li>\n\n\n\n<li><strong>Reaction Quotient (Qc):<\/strong><strong><br><\/strong>Qc = [C]^c [D]^d \/ [A]^a [B]^b<\/li>\n\n\n\n<li><strong>Acid Ionisation Constant:<\/strong><strong><br><\/strong>Ka = [H\u207a][A\u207b] \/ [HA]<\/li>\n\n\n\n<li><strong>Base Ionisation Constant:<\/strong><strong><br><\/strong>Kb = [BH\u207a][OH\u207b] \/ [B]<\/li>\n\n\n\n<li><strong>Ionic Product of Water:<\/strong><strong><br><\/strong>Kw = [H\u207a][OH\u207b] = 1.0 \u00d7 10\u207b\u00b9\u2074 mol\u00b2 L\u207b\u00b2 at 298 K<\/li>\n\n\n\n<li><strong>Henderson-Hasselbalch Equation:<\/strong><strong><br><\/strong>pH = pKa + log([Salt] \/ [Acid])<\/li>\n\n\n\n<li><strong>Solubility Product Constant (for AB \u21cc A\u207a + B\u207b):<\/strong><strong><br><\/strong>Ksp = [A\u207a][B\u207b]<\/li>\n<\/ul>\n\n\n\n<p><strong>Explore Notes of Class 11 Chemistry<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#f1c9d3\"><tbody><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-1-some-basic-concepts-of-chemistry\/\"><strong>Chapter 1<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-2-structure-of-atom\/\"><strong>Chapter 2<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-3-classification-of-elements-and-periodicity-in-properties\/\"><strong>Chapter 3<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-class-11-chemistry-part-i-chapter-iv-chemical-bonding-and-molecular-structure\/\"><strong>Chapter 4<\/strong><\/a><\/td><td><strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-5-thermodynamics-free-pdf\/\">Chapter 5<\/a><\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Download the Solutions of Other Chapters of Class 11 Chemistry<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#d0a0f9\"><tbody><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-1-chapter-1-some-basic-concepts-of-chemistry\/\"><strong>Chapter 1<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-i-chapter-2-structure-of-atom-free-pdf\/\"><strong>Chapter 2<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-1-chapter-3-classification-of-elements-and-periodicity-in-properties\/\"><strong>Chapter 3<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-1-chapter-iv-chemical-bonding-molecular-structure\/\"><strong>Chapter 4<\/strong><\/a><\/td><td><strong>Chapter 5<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><br><strong>Related Reads<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table class=\"has-pale-ocean-gradient-background has-background has-fixed-layout\"><tbody><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-political-science-indian-constitution-at-work-chapter-7-federalism\/\"><strong>NCERT Solutions Class 11 Political Science Indian Constitution at Work Chapter 7: Federalism (Free PDF)<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-political-science-indian-constitution-at-work-chapter-6-judiciary\/\"><strong>NCERT Solutions Class 11 Political Science Indian Constitution at Work Chapter 6: Judiciary (Free PDF)<\/strong><\/a><\/td><\/tr><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-political-science-indian-constitution-at-work-chapter-5-legislature\/\"><strong>NCERT Solutions Class 11 Political Science Indian Constitution at Work Chapter 5: Legislature (Free PDF)<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-english-snapshots-chapter-2-the-address\/\"><strong>NCERT Solutions Class 11 English Snapshots Chapter 2 The Address (Free PDF)<\/strong><\/a><\/td><\/tr><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-sociology-understanding-society-chapter-4-introducing-western-sociologists\/\"><strong>NCERT Solutions Class 11 Sociology Understanding Society Chapter 4: Introducing Western Sociologists (Free PDF)<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-political-science-chapter-6-political-theory-citizenship\/\"><strong>NCERT Class 11 Political Science Chapter 6 Political Theory: Citizenship Solutions (Free PDF)<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<figure class=\"wp-block-embed is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe loading=\"lazy\" title=\"equilibrium chemistry class 11 one shot revision complete chapter\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/CVPh5tnR6Jk?