{"id":1203,"date":"2025-11-21T11:38:25","date_gmt":"2025-11-21T16:38:25","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/?post_type=chapter&p=1203"},"modified":"2026-03-23T15:16:13","modified_gmt":"2026-03-23T19:16:13","slug":"1-chemical-equilibrium","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/chapter\/1-chemical-equilibrium\/","title":{"raw":"1. Chemical Equilibrium","rendered":"1. Chemical Equilibrium"},"content":{"raw":"
\r\n

1.1 The Concept of Equilibrium.<\/h2>\r\nAll chemical reactions have a natural spontaneous tendency to move toward equilibrium, however fast or slow the reaction may be. Chemical equilibrium is the state where no net change in composition occurs under specified conditions. More specifically, at equilibrium, all species have the same thermodynamic chemical potential. The hypothetical reaction,\r\n\r\n\\[\\ce{A + B <--> C + D}\\tag{1}\\]\r\n\r\nis at equilibrium when the composition of a mixture containing A, B, C and D cannot change. (Of course if conditions change, such as temperature or addition of one of A or B etc, then change will occur.) The state of equilibrium does not imply that all chemical processes have stopped. Rather, the rates of the forward and the reverse reactions at equilibrium are equal. Hence there is no net change in composition with time. Equilibrium is a dynamic balance of forward and reverse reactions rates. Phenomenologically, this is how equilibrium is defined.\r\n\r\nFor reaction (1), the reactants A and B react to form C and D at a rate that can be stated as the change in concentration of A or B with time, i.e. rate = -d[A]\/dt = -d[B]\/dt\u00a0 (where [A] is concentration of A, etc. and t is time; negative because the reactants are decreasing in concentration). But C and D also react to form A and B in a reverse reaction, at a rate = -d[C]\/dt = -d[D]\/dt, as indicated by the double-headed arrow. If the rate of the forward reaction is fast relative to that of the back reaction, C and D will predominate when a stoichiommetric mixture of A and B are combined. If the forward reaction is slow relative to the back reaction, little of C and D will form. At equilibrium the forward and backward reaction rates are equal. (Not equal rate constants, but equal actual rates in concentration per unit time.) Some reactions will go virtually to completion. For example, when an equimolar mixture of H2<\/sub>\u00a0and O2 <\/sub>gas\u00a0is ignited, the reaction is,\r\n\r\n\\[\\ce{H2_{(g)} + O2_{(g)} = H2O_{(l)}} \\tag{2}\\]\r\n\r\nIn the case of reactions that go to completion, as in the H2<\/sub> + O2<\/sub> case, the forward reaction is so fast compared to the reverse reaction that virtually no net reverse reaction occurs. Then the equilibrium constant is immense; e.g. 3.5 x 1041<\/sup> for reaction (2).\r\n
Note:<\/strong> That a mixture of A, B, C and D is not undergoing any compositional change is not necessarily indicative of equilibrium having been attained. For instance, a mixture of H2<\/sub> and O2 <\/sub>in a container will persist without change for a long time. But, we know that this is not the equilibrium situation: provide a spark or a suitable catalytic surface, and water will form until one or the other of H2<\/sub> or O2<\/sub> is consumed. The equilibrium state is water plus left over excess O2 <\/sub>or H2<\/sub>, whichever was present in greater amount to start with.<\/blockquote>\r\nThis illustrates a key point, which is the distinction between kinetics and thermodynamics. In the limit of infinite time, equilibrium will be attained. Thermodynamics will prevail. It\u2019s just that thermodynamics cannot tell us how fast the reactions will be. That is the province of kinetics. Many hydrometallurgical processes are too slow to attain equilibrium within the residence time provided because the temperatures are typically quite low.\r\n

