{"id":1348,"date":"2024-08-17T11:02:12","date_gmt":"2024-08-17T11:02:12","guid":{"rendered":"https:\/\/www.meniit.com\/study-material\/?p=1348"},"modified":"2024-08-21T07:34:57","modified_gmt":"2024-08-21T07:34:57","slug":"equilibrium","status":"publish","type":"post","link":"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/equilibrium","title":{"rendered":"Chemical Equilibrium"},"content":{"rendered":"<h2>CHARACTERISTICS OF EQUILIBRIUM STATE<\/h2>\n<ul>\n<li style=\"text-align: justify;\">Chemical equilibrium is dynamic in nature.<\/li>\n<li style=\"text-align: justify;\">At constant temperature, certain properties such as pressure, concentration, density or colour<br \/>\nbecome constant.<\/li>\n<li style=\"text-align: justify;\">Equilibrium can be attained from either side, i.e., from the side of reactants or products.<\/li>\n<\/ul>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1349 size-full\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/equilibrium-1.png\" alt=\"equilibrium\" width=\"351\" height=\"63\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/equilibrium-1.png 351w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/equilibrium-1-300x54.png 300w\" sizes=\"auto, (max-width: 351px) 100vw, 351px\" \/><\/p>\n<ul>\n<li style=\"text-align: justify;\">Catalyst does not change the equilibrium state but helps in attaining it rapidly.<\/li>\n<\/ul>\n<h2>LAW OF MASS ACTION<\/h2>\n<p style=\"text-align: justify;\">The law of mass action (given by Guldberg and Waage) states that the rate of a chemical reaction is proportional to the product of effective concentrations (active masses) of the reacting species, each raised to a power that is equal to the corresponding stoichiometric number of the substance appearing in the chemical reaction.<\/p>\n<p style=\"text-align: justify;\">By the rate of a chemical reaction, we mean the amount of reactant transformed into products in unit time. It is represented by dx\/dt.<\/p>\n<p style=\"text-align: justify;\"><strong>Active mass means the molar concentration,<\/strong> i.e., the number of moles in 1 litre. Suppose 3 moles of nitrogen are present in a 4-litre vessel, the active mass of nitrogen = 3\/4= 0.75 mole\/litre. Active mass of a substance is represented by writing molar concentration in square brackets.<\/p>\n<p>Active mass of reactant \u03b1 molarity<\/p>\n<p style=\"text-align: center;\">Active mass of reactant = \u03b3 \u00d7 molarity<\/p>\n<p style=\"text-align: center;\">where \u03b3 is the activity coefficient.<\/p>\n<p style=\"text-align: center;\">\u2234\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Active Mass = \u03b3 \u00d7 molarity<\/p>\n<p style=\"text-align: center;\">For very dilute solutions, the value of activity coefficient is unity.<\/p>\n<p style=\"text-align: center;\">Active Mass can be taken to be same as Molarity.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_69_1 counter-hierarchy ez-toc-counter ez-toc-light-blue ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/equilibrium\/#EQUILIBRIUM-CONSTANT-KC-AND-KP\" title=\"EQUILIBRIUM CONSTANT KC AND KP\">EQUILIBRIUM CONSTANT KC AND KP<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/equilibrium\/#PROPERTIES-OF-EQUILIBRIUM-CONSTANT\" title=\"PROPERTIES OF EQUILIBRIUM CONSTANT\">PROPERTIES OF EQUILIBRIUM CONSTANT<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/equilibrium\/#UNITS-OF-KC-AND-KP\" title=\"UNITS OF KC AND KP\">UNITS OF KC AND KP<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/equilibrium\/#SECTION-6-LE-CHATELIERS-PRINCIPLE\" title=\"SECTION 6 : LE CHATELIER&#8217;S PRINCIPLE\">SECTION 6 : LE CHATELIER&#8217;S PRINCIPLE<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/equilibrium\/#EFFECT-OF-TEMPERATURE\" title=\"EFFECT OF TEMPERATURE\">EFFECT OF TEMPERATURE<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/equilibrium\/#SECTION-7-THERMODYNAMICS-OF-EQUILIBRIUM-CONSTANT\" title=\"SECTION 7 : THERMODYNAMICS OF EQUILIBRIUM CONSTANT\">SECTION 7 : THERMODYNAMICS OF EQUILIBRIUM CONSTANT<\/a><\/li><\/ul><\/nav><\/div>\n<h3><span class=\"ez-toc-section\" id=\"EQUILIBRIUM-CONSTANT-KC-AND-KP\"><\/span>EQUILIBRIUM