{"id":90,"date":"2019-03-06T09:23:08","date_gmt":"2019-03-06T09:23:08","guid":{"rendered":"http:\/\/esp16.epgpbooks.inflibnet.ac.in\/?post_type=chapter&p=90"},"modified":"2019-04-25T07:02:02","modified_gmt":"2019-04-25T07:02:02","slug":"environmental-chemical-and-physical-equilibria","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/esp16\/chapter\/environmental-chemical-and-physical-equilibria\/","title":{"rendered":"Environmental Chemical and Physical Equilibria"},"content":{"raw":"
\"epgp<\/a>\r\n<\/span><\/div>\r\n \r\n\r\n \r\n\r\nContents<\/strong>\r\n
    \r\n \t
  1. Stoichiometry<\/li>\r\n \t
  2. Limiting Reagent<\/li>\r\n \t
  3. What is Equilibrium?<\/li>\r\n \t
  4. Chemical Equilibrium<\/li>\r\n \t
  5. Equilibrium Constant<\/li>\r\n \t
  6. Significance of Equilibrium Constants<\/li>\r\n \t
  7. Homogeneous Chemical Equilibia<\/li>\r\n \t
  8. Heterogeneous Chemical Equilibia<\/li>\r\n \t
  9. Equilibrium Constant and Free Energy Change<\/li>\r\n \t
  10. Effect of Temperature on Equilibrium Constant<\/li>\r\n \t
  11. Le-Chatelier\u2019s Principle<\/li>\r\n \t
  12. Reference<\/li>\r\n<\/ol>\r\n\u00a0 \u00a0 \u00a0Stoichiometry<\/strong>\r\n

    \u00a0 \u00a0Stoichiometry of a chemical reaction indicates the relative number of moles and relative masses of reactants and products in a balanced chemical equation. For example, Eq. 1 shows that in the reaction of N2 with O2 to form nitrogen oxide, one mole of N2 reacts with one mole of O2 to form two moles of NO. Thus, N2 and O2 react in 1:1 stoichiometric ratio to form NO. The ratio of N2, O2 and NO in Eq. 1 is 1:1:2.<\/p>\r\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 N2 + O2\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 2NO\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 \u00a0 \u00a0 \u00a0 (1)\r\n\r\n \r\n

    In terms of masses, Eq. 1 also shows that 28 g N2 reacts with 32 g O2 to produce 60 g NO.<\/p>\r\n

    Consider a familiar reaction (Eq. 2) in an internal combustion engine of a petrol driven<\/p>\r\n

    vehicle.<\/p>\r\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C8H18 + 12.5 O2\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0CO2 + H2O\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 \u00a0 (2)\r\n\r\n \r\n

    \u00a0 \u00a0 The air\u2013fuel ratio (mass of air\/ mass of fuel<\/em>) is the most common reference term used for mixtures in internal combustion engines. It is the ratio between the mass<\/em> of air and the mass of fuel in the fuel\u2013air mix at any given moment. The mass used is the mass of all constituents that compose the fuel and air whether the constituents are combustible or not. For example, in calculating the mass of natural gas which often contains carbon dioxide(CO2) and nitrogen(N2)as well as various alkanes, the masses of the carbon dioxide and nitrogen are included in addition to those of all alkanes to determine the value of mass of fuel. For pure octane, the stoichiometric mixture is approximately 14.7:1(air\/<\/strong>fuel).<\/p>\r\n \r\n

    \u00a0 \u00a0 According to Eq. 2, the air\/ fuel ratio for complete combustion of petrol should be 14.7. If the ratio is less than this, complete combustion of carbon to CO2 will not take place and due to this some CO - a partial oxidation product of carbon- would be formed. This is the reason for CO pollution in air.<\/p>\r\n \r\n

    Problem 1.<\/em><\/strong>H2 and O2 react to produce water: H2+ 0.5O2\u00a0 \u00a0=\u00a0 H2O.<\/p>\r\n

    How many moles of H2 would be needed to react completely with 0.5 mole O2?<\/p>\r\nSolution <\/em><\/strong>.Since H2 and O2 react in 2:1 ratio we have,\r\n\r\n\"\"\r\n\r\nHydrogen reacted = 2 x 0.5\/1 = 1.0 mol.\r\n\r\n \r\n