1.2 The Equilibrium Constant<\/h2>\r\nEquilibrium constants are defined in terms of the activity of each species involved in the reaction. Activity is a modified concentration. It is observed that as concentration of a solute increases, the solution\u2019s apparent concentration is not equivalent to its actual concentration. For example, the rate of reaction (1) was expressed as -d[A]\/dt, where [A] is the actual concentration of reactant A. Over a limited range of concentration this holds true, especially at low concentration. But, as concentration increases it becomes apparent that there are deviations from this simple rule. The rate may appear to be slower than would be expected; reactant A is behaving as if its concentration is less than its true concentration. We refer to the apparent concentration as the activity. The effect is much more pronounced with ionic solutes than with neutral ones.\r\n\r\nThe origins of the activity have to do with interactions between the solvent and the solute, and between the solutes themselves. When the solution is very dilute the solvent molecules keep the solutes ions quite far apart, and the solutes interact only weakly. But as the solution gets more concentrated, the solutes get closer together, and the interaction starts to become strong. This makes them less readily able to participate in chemical reactions; they behave as if their concentration is actually less than what it actually is. Their activity is lower than their concentration. In cases where very high concentration solutions can be obtained the activity may become significantly greater than the concentration.\r\n\r\nThe activity takes into account the fact that many solutions are non-ideal. For ideal solutions the solute and the solvent do not interact significantly. There are numerous cases where solutions of molecular solutes do behave ideally over a wide range of concentrations. But aqueous solutions, particularly of ionic solutes are highly non-ideal.\r\n\r\nFor the general reaction,\r\n\r\n\\[\\ce{aA + bB = cC + dD}\\tag{3}\\]\r\n\r\nwhere a, b, c and d are the stoichiommetric coefficients, the equilibrium constant is defined as,\r\n\r\n\\[\r\nK = \\frac{a_C^c\\, a_D^d}{a_A^a\\, a_B^b}\\tag{4}\r\n\\]\r\n\r\nwhere aC<\/sub>c<\/sup> etc. are the activities of the various species raised to the powers of their stoichiommetric coefficients. For a solution that behaves ideally the activities are equal to the concentrations. Aqueous solutions of ionic solutes will behave ideally at quite low concentrations, i.e. <10-3<\/sup> M for singly charged anions; lower still for more highly charged ions. Generally though, solutions of practical interest in hydrometallurgy are quite concentrated and hence far from ideal. Unfortunately activities are usually quite difficult to measure.\r\n\r\nFor the gas phase we can often assume the ideal gas law. Pressure is a proxy for concentration when dealing with gases. For an ideal gas,\r\n\r\n\\[ P = \\frac{nRT}{V} = \\frac{n(RT)}{V} \\tag{5}\\]\r\n\r\nwhere n\/V (moles\/L) is equal to concentration. Again, however, interactions between gas molecules lead to deviations from ideality, and then in general we need an activity for gases in order to take attractive and repulsive interactions into account. For gases we call this activity the fugacity. It can be thought of as an effective pressure. There are good methods for estimating or measuring fugacities. At room temperature, typically 20\u00b0C, and 1 atm pressure the ideal gas law is a moderately good approximation for many gases. As pressure increases and\/or temperature decreases the ideal gas law becomes a poorer approximation.