CONSTANT K<sub>C<\/sub> AND K<sub>P<\/sub><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\">Let us have an equilibrium reaction as :<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1350 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/equilibrium-2.png\" alt=\"equilibrium\" width=\"192\" height=\"37\" \/><\/p>\n<p style=\"text-align: justify;\">For this reaction, which is in equilibrium, there exist an equilibrium constant (K<sub>eq<\/sub> ) represented as :<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1351 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-29.png\" alt=\"formula\" width=\"153\" height=\"61\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-29.png 153w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-29-150x61.png 150w\" sizes=\"auto, (max-width: 153px) 100vw, 153px\" \/><\/p>\n<p style=\"text-align: justify;\">For the given equilibrium, irrespective of the reacting species (i.e, either X + Y or Z or X + Z or Y + Z or X + Y + Z) and their amount we start with, the ratio, <img decoding=\"async\" class=\"alignnone\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-30.png\" \/>is always constant at a given temperature. This really looks amazing. Isn\u2019t it? Let us see, how such a thing is possible.<\/p>\n<p style=\"text-align: justify;\">We have learnt that at the equilibrium, rate of forward and reverse reactions are equal and we also know the law of mass action. Using this, we can write<\/p>\n<p style=\"text-align: center;\">Rate of forward reaction \u03b1 [X] [Y]<\/p>\n<p style=\"text-align: center;\">Rate of forward reaction = k<sub>f<\/sub> [X] [Y]<\/p>\n<p style=\"text-align: justify;\">where k<sub>f<\/sub> is the rate constant for the forward reaction.<\/p>\n<p style=\"text-align: justify;\">Similarly, rate of reverse reaction \u03b1 [Z]<\/p>\n<p style=\"text-align: justify;\">Rate of reverse reaction = k<sub>r<\/sub>[Z]<\/p>\n<p style=\"text-align: justify;\">where k<sub>r<\/sub> is the rate constant for the reverse reaction.<\/p>\n<p style=\"text-align: justify;\">At equilibrium,<\/p>\n<p style=\"text-align: center;\">Rate of forward reaction = Rate of reverse reaction.<\/p>\n<p style=\"text-align: center;\">\u2234\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0k f [X] [Y] = k r[Z]<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-31.png\" \/><\/p>\n<p style=\"text-align: justify;\">Since, k<sub>f<\/sub> and k<sub>r<\/sub> are constants at a given temperature, so their ratio\u00a0<img decoding=\"async\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-32.png\" \/> would also be a constant, referred as K<sub>eq.<\/sub><\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1470 size-full\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/equilibrium-3.png\" alt=\"equilibrium\" width=\"441\" height=\"58\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/equilibrium-3.png 441w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/equilibrium-3-300x39.png 300w\" sizes=\"auto, (max-width: 441px) 100vw, 441px\" \/><\/p>\n<p style=\"text-align: justify;\">As K<sub>eq<\/sub> is the ratio of rate constants for forward and reverse reaction, so the value of K<sub>eq<\/sub> would always be a constant and will not depend on the species we have started with and their initial concentrations.<\/p>\n<p style=\"text-align: justify;\">The given expression involves all variable terms (variable term means the concentration of the involved species changes from the start of the reaction to the stage when equilibrium is reached), so the ratio <img decoding=\"async\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-34.png\" \/> can also be referred as K<sub>C<\/sub> .<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1471 size-full aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/equilibrium2.png\" alt=\"equilibrium2\" width=\"470\" height=\"64\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/equilibrium2.png 470w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/equilibrium2-300x41.png 300w\" sizes=\"auto, (max-width: 470px) 100vw, 470px\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"PROPERTIES-OF-EQUILIBRIUM-CONSTANT\"><\/span>PROPERTIES OF EQUILIBRIUM CONSTANT<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"text-align: justify;\">It does not depend on initial concentration, because product will be formed in more amount when concentration of the ractants is increased, since the value of <strong>K<\/strong> remains constant.