    \r\n

    Problem 2. <\/em><\/strong>According to balanced Eq.: 3Fe + 4H2O = Fe3O4+ 4H2, how many gram of Fe\u00a0will be required to react completely with 2.0 mole of water?<\/span><\/p>\r\n\r\n<\/div>\r\n

    Solution<\/em><\/strong>. Fe and H2O react in 3:4 mole ratio. So if x <\/em>mole of Fe react with 2.0 mole of water then,<\/p>\r\n\"\"\r\n

    Atomic weight of Fe = 55.8, 1 mole Fe = 55.8 g Fe. To obtain Fe in gram proceed as follows.<\/p>\r\n\"\"\r\n\r\nFe in gram = 58.5 x 1.5\/1 = 87.75.\r\n\r\n \r\n\r\nLimiting Reagent<\/strong>\r\n

    \u00a0 \u00a0The amount of the product, which is obtained from a chemical reaction, is determined by the amount of the reactant present in deficit (i. e., less than the stoichiometric ratio).<\/p>\r\n \r\n\r\n\"\"\r\n\r\nSo oxygen required = 32 x 20.2 \/4.04 = 160 g.\r\n

    \u00a0 \u00a0Since the available O2 is 90 g so it is less than the required amount of 160 g, the limiting reagent is O2. The amount of H2O formed shall be determined by the amount of O2 available. So,<\/p>\r\n\"\"\r\n\r\nWater formed = 90 \u00d7 36\/<\/strong>32 = 101.25 g.\r\n\r\n \r\n\r\nWhat is Equilibrium?<\/strong>\r\n

    \u00a0 \u00a0All physical and chemical processes have a natural tendency to reach the state of maximum stability, which is called the state of equilibrium. Once the equilibrium has been attained under specified conditions, the system loses the tendency for any further change. Consider that the water is filled in a container at a specified temperature. The water will continue to evaporate until the achievement of a constant vapor pressure, called saturated vapor pressure. This is the state of equilibrium. Liquid-vapor system will remain in this state, if the temperature is not changed.<\/p>\r\n \r\n

    Now, consider the following well- known chemical reaction in a closed vessel under specified conditions.<\/p>\r\n\"\"\r\n

    \u00a0 \u00a0The reaction proceeds to a certain point at which the concentrations of reactants and products become constant. The reaction does not go to completion. Under the given set of conditions of P, T and composition, the reaction always stops at this point, and the concentrations of N2, H2 and NH3 always remain the same.<\/p>\r\n \r\n\r\nChemical Equilibrium<\/strong>\r\n

    \u00a0 \u00a0Once the reaction has reached the state of equilibrium, the concentrations of reactants and the products attain certain fixed values, and do not change with time, provided the conditions do not change. These equilibria are dynamic in nature in that although, the concentrations do not change but the reaction itself does not cease to occur. The rates of the forward and backward reactions have the same value and so the extent of reaction in both directions remains the same. Hence there is no in change in the concentrations of reactants and products. Following are the features of chemical equilibria.<\/p>\r\n\r\n

      \r\n \t
    1. These reactions do not go the completion.<\/li>\r\n \t
    2. If a reaction is initiated by taking reactants only, the concentrations of the reactants would decrease with the progress of the reaction and those of products increase until the equilibrium is established. If the reaction in Eq. 3 is started by taking N2 and H2, the concentrations of both N2 and H2 will decrease and that of product NH3 will increase. Likewise, if the reaction is started by taking NH3 alone, the concentration of NH3 will decrease and that of N2 and H2 will increase.<\/li>\r\n \t
    3. At equilibrium the rate of forward reaction:\"\"will become equal to the rate of the backward reaction:\u00a0\"\"<\/li>\r\n \t
    4. Despite the establishment of the equilibrium, the reaction in both directions will continue to occur at equal speed.<\/li>\r\n<\/ol>\r\nEquilibrium Constant<\/strong>\r\n\r\nConsider a general reaction:\r\n\r\n\"\"\r\n\r\nWhen the reaction (4) is in equilibrium:\r\n\r\nK = {[L]l \u00d7 [M]m--------}\/{<\/strong>[A}a \u00d7 [B]b--------\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\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (5)\r\n