\r\n\r\nWhere a gas is involved in a reaction, we include it in the equilibrium constant expression by using its fugacity, e.g. for:\r\n\r\n\\[\\ce{aA + bB = cC + eE_{(g)}}\\tag{6}\\]\r\n\r\n\\[ K = \\frac{a_C^c \\, f_E^e}{a_A^a \\, a_B^b}\\tag{7}\\]\r\n\r\nSolution concentration is commonly expressed in terms of molarity (M), molality (m) or mole fraction (X). Molality has units of moles of solute per kg of solvent (not kg of solution!). It is commonly used in thermodynamics because it is independent of solution density, whereas molarity varies with density; the volume of the solution changes with temperature. Mole fraction is less commonly used. It is equal to moles of a species divided by the total moles of all species. To illustrate, a 1.032 M NaCl solution at 20\u00b0C has a molality of 1.054 and a mole fraction of 0.01863. For dilute aqueous solutions with densities close to 1 g\/mL, molality and molarity do not differ greatly. For concentrated solutions, with densities significantly different from that of pure water, molality and molarity diverge.\r\n\r\nThe activity is defined so as to be unitless, which is helpful because it is used heavily in logarithmic functions. To make this work the activity is defined as the ratio:\r\n\r\n\\[ a = \\frac{\\text{absolute activity}}{\\text{standard state activity}}\\tag{8}\\]\r\n\r\nAbsolute activity is the effective concentration; it has concentration units. The standard state is a reference point which we define so as to have unit activity. The common standard state for solutions is 1 molal and behaving as if ideal. It is a hypothetical state; most real solutions at unit molality are far from ideal. But, the construct is helpful. The activity of a solute in a solution as defined in equation [4] is numerically equal to its absolute activity. The standard state for gases is 1 bar pressure and acting as if ideal. (1 bar = 0.987 atm.) For liquids and gases the standard state is the pure liquid or solid. (A fuller discussion of these ideas is usually presented in thermodynamics courses and texts.)\r\n\r\nWhere pure solids, for example, are involved in a reaction the activity as defined by equation (8) is,\r\n\r\n\\[\\frac{\\text(activity of pure solid)}{\\text(activity of pure solid)} =\u00a0 a\u00a0 = 1\\tag{9}\\]\r\n\r\nIn other words, the pure solid is in its standard state. If the solid is not pure, i.e. it is a solid solution, or a mixture, or it is in a chemical form that is less stable than the standard state form, then the activity is not equal to 1. (Many compounds can exist in a number of different crystalline forms or compositions; elemental sulfur might be formed not as the stable S8<\/sub>, but as polymeric sulfur, for example.)\r\n\r\nTo a first approximation the equilibrium constant is often expressed in terms of the concentrations and pressures of solutes and gases, respectively. The approximation can be rather poor at high concentrations, but for the purposes of this course the important principles can be presented without needing to develop the concept of activity further than this. For accurate work, activities and fugacities are required.)\r\n\r\nWe will take the equilibrium constant to involve concentrations in molal or molar and gas pressures in atm. (Strictly speaking we should use pressure in bars, but much older and still useful data is referenced to pressure in atm.) As above, the activities of pure solids and liquids are taken to be 1. Where water is involved in the chemical reaction, we will treat it as if it is a pure liquid. (Properly one would include its activity; in dilute solution this is very close to 1, but in concentrated solutions it may actually be considerably lower than 1.) The concentration of water can be found from its density (0.998 g\/mL at 20\u00b0C) and molecular weight:\r\n\r\n\\[\r\n\\frac{998\\ \\text{g H2O}}{1\\ \\text{L}}\\times\r\n\\frac{1\\ \\text{mol H2O}}{18.015\\ \\text{g H2O}}\r\n= 55.4\\ \\text{mol\/L}\r\n\\tag{10}\\]\r\n\r\nHence even a 1 M solution is still roughly 98% water.\r\n

1.3 Understanding and Using Equilibrium Constants.<\/h2>\r\n