<\/li>\n<li style=\"text-align: justify;\">The value of <strong>K<\/strong> is constant at a fixed temperature but on increasing the temperature, value of <strong>K<\/strong> will change because concentration of reactants and products undergoing chemical reaction increases on increasing temperature.<\/li>\n<li style=\"text-align: justify;\">When the value of <strong>\u2018K<sub>C<\/sub>\u2019<\/strong> is more than 1.0 then product is formed in greater amount, i.e., forward reaction is faster but when it is less than 1.0 the product is formed in lower amount, i.e., backward reaction is faster.<\/li>\n<li style=\"text-align: justify;\">Value of K<sub>C<\/sub> remains unaffected by the presence of catalyst, because catalyst has equal effect on the rates of forward and backward reactions.<\/li>\n<li style=\"text-align: justify;\">If the same reaction is brought to equilibrium by two different methods the values of equilibrium constant will be different under the two conditions.<\/li>\n<li style=\"text-align: justify;\">Some mathematical operations that can change the value of equilibrium constant are as given below :<\/li>\n<\/ol>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1360 size-full\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-36.png\" alt=\"formula\" width=\"425\" height=\"358\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-36.png 425w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-36-300x253.png 300w\" sizes=\"auto, (max-width: 425px) 100vw, 425px\" \/><\/p>\n<ol style=\"list-style-type: lower-alpha;\" start=\"7\">\n<li>The equilibrium constants K<sub>c<\/sub> and K<sub>p<\/sub> can be related as<\/li>\n<\/ol>\n<p style=\"text-align: center;\">K<sub>P<\/sub> = K<sub>C<\/sub> (RT)<sup>\u2206n<\/sup><\/p>\n<p style=\"text-align: left;\">where \u2206n = sum of the number of moles of gaseous products \u2013 sum of the number of moles of gaseous reactants.<\/p>\n<p style=\"text-align: center;\">R = gas constant<\/p>\n<p style=\"text-align: center;\">and\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0T = absolute or Kelvin temperature at which equilibrium is established.<\/p>\n<p style=\"text-align: justify;\">Since, partial pressures are generally noted in atm and concentrations are measured in <img decoding=\"async\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-37.png\" \/> so the value of R used in the given expression should be in litre-atm per mole per Kelvin.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"UNITS-OF-KC-AND-KP\"><\/span>UNITS OF K<sub>C<\/sub> AND K<sub>P<\/sub><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\">Although it is not customary to mention the units of equilibrium constants K<sub>P<\/sub> and K<sub>C<\/sub> but when required, the unit of K<sub>C<\/sub> would be\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1362\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-38.png\" alt=\"formula\" width=\"86\" height=\"55\" \/>as the concentration of a species is generally expressed in moles\/litre and the unit of K<sub>P<\/sub><\/p>\n<p style=\"text-align: justify;\">would be (atm)<sup>\u2206n<\/sup> as the partial pressure is generally measured in atm.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"SECTION-6-LE-CHATELIERS-PRINCIPLE\"><\/span>SECTION 6 : LE CHATELIER&#8217;S PRINCIPLE<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\">It states that if a change is done to a system in equilibrium, the equilibrium condition is altered. A net reaction occurs in that direction which tends to relieve the external stress and finally a new equilibrium is attained.<\/p>\n<p style=\"text-align: justify;\">To understand its application to a system, let us consider following example :<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-39.png\" \/> for forward reaction, \u2206H = +92 kJ for backward reaction.