      \u00a0 \u00a0where K is the equilibrium constant and [L], [M], [A] and [B] represent the activities of concerned chemical species. In environmental chemistry, the systems dealt with are dilute aqueous solutions or gases at ambient pressure, so in place of activities, concentrations in mole L-1 or partial pressures or mole fractions can be used, and are generally used. Thus, the expressions(6 and 7) for equilibrium constants can be written as follows.<\/p>\r\nKc = [CL]l \u00d7 [CM]m \u00d7 ---------\/[CA]a \u00d7 [CB]b \u00d7-----------\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (6)\r\n\r\n \r\n\r\nKp\u00a0 = [pL]l \u00d7 [pM]m \u00d7 ---------\/[pA]a \u00d7 [pB]b \u00d7-----------\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (7)\r\n\r\n \r\n

      \u00a0 \u00a0where K<\/strong>c<\/strong> is called concentration equilibrium constant and K<\/strong>p<\/strong> is called pressure equilibrium constant.<\/p>\r\nEquilibrium constant is a dimensionless quantity.\r\n\r\n \r\n\r\nRelationship between K<\/em><\/strong>p<\/em><\/strong> and K<\/em><\/strong>c<\/em><\/strong>\r\n

      \u00a0 \u00a0Since partial pressure, p<\/strong>i<\/strong>, is related to concentration, c<\/strong>i<\/strong>, in mol L-1 by relationship: p<\/strong>i<\/strong> = c<\/strong>i<\/strong>RT<\/strong>, the relationship between Kp and Kc can be shown to be Eq. 8.<\/p>\r\n\"\"\r\n

      \u00a0 \u00a0where n = ( number of product molecules \u2013 number of reactant molecules), R is universal gas constant ant T is temperature in Kelvin.<\/p>\r\n \r\n\r\nSignificance of Equilibrium Constants<\/strong>\r\n

        \r\n \t
      1. The values of K are the indicator of the fact that how far the reaction will proceed in any direction. If K value is low, say for example 1\u00d7 10-5 , as in the case of dissociation of aqueous ammonia, the reaction will lie far to the left and dissociation of NH4OH would be very small<\/li>\r\n<\/ol>\r\n\"\"\r\n\r\n \r\n

        \u00a0 \u00a02. If K has a large value as in the case of aqueous phase dissociation of HNO3 (K = 15.4), the reaction will lie far to the\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 right and HNO3 would be almost 100% dissociated.<\/span><\/p>\r\n\"\"\r\n\r\n \r\n

          \r\n \t
        1. If the value is of K is very high, then the reaction would be virtually complete and hardly any unreacted reactant would remain. Such reactions are Irreversible Reactions<\/em><\/strong>. For example for the well- known atmospheric reaction of methane, the value of K is ~1\u00d710140. Hence, this reaction will be irreversible.<\/li>\r\n<\/ol>\r\n\"\"\r\n\r\n \r\n\r\nHomogeneous Chemical Equilibia<\/strong>\r\n

          \u00a0 \u00a0In homogeneous reactions, all the reactants and products are in the same phase\/state.<\/strong> For example, reaction in Eq. 11 is an example of this type because all the reactants and products are in the gaseous state.<\/p>\r\n\"\"\r\n\r\nFor this reaction, equilibrium constants have the values:\r\n\r\n\"\"\r\n

          where p-terms<\/em> denote the partial pressures and C-terms<\/em> denote the concentrations of respective species.<\/p>\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\nHeterogeneous Chemical Equilibia<\/strong>\r\n

          \u00a0 \u00a0In these reactions, the all of the reactants and the products are not in the same phase\/state. For example the decomposition of CaCO3 in nature is an example of this type ( Eq. 13).<\/p>\r\n\"\"\r\n

          Equilibrium Constant and Free Energy Change<\/strong><\/p>\r\nConsider a general reaction:\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\n\u0394G<\/em><\/strong>O<\/em><\/strong> and K Values<\/em><\/strong>\r\n