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><\/figure>\n\n\n\n<p><strong>Source- Munil Sir<\/strong><\/p>\n\n\n\n<p><strong>Explore notes on other subjects in the NCERT Class 11<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#ffc7c7\"><tbody><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-and-solutions-class-11-english\/\"><strong>English<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-and-solutions-class-11-sociology\/\"><strong>Sociology<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-and-solutions-class-11-geography\/\"><strong>Geography<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-and-solutions-class-11-political-science\/\"><strong>Political Science<\/strong><\/a><\/td><td><strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-and-solutions-class-11-psychology\/\">Psychology<\/a><\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-faqs\">FAQs<\/h2>\n\n\n\n<div class=\"schema-faq wp-block-yoast-faq-block\"><div class=\"schema-faq-section\" id=\"faq-question-1754653167100\"><strong class=\"schema-faq-question\"><strong>Q.1 What is the difference between physical and chemical equilibrium?<\/strong><\/strong> <p class=\"schema-faq-answer\"><strong>Answer: <\/strong>Physical equilibrium involves physical changes such as phase transitions (e.g., liquid \u21cc vapour, solid \u21cc liquid), where the substance remains the same but changes state. Chemical equilibrium involves reversible chemical reactions where reactants form products and vice versa, with no net change in concentrations once equilibrium is established.<\/p> <\/div> <div class=\"schema-faq-section\" id=\"faq-question-1754653191880\"><strong class=\"schema-faq-question\"><strong>Q.2 How does Le Chatelier\u2019s Principle help in industrial processes?<\/strong><\/strong> <p class=\"schema-faq-answer\"><strong>Answer: <\/strong>Le Chatelier\u2019s Principle helps predict how a change in conditions (concentration, pressure, or temperature) will shift the equilibrium position.<br\/>In industries, it is used to:<br\/>Maximise product formation (e.g., ammonia in the Haber process).<br\/>Minimise waste.<br\/>Optimize temperature and pressure conditions for better yield and efficiency.<\/p> <\/div> <div class=\"schema-faq-section\" id=\"faq-question-1754653213379\"><strong class=\"schema-faq-question\"><strong>Q.3 What is the significance of the equilibrium constant (K)?<\/strong><\/strong> <p class=\"schema-faq-answer\"><strong>Answer: <\/strong>The spontaneity of a chemical reaction is determined by the Gibbs free energy change (\u0394G), which is given by the equation:<br\/>\u0394G = \u0394H \u2013 T\u0394S, where \u0394H is the enthalpy change, T is the temperature, and \u0394S is the entropy change.<br\/>If \u0394G &lt; 0, the reaction is spontaneous.<br\/>If \u0394G > 0, the reaction is non-spontaneous.<br\/>If \u0394G = 0, the system is at equilibrium.<\/p> <\/div> <\/div>\n\n\n\n<p>For more topics, follow LeverageEdu <a href=\"https:\/\/leverageedu.com\/discover\/category\/school-education\/ncert-study-material\/\"><strong>NCERT Study Material<\/strong><\/a> today!<\/p>\n","protected":false},"excerpt":{"rendered":"NCERT Notes Class 11 Chemistry (Part-1) Chapter 6: Equilibrium introduces the concept of equilibrium in both physical and&hellip;\n","protected":false},"author":133,"featured_media":866411,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"editor_notices":[],"footnotes":""},"categories":[477,389],"tags":[],"class_list":{"0":"post-866386","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ncert-study-material","8":"category-school-education"},"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.3 (Yoast SEO v27.3) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Notes Class 11 Chemistry (Part-I) Chapter 6: Equilibrium (Free PDF) - Leverage Edu Discover<\/title>\n<meta name=\"description\" content=\"NCERT Notes for Class 11 Chemistry Chapter 6: Equilibrium. 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