<\/p>\n<p style=\"text-align: justify;\">Note that in above reaction,<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"text-align: justify;\">Forward reaction is exothermic (towards formation of NH<sub>3<\/sub>) and backward reaction is endothermic (favours decomposition of NH<sub>3<\/sub>).<\/li>\n<li style=\"text-align: justify;\">Formation of NH<sub>3<\/sub> result in decrease in number of moles, i.e., decrease in volume to right.<\/li>\n<li style=\"text-align: justify;\">Both reactants and products are gases, i.e., they will be influenced by changes in P, T and changing concentration.<\/li>\n<\/ol>\n<h3><span class=\"ez-toc-section\" id=\"EFFECT-OF-TEMPERATURE\"><\/span>EFFECT OF TEMPERATURE<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\">Temperature can be increased by adding heat and can be decreased by taking out heat from the system. Increase or decrease in temperature reaction either shifts in forward or backward direction and it also changes the equilibrium constant.<\/p>\n<p><strong>Endothermic Reaction\u00a0<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1366 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-40.png\" alt=\"formula\" width=\"118\" height=\"31\" \/><\/p>\n<ol>\n<li style=\"text-align: justify;\">Increase in temperature shifts the reaction in the forward direction.<\/li>\n<li style=\"text-align: justify;\">Decrease in temperature shifts the reaction in the backward direction.<\/li>\n<\/ol>\n<p><strong>Exothermic Reaction<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1367 size-full\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-41.png\" alt=\"formula\" width=\"124\" height=\"34\" \/><\/p>\n<ol>\n<li style=\"text-align: justify;\">Increase in temperature, shifts the reaction in the backward direction.<\/li>\n<li style=\"text-align: justify;\">Decrease in temperature, shifts the reaction in the forward direction.<\/li>\n<\/ol>\n<p style=\"text-align: justify;\">Effect of temperature on equilibrium constant by using <strong>van&#8217;t Hoff equation.<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1368 size-full\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-42.png\" alt=\"formula\" width=\"190\" height=\"68\" \/><\/p>\n<p style=\"text-align: justify;\">For Endothermic Reaction :<\/p>\n<p style=\"text-align: justify;\">If T<sub>2<\/sub> &gt; T<sub>1<\/sub> then\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>2&lt;<\/sub> &gt; K<sub>1<\/sub><\/p>\n<p style=\"text-align: justify;\">For exothermic reaction if T<sub>2<\/sub> &gt; T<sub>1<\/sub> then K<sub>2<\/sub> &lt; K<sub>1<\/sub>.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"SECTION-7-THERMODYNAMICS-OF-EQUILIBRIUM-CONSTANT\"><\/span>SECTION 7 : THERMODYNAMICS OF EQUILIBRIUM CONSTANT<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\">Although the equilibrium constant is not dependent on rate, but it is directly related to Gibb\u2019s Energy (\u2206G). The relationship between K<sub>eq<\/sub> and \u2206G is well established.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"text-align: justify;\">If \u2206G &lt; 0, then the reaction is spontaneous and proceeds forward.<\/li>\n<li style=\"text-align: justify;\">If \u2206G &gt; 0, then the reaction is non-spontaneous.<\/li>\n<li style=\"text-align: justify;\">If \u2206G = 0, then the reaction has achieved equilibrium. At this state no free energy is available to run the reaction.<\/li>\n<\/ol>\n<p style=\"text-align: justify;\">Mathematically, for a reaction we should know that<\/p>\n<p style=\"text-align: center;\">\u2206G = \u2206G<sup>o<\/sup> + 2.303RT log Q<\/p>\n<p style=\"text-align: justify;\">Here, \u2206G<sup>o<\/sup> is standard free energy change and Q is the reaction quotient of the reaction not in equilibrium. When equilibrium is achieved \u2206G = 0 and Q = K<sub>C<\/sub><\/p>\n<p style=\"text-align: justify;\">So,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 0 = \u2206G<sup>o<\/sup> + 2.303RT log K<sub>C<\/sub> is what we get<\/p>\n<p style=\"text-align: justify;\">\u21d2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2206G<sup>o<\/sup> = \u20132.303RT log K<sub>C<\/sub> = \u2013RT ln K<sub>C<\/sub><\/p>\n<p>In other words,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <img decoding=\"async\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/formula-43.png\" \/><\/p>\n<p style=\"text-align: justify;\">This equation can be used to determine the spontaneity of a reaction that is if \u2206G <sup>o<\/sup> &lt; 0 then e<sup>\u2013\u2206Go\/RT<\/sup> &gt; 1 making K<sub>C<\/sub> &gt; 1 which implies a spontaneous forward reaction.