          \u00a0 \u00a0Magnitude of \u0394GO is the indicator of magnitude of K values and also of the extent of completion of the reaction, for example, if \u0394G<\/em>O<\/em> < 0 then K > 1<\/em> and if \u0394G<\/em>O<\/em> > 0 then K < 1.<\/em> If at 25oC, \u0394GO = - 100 kJ mol-1 then K= 3.4 x 1017, which is very high.<\/p>\r\nOn the other hand, if at the same temperature, \u0394GO = 100 kJ mol-1 then K= 3.0 x 10-18, which is very low.\r\n\r\n \r\n\r\nEffect of Temperature on Equilibrium Constant<\/strong>\r\n

          \u00a0 \u00a0The value of the equilibrium constant K of a reaction does depend on the concentration, pressure or catalyst but depends on the temperature in the following way. For<\/p>\r\n\"\"\r\n\r\nReactions for which has a positive value, i. e., endothermic reactions, the value of K increases on increasing temperature.\r\n\r\n \r\n

          \u00a0 \u00a0For example the reaction: N2 + O2\u00a0 \u00a0=\u00a0 2NO, is an endothermic reaction (\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = +180.7 kJ mol-1). So on increasing temperature, the value of K increases and more of the pollutant, NO, forms. Actually, K has very low value at ambient temperature and so this reaction is virtually negligible but at higher temperatures (2500oC in ignition chambers of automobiles, ~ 600oC of kitchen gas burner, the value of K becomes very high and so large amounts of NO forms under these conditions causing indoor and outdoor air pollution.<\/p>\r\n \r\n\r\nLe-Chatelier\u2019s Principle<\/strong>\r\n

          \u00a0 \u00a0This principle states that if a system at equilibrium is constrained, the system will move in that direction in which the effect of the constraint is the minimum. Generally, the constraints applied are change in concentration, pressure, volume or temperature. We shall consider these briefly.<\/p>\r\n\r\n

            \r\n \t
          1. Temperature change<\/em><\/strong>. If the temperature is increased( i. e., more heat is added), the system will move in that direction in which the heat is consumed. Consider two reactions:\"\"In order to consume heat supplied, when the temperature is raised, reaction (21) moves in the backward direction, and reaction (22) in forward direction. On the other hand, on decreasing temperature (i. e., cooling), the reactions (21) and (22) will move in forward and backward directions, respectively to produce heat in order to minimize the effect of cooling.<\/li>\r\n<\/ol>\r\n
              \r\n \t
            1. Concentration change. <\/em><\/strong>If the concentration of reactant(s) is increased, the reaction moves in the forward direction to consume added reactant(s).In case of both reactions (21) and (22) addition of reactants(N2 and H2 in (21) and NO2 in (22))will make the reaction to proceed in forward direction to produce more of products in order to consume added reactants. The effect would be opposite, if the products are removed from the system.<\/li>\r\n<\/ol>\r\n

              \u00a0 \u00a0 4. Pressure\/Volume change.<\/strong> <\/em>Pressure is inversely related to volume, and volume is directly related to the number of molecules. If the pressure is increased at constant volume, the reaction will move in that direction in which the pressure decreases, that is, lesser number of molecules are produced. Thus on increasing pressure in reaction (21), the reaction will move in forward direction as four molecules (one N2 and three H2) produce only three molecules( of NH3). The effect of increase of volume would be same as that of decreasing pressure as they are inversely related.<\/p>\r\n \r\n

              \u00a0 \u00a0 The effect of pressure on melting point of water is interesting. The same mass of water occupies more volume as ice than as liquid water. Hence, at 0oC when pressure on water is increased, the equilibrium: water(solid) = water(liquid) <\/em>moves in the direction of decreasing volume and so the melting point of ice decreases.<\/p>\r\n\r\n\r\n\r\n
              you can view video on Environmental Chemical and Physical Equilibria<\/strong><\/td>\r\n\"\"<\/a><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
              \r\n

              References<\/strong><\/h3>\r\n
                \r\n \t
              1. P. V. Hobbs(2000), Basic Physical Chemistry for the Atmospheric Sciences<\/em>, Cambridge, UK<\/li>\r\n \t
              2. Chemistry I for Class XI<\/em>, NCERT, New Delhi<\/li>\r\n \t
              3. A Bahl, B. S. Bahl and G. D. Tuli(2012), Essentials of Physical Chemistry<\/em>, S. Chand, New Delhi<\/li>\r\n \t
              4. Philip Matthews(2013), Advanced Chemistry<\/em>, Cambridge, New Delhi<\/li>\r\n<\/ol>\r\n<\/div>\r\n ","rendered":"
                \"epgp<\/a>
                \n<\/span><\/div>\n