1. Numerical Significance.<\/h3>\r\nWhere equilibrium constants are concerned, size matters. A large equilibrium constant indicates that a reaction is very favourable and will proceed to the right to a very high degree. A very small equilibrium constant indicates a very unfavourable reaction, which will proceed to the right only to a very small extent. When the equilibrium constant is near one, similar proportions of reactants and products will result. For example, water dissociates weakly into H+<\/sup> and OH-<\/sup>:\r\n\r\n\\[\\ce{H2O_{(l)} = H+_{(aq)} + OH-_{(aq)}}\\tag{11}\\]\r\n\r\n\\[\\ce{K_{w25\u00b0C} = [H+][OH-] = 1 x 10^{-14}} \\tag{12}\\]\r\n\r\nOnly about 2 H2<\/sub>O molecules per billion dissociate on average. On the other hand, if we added solid NaOH (strong base) to a solution of HCl (strong acid), the reaction would be:\r\n\r\n\\[\\ce{NaOH_{(s)} = Na+_{(aq)} + OH-_{(aq)}}\\tag{13}\\]\r\n\r\n\\[\\ce{H+_{(aq)} + OH-_{(aq)} = H2O_{(l)}}\\tag{14}\\]\r\n\r\nClearly, if reaction (11) is unfavourable, reaction (12) is very favourable. It is important to remember that in principle reactions can proceed in either direction. This is the basis for equilibrium.\r\n

2. Manipulating Equilibrium Constants.<\/h3>\r\n

(a). Reversing the reaction.<\/h5>\r\nWhat is the equilibrium constant for reaction (14)?\r\n\r\n\\[K = \\frac{1}{[H^+][OH^-]}\r\n= \\frac{1}{1 \\times 10^{-14}}\r\n= 1 \\times 10^{14}\\tag{15}\\]\r\n\r\nIn other words, when the reaction is reversed, the equilibrium constant is inverted. For,\r\n\r\n\\[aA + bB = cC\r\n\\qquad\r\nK = \\frac{[C]^c}{[A]^a\\, [B]^b}\r\n\\tag{16}\\]\r\n\r\nand for the reverse reaction,\r\n\r\n\\[\r\ncC = bB + aA\r\n\\qquad\r\nK' = \\frac{[A]^a [B]^b}{[C]^c} = \\frac{1}{K}\\tag{17}\\]\r\n
(b). Multiplying the Reaction by a Factor.<\/h5>\r\nThe only constraint on the stoichiommetric coefficients is that the reaction must be balanced. Other than that they can be any numbers. When a reaction is written with different stoichiommetric coefficients the equilibrium constant changes. The same is true if a reaction is multiplied by some factor. Consider the reaction,\r\n\r\n\\[ \\ce{NH3 + H2O = NH4+ + OH-} \\tag{18}\\]\r\n\r\n\\[ K = \\frac{\\ce{[NH4+][OH-]}}{\\ce{[NH3]^2}} = 1.8 \\times 10^{-5} \\tag{19}\\]\r\n\r\nWe can also write the reaction as,\r\n\r\n\\[ \\ce{2NH3 + 2H2O = 2NH4+ + 2OH-} \\tag{20}\\]\r\n\r\n\\[ K = \\frac{\\ce{[NH4+]^2[OH-]^2}}{\\ce{[NH3]^2}} = (1.8 \\times 10^{-5})^2 = 3.2 \\times 10^{-10} \\tag{21}\\]\r\n\r\nIn general, multiply a reaction by n, raise its equilibrium constant to the power n.\r\n
(c). Adding Reactions.<\/h5>\r\nBalanced reactions are statements of mass balance. Just like any arithmetic equation, they can be added together in the same way. When two or more reactions are added together, their equilibrium constants are multiplied. An example is the series of reactions listed below. Mercury oxide will react readily with thiosulfate ion to form a very stable complex. This could afford an important way to control mercury in some hydrometallurgical processes. This reaction can be written as:\r\n\r\n\\[ \\ce{HgO_{(s)} + 2S2O3^2- + H2O_{(l)} = [Hg(S2O3)2]^2- + 2OH-} \\tag{22}\\]\r\n\r\nThis reaction can be split into two, the sum of which is the preceding reaction:\r\n\r\nFirst Half\r\n\r\n\\[ \\ce{HgO_{(s)} + H2O_{(l)} = Hg^2+ + 2OH-} \\tag{23}\\]\r\n\r\n\\[K_{\\mathrm{sp}} = [\\mathrm{Hg}^{2+}][\\mathrm{OH}^{-}]^{2} = 3.6 \\times 10^{-26}\\tag{24}\\]\r\n\r\nSecond Half:\r\n\r\n\\[ \\ce{Hg^2+ + 2S2O3^2- = [Hg(S2O3)2]^2-} \\tag{25}\\]\r\n\r\n\\[K = \\frac{[\\text{Hg(S2O3)2}^{2-}]}{[\\text{Hg}^{2+}]\\,[\\text{S2O3}^{2-}]^{2}}= 1 \\times 10^{32}\\tag{26}\\]\r\n\r\nCombine:\r\n\r\n\\[ \\ce{HgO_{(s)} + H2O_{(l)} = Hg^2+_{(aq)} + 2OH-_{(aq)}} \\tag{27}\\]\r\n\r\n\\[ \\ce{Hg^2+_{(aq)} + 2S2O3^2-_{(aq)} = [Hg(S2O3)2]^2-_{(aq)}} \\tag{28}\\]\r\n\r\n\\[ \\ce{HgO_{(s)} + 2S2O3^2-_{(aq)} + H2O_{(l)} = [Hg(S2O3)2]^2-_{(aq)} + 2OH-_{(aq)}} \\tag{29}\\]\r\n\r\nThe equilibrium constant for reaction (22) is,\r\n\r\n\\[K' = \\frac{[\\mathrm{Hg(S_2O_3)_2}^{2-}][\\mathrm{OH^-}]^{2}}\r\n{[\\mathrm{S_2O_3}^{2-}]^{2}}\r\n= \\frac{[\\mathrm{Hg(S_2O_3)_2}^{2-}][\\mathrm{Hg^{2+}}][\\mathrm{OH^-}]^{2}}\r\n{[\\mathrm{S_2O_3}^{2-}]^{2}[\\mathrm{Hg^{2+}}]}\\tag{30}\\]\r\n\r\nThe latter part of equation [15] can be clearly seen to be equal to K Ksp:\r\n\r\n\\[ K' = K \\cdot K_{sp} = 3.6 \\times 10^6 \\tag{31}\\]\r\n