<\/p>\n<p style=\"text-align: justify;\">If \u2206G<sup>o<\/sup> &lt; 0 then e<sup>\u2013\u2206Go\/RT<\/sup> &lt; 1, i.e., K &lt; 1 which implies a backward reaction spontaneity. Such reaction proceed forward to such a small extent that a very small amount of product is formed.<\/p>\n<p style=\"text-align: justify;\"><strong>Relationship of \u2206G\u00b0 to the equilibrium constant K.<\/strong><\/p>\n<p style=\"text-align: justify;\">\u2206G for a reaction under any set of conditions is related to its value for standard conditions, that is, \u2206G\u00b0 by equation.<\/p>\n<p style=\"text-align: center;\">\u2206G = \u2206G\u00b0 + 2.303 RT log Q<\/p>\n<p style=\"text-align: left;\">Here, Q refers to reaction quotient of the reaction that is not in equilibrium.<\/p>\n<p style=\"text-align: left;\">Under equilibrium condition<\/p>\n<p style=\"text-align: center;\">Q = K<sub>p<\/sub> = K<sub>c<\/sub> = K<\/p>\n<p style=\"text-align: center;\">\u2206G = 0<\/p>\n<p style=\"text-align: center;\">\u2206G\u00b0 = \u20132.303 RT log K<\/p>\n<p style=\"text-align: left;\">K is called thermodynamic equilibrium constant.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"newspaper-x-tags\"><strong>TAGS: <\/strong><span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/chemical-equilibrium\" rel=\"tag\">Chemical Equilibrium<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/effect-of-temperature\" rel=\"tag\">Effect of Temperature<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/endothermic-reaction\" rel=\"tag\">Endothermic Reaction<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/equilibrium-constant\" rel=\"tag\">Equilibrium Constant<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/exothermic-reaction\" rel=\"tag\">Exothermic Reaction<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/law-of-mass-action\" rel=\"tag\">Law of Mass Action<\/a> <\/div>\n","protected":false},"excerpt":{"rendered":"<p>CHARACTERISTICS OF EQUILIBRIUM STATE Chemical equilibrium is dynamic in nature. At constant temperature, certain properties such as pressure, concentration, density&nbsp;&nbsp;&#8230;.<a class=\"read_more\" href=\"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/equilibrium\" rel=\"nofollow\">Read More >><\/a><\/p>\n","protected":false},"author":5,"featured_media":1483,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"om_disable_all_campaigns":false,"rank_math_lock_modified_date":false,"footnotes":""},"categories":[268,241,240],"tags":[452,455,456,454,457,453],"class_list":["post-1348","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-chemistry","category-class-11th","category-neet","tag-chemical-equilibrium","tag-effect-of-temperature","tag-endothermic-reaction","tag-equilibrium-constant","tag-exothermic-reaction","tag-law-of-mass-action"],"acf":[],"_links":{"self":[{"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/posts\/1348","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/comments?post=1348"}],"version-history":[{"count":9,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/posts\/1348\/revisions"}],"predecessor-version":[{"id":1467,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/posts\/1348\/revisions\/1467"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/media\/1483"}],"wp:attachment":[{"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/media?parent=1348"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/categories?post=1348"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/tags?post=1348"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}