                 <\/p>\n

                 <\/p>\n

                Contents<\/strong><\/p>\n

                  \n
                1. Stoichiometry<\/li>\n
                2. Limiting Reagent<\/li>\n
                3. What is Equilibrium?<\/li>\n
                4. Chemical Equilibrium<\/li>\n
                5. Equilibrium Constant<\/li>\n
                6. Significance of Equilibrium Constants<\/li>\n
                7. Homogeneous Chemical Equilibia<\/li>\n
                8. Heterogeneous Chemical Equilibia<\/li>\n
                9. Equilibrium Constant and Free Energy Change<\/li>\n
                10. Effect of Temperature on Equilibrium Constant<\/li>\n
                11. Le-Chatelier\u2019s Principle<\/li>\n
                12. Reference<\/li>\n<\/ol>\n

                  \u00a0 \u00a0 \u00a0Stoichiometry<\/strong><\/p>\n

                  \u00a0 \u00a0Stoichiometry of a chemical reaction indicates the relative number of moles and relative masses of reactants and products in a balanced chemical equation. For example, Eq. 1 shows that in the reaction of N2 with O2 to form nitrogen oxide, one mole of N2 reacts with one mole of O2 to form two moles of NO. Thus, N2 and O2 react in 1:1 stoichiometric ratio to form NO. The ratio of N2, O2 and NO in Eq. 1 is 1:1:2.<\/p>\n

                  \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 N2 + O2\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 2NO\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 \u00a0 \u00a0 \u00a0 (1)<\/p>\n

                   <\/p>\n

                  In terms of masses, Eq. 1 also shows that 28 g N2 reacts with 32 g O2 to produce 60 g NO.<\/p>\n

                  Consider a familiar reaction (Eq. 2) in an internal combustion engine of a petrol driven<\/p>\n

                  vehicle.<\/p>\n

                  \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C8H18 + 12.5 O2\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0CO2 + H2O\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 \u00a0 (2)<\/p>\n

                   <\/p>\n

                  \u00a0 \u00a0 The air\u2013fuel ratio (mass of air\/ mass of fuel<\/em>) is the most common reference term used for mixtures in internal combustion engines. It is the ratio between the mass<\/em> of air and the mass of fuel in the fuel\u2013air mix at any given moment. The mass used is the mass of all constituents that compose the fuel and air whether the constituents are combustible or not. For example, in calculating the mass of natural gas which often contains carbon dioxide(CO2) and nitrogen(N2)as well as various alkanes, the masses of the carbon dioxide and nitrogen are included in addition to those of all alkanes to determine the value of mass of fuel. For pure octane, the stoichiometric mixture is approximately 14.7:1(air\/<\/strong>fuel).<\/p>\n

                   <\/p>\n

                  \u00a0 \u00a0 According to Eq. 2, the air\/ fuel ratio for complete combustion of petrol should be 14.7. If the ratio is less than this, complete combustion of carbon to CO2 will not take place and due to this some CO – a partial oxidation product of carbon- would be formed. This is the reason for CO pollution in air.<\/p>\n

                   <\/p>\n

                  Problem 1.<\/em><\/strong>H2 and O2 react to produce water: H2+ 0.5O2\u00a0 \u00a0=\u00a0 H2O.<\/p>\n

                  How many moles of H2 would be needed to react completely with 0.5 mole O2?<\/p>\n

                  Solution <\/em><\/strong>.Since H2 and O2 react in 2:1 ratio we have,<\/p>\n

                  \"\"<\/p>\n

                  Hydrogen reacted = 2 x 0.5\/1 = 1.0 mol.<\/p>\n

                   <\/p>\n

                  \n

                  Problem 2. <\/em><\/strong>According to balanced Eq.: 3Fe + 4H2O = Fe3O4+ 4H2, how many gram of Fe\u00a0will be required to react completely with 2.0 mole of water?<\/span><\/p>\n<\/div>\n

                  Solution<\/em><\/strong>. Fe and H2O react in 3:4 mole ratio. So if x <\/em>mole of Fe react with 2.0 mole of water then,<\/p>\n

                  \"\"<\/p>\n

                  Atomic weight of Fe = 55.8, 1 mole Fe = 55.8 g Fe. To obtain Fe in gram proceed as follows.<\/p>\n

                  \"\"<\/p>\n

                  Fe in gram = 58.5 x 1.5\/1 = 87.75.<\/p>\n

                   <\/p>\n

                  Limiting Reagent<\/strong><\/p>\n

                  \u00a0 \u00a0The amount of the product, which is obtained from a chemical reaction, is determined by the amount of the reactant present in deficit (i. e., less than the stoichiometric ratio).<\/p>\n

                   <\/p>\n

                  \"\"<\/p>\n

                  So oxygen required = 32 x 20.2 \/4.04 = 160 g.<\/p>\n

                  \u00a0 \u00a0Since the available O2 is 90 g so it is less than the required amount of 160 g, the limiting reagent is O2. The amount of H2O formed shall be determined by the amount of O2 available. So,<\/p>\n

                  \"\"<\/p>\n

                  Water formed = 90 \u00d7 36\/<\/strong>32 = 101.25 g.<\/p>\n

                   <\/p>\n

                  What is Equilibrium?<\/strong><\/p>\n

                  \u00a0 \u00a0All physical and chemical processes have a natural tendency to reach the state of maximum stability, which is called the state of equilibrium. Once the equilibrium has been attained under specified conditions, the system loses the tendency for any further change. Consider that the water is filled in a container at a specified temperature. The water will continue to evaporate until the achievement of a constant vapor pressure, called saturated vapor pressure. This is the state of equilibrium. Liquid-vapor system will remain in this state, if the temperature is not changed.<\/p>\n

                   <\/p>\n

                  Now, consider the following well- known chemical reaction in a closed vessel under specified conditions.<\/p>\n

                  \"\"<\/p>\n

                  \u00a0 \u00a0The reaction proceeds to a certain point at which the concentrations of reactants and products become constant. The reaction does not go to completion. Under the given set of conditions of P, T and composition, the reaction always stops at this point, and the concentrations of N2, H2 and NH3 always remain the same.<\/p>\n

                   <\/p>\n

                  Chemical Equilibrium<\/strong><\/p>\n

                  \u00a0 \u00a0Once the reaction has reached the state of equilibrium, the concentrations of reactants and the products attain certain fixed values, and do not change with time, provided the conditions do not change. These equilibria are dynamic in nature in that although, the concentrations do not change but the reaction itself does not cease to occur. The rates of the forward and backward reactions have the same value and so the extent of reaction in both directions remains the same. Hence there is no in change in the concentrations of reactants and products. Following are the features of chemical equilibria.<\/p>\n

                    \n
                  1. These reactions do not go the completion.<\/li>\n
                  2. If a reaction is initiated by taking reactants only, the concentrations of the reactants would decrease with the progress of the reaction and those of products increase until the equilibrium is established. If the reaction in Eq. 3 is started by taking N2 and H2, the concentrations of both N2 and H2 will decrease and that of product NH3 will increase. Likewise, if the reaction is started by taking NH3 alone, the concentration of NH3 will decrease and that of N2 and H2 will increase.<\/li>\n
                  3. At equilibrium the rate of forward reaction:\"\"will become equal to the rate of the backward reaction:\u00a0\"\"<\/li>\n
                  4. Despite the establishment of the equilibrium, the reaction in both directions will continue to occur at equal speed.<\/li>\n<\/ol>\n

                    Equilibrium Constant<\/strong><\/p>\n

                    Consider a general reaction:<\/p>\n

                    \"\"<\/p>\n

                    When the reaction (4) is in equilibrium:<\/p>\n

                    K = {[L]l \u00d7 [M]m——–}\/{<\/strong>[A}a \u00d7 [B]b——–\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\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (5)<\/p>\n

                    \u00a0 \u00a0where K is the equilibrium constant and [L], [M], [A] and [B] represent the activities of concerned chemical species. In environmental chemistry, the systems dealt with are dilute aqueous solutions or gases at ambient pressure, so in place of activities, concentrations in mole L-1 or partial pressures or mole fractions can be used, and are generally used. Thus, the expressions(6 and 7) for equilibrium constants can be written as follows.<\/p>\n

                    Kc = [CL]l \u00d7 [CM]m \u00d7 ———\/[CA]a \u00d7 [CB]b \u00d7———–\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (6)<\/p>\n

                     <\/p>\n

                    Kp\u00a0 = [pL]l \u00d7 [pM]m \u00d7 ———\/[pA]a \u00d7 [pB]b \u00d7———–\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (7)<\/p>\n

                     <\/p>\n

                    \u00a0 \u00a0where K<\/strong>c<\/strong> is called concentration equilibrium constant and K<\/strong>p<\/strong> is called pressure equilibrium constant.<\/p>\n

                    Equilibrium constant is a dimensionless quantity.<\/p>\n

                     <\/p>\n

                    Relationship between K<\/em><\/strong>p<\/em><\/strong> and K<\/em><\/strong>c<\/em><\/strong><\/p>\n

                    \u00a0 \u00a0Since partial pressure, p<\/strong>i<\/strong>, is related to concentration, c<\/strong>i<\/strong>, in mol L-1 by relationship: p<\/strong>i<\/strong> = c<\/strong>i<\/strong>RT<\/strong>, the relationship between Kp and Kc can be shown to be Eq. 8.<\/p>\n

                    \"\"<\/p>\n

                    \u00a0 \u00a0where n = ( number of product molecules \u2013 number of reactant molecules), R is universal gas constant ant T is temperature in Kelvin.<\/p>\n

                     <\/p>\n

                    Significance of Equilibrium Constants<\/strong><\/p>\n

                      \n
                    1. The values of K are the indicator of the fact that how far the reaction will proceed in any direction. If K value is low, say for example 1\u00d7 10-5 , as in the case of dissociation of aqueous ammonia, the reaction will lie far to the left and dissociation of NH4OH would be very small<\/li>\n<\/ol>\n

                      \"\"<\/p>\n

                       <\/p>\n

                      \u00a0 \u00a02. If K has a large value as in the case of aqueous phase dissociation of HNO3 (K = 15.4), the reaction will lie far to the\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 right and HNO3 would be almost 100% dissociated.<\/span><\/p>\n

                      \"\"<\/p>\n

                       <\/p>\n

                        \n
                      1. If the value is of K is very high, then the reaction would be virtually complete and hardly any unreacted reactant would remain. Such reactions are Irreversible Reactions<\/em><\/strong>. For example for the well- known atmospheric reaction of methane, the value of K is ~1\u00d710140. Hence, this reaction will be irreversible.<\/li>\n<\/ol>\n

                        \"\"<\/p>\n

                         <\/p>\n

                        Homogeneous Chemical Equilibia<\/strong><\/p>\n

                        \u00a0 \u00a0In homogeneous reactions, all the reactants and products are in the same phase\/state.<\/strong> For example, reaction in Eq. 11 is an example of this type because all the reactants and products are in the gaseous state.<\/p>\n

                        \"\"<\/p>\n

                        For this reaction, equilibrium constants have the values:<\/p>\n

                        \"\"<\/p>\n

                        where p-terms<\/em> denote the partial pressures and C-terms<\/em> denote the concentrations of respective species.<\/p>\n

                        \"\"<\/p>\n

                        \"\"<\/p>\n

                         <\/p>\n

                        Heterogeneous Chemical Equilibia<\/strong><\/p>\n

                        \u00a0 \u00a0In these reactions, the all of the reactants and the products are not in the same phase\/state. For example the decomposition of CaCO3 in nature is an example of this type ( Eq. 13).<\/p>\n

                        \"\"<\/p>\n

                        Equilibrium Constant and Free Energy Change<\/strong><\/p>\n

                        Consider a general reaction:<\/p>\n

                        \"\"<\/p>\n

                        \"\"<\/p>\n

                         <\/p>\n

                        \u0394G<\/em><\/strong>O<\/em><\/strong> and K Values<\/em><\/strong><\/p>\n

                        \u00a0 \u00a0Magnitude of \u0394GO is the indicator of magnitude of K values and also of the extent of completion of the reaction, for example, if \u0394G<\/em>O<\/em> < 0 then K > 1<\/em> and if \u0394G<\/em>O<\/em> > 0 then K < 1.<\/em> If at 25oC, \u0394GO = – 100 kJ mol-1 then K= 3.4 x 1017, which is very high.<\/p>\n

                        On the other hand, if at the same temperature, \u0394GO = 100 kJ mol-1 then K= 3.0 x 10-18, which is very low.<\/p>\n

                         <\/p>\n

                        Effect of Temperature on Equilibrium Constant<\/strong><\/p>\n

                        \u00a0 \u00a0The value of the equilibrium constant K of a reaction does depend on the concentration, pressure or catalyst but depends on the temperature in the following way. For<\/p>\n

                        \"\"<\/p>\n

                        Reactions for which has a positive value, i. e., endothermic reactions, the value of K increases on increasing temperature.<\/p>\n

                         <\/p>\n

                        \u00a0 \u00a0For example the reaction: N2 + O2\u00a0 \u00a0=\u00a0 2NO, is an endothermic reaction (\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = +180.7 kJ mol-1). So on increasing temperature, the value of K increases and more of the pollutant, NO, forms. Actually, K has very low value at ambient temperature and so this reaction is virtually negligible but at higher temperatures (2500oC in ignition chambers of automobiles, ~ 600oC of kitchen gas burner, the value of K becomes very high and so large amounts of NO forms under these conditions causing indoor and outdoor air pollution.<\/p>\n

                         <\/p>\n

                        Le-Chatelier\u2019s Principle<\/strong><\/p>\n

                        \u00a0 \u00a0This principle states that if a system at equilibrium is constrained, the system will move in that direction in which the effect of the constraint is the minimum. Generally, the constraints applied are change in concentration, pressure, volume or temperature. We shall consider these briefly.<\/p>\n

                          \n
                        1. Temperature change<\/em><\/strong>. If the temperature is increased( i. e., more heat is added), the system will move in that direction in which the heat is consumed. Consider two reactions:\"\"In order to consume heat supplied, when the temperature is raised, reaction (21) moves in the backward direction, and reaction (22) in forward direction. On the other hand, on decreasing temperature (i. e., cooling), the reactions (21) and (22) will move in forward and backward directions, respectively to produce heat in order to minimize the effect of cooling.<\/li>\n<\/ol>\n
                            \n
                          1. Concentration change. <\/em><\/strong>If the concentration of reactant(s) is increased, the reaction moves in the forward direction to consume added reactant(s).In case of both reactions (21) and (22) addition of reactants(N2 and H2 in (21) and NO2 in (22))will make the reaction to proceed in forward direction to produce more of products in order to consume added reactants. The effect would be opposite, if the products are removed from the system.<\/li>\n<\/ol>\n

                            \u00a0 \u00a0 4. Pressure\/Volume change.<\/strong> <\/em>Pressure is inversely related to volume, and volume is directly related to the number of molecules. If the pressure is increased at constant volume, the reaction will move in that direction in which the pressure decreases, that is, lesser number of molecules are produced. Thus on increasing pressure in reaction (21), the reaction will move in forward direction as four molecules (one N2 and three H2) produce only three molecules( of NH3). The effect of increase of volume would be same as that of decreasing pressure as they are inversely related.<\/p>\n

                             <\/p>\n

                            \u00a0 \u00a0 The effect of pressure on melting point of water is interesting. The same mass of water occupies more volume as ice than as liquid water. Hence, at 0oC when pressure on water is increased, the equilibrium: water(solid) = water(liquid) <\/em>moves in the direction of decreasing volume and so the melting point of ice decreases.<\/p>\n\n\n\n
                            you can view video on Environmental Chemical and Physical Equilibria<\/strong><\/td>\n\"\"<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                            \n

                            References<\/strong><\/h3>\n
                              \n
                            1. P. V. Hobbs(2000), Basic Physical Chemistry for the Atmospheric Sciences<\/em>, Cambridge, UK<\/li>\n
                            2. Chemistry I for Class XI<\/em>, NCERT, New Delhi<\/li>\n
                            3. A Bahl, B. S. Bahl and G. D. Tuli(2012), Essentials of Physical Chemistry<\/em>, S. Chand, New Delhi<\/li>\n
                            4. Philip Matthews(2013), Advanced Chemistry<\/em>, Cambridge, New Delhi<\/li>\n<\/ol>\n<\/div>\n

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