(d) Le Chatelier\u2019s Principle.<\/h5>\r\nHenri Louis Le Chatelier (born 1850) was an influential chemist and engineer. His landmark chemical principle states that for any change made to a chemical system at equilibrium, the system will respond to minimize the change. In other words, add a reagent to a reaction at equilibrium and the reaction will tend to be driven to the right, i.e. to form products, consuming some of the reactant, and lessening the added concentration of it. Likewise, add a product of a reaction and the reaction will tend to be driven toward the left, i.e. to form reactants. Similarly for heat, if a reaction is endothermic (\u0394H > 0) then raising the temperature (adding heat) will usually drive the reaction to the right. And if the reaction is exothermic (\u0394H < 0), raising the temperature will usually drive the reaction toward the left, or lessen its tendency to proceed to the right. The extent of these effects depends on the concentration of substance added, \u0394H, the equilibrium constant and the composition of the reaction mixture. The value of this simple principle is that it allows one to make qualitative (and quantitative) predictions of the effects of changes on a reaction system.\r\n
(e) Relationship to Thermodynamics.<\/h5>\r\nThe equilibrium constant for a reaction is related to the Gibbs free energy change for the reaction as follows:\r\n\r\n\\[ \\Delta G^\\circ = -RT \\ln K = -2.303RT \\log K \\tag{32}\\]\r\n\r\nwhere R is the ideal gas constant (8.31441 J\/mol K), T is the temperature in Kelvin and \u0394G\u00b0 is the standard Gibbs free energy change for the reaction. The term standard refers to standard state, i.e. 1 atm pressure, pure compounds in their most stable states (for gases, liquids and solids) and 1 molal solutions behaving as if ideal. If one knows the standard free energy change for a reaction then the equilibrium constant can be calculated, or vice versa. The procedure in part (c), involving adding reactions and calculating new equilibrium constants may also be used to determine the equilibrium constant of a new reaction. For the general reaction.\r\n\r\n\\[ \\ce{aA + bB <=> cC + dD} \\tag{33}\\]\r\n\r\nthe \u0394G\u00b0 for the reaction is given by the appropriate combination of \u0394G\u00b0f values for the species involved:\r\n\r\n\\[ \\Delta G^\\circ = c\\Delta G^\\circ_f(C) + d\\Delta G^\\circ_f(D) - a\\Delta G^\\circ_f(A) - b\\Delta G^\\circ_f(B) \\tag{34}\\]\r\n\r\nThe \u0394G\u00b0f<\/sub> is called the free energy of formation. These refer to \u0394G\u00b0 for reactions which form a species from its elements, in the case of neutral compounds and elements plus H+<\/sup> and H2<\/sub> in the case of ions. Reactants and products are understood to be in their standard states (as above). The usual standard state for a solution is 1 atm pressure, unit molality and treated as if ideal. Finally, the elements in their standard states are assigned \u0394G\u00b0f<\/sub> = 0. (A detailed explanation of these considerations can be found in any physical chemistry or thermodynamics text.) The following are some examples:\r\n\r\n\\[ \\ce{H2_{(g)} + 1\/8S8_{(s)} <=> H2S_{(g)}} \\quad \\Delta G^\\circ = \\Delta G^\\circ_f(\\ce{H2S_{(g)}}) \\tag{35}\\]\r\n\r\n\\[ \\ce{H2_{(g)} + 1\/8S8_{(s)} <=> H2S_{(aq)}} \\quad \\Delta G^\\circ = \\Delta G^\\circ_f(\\ce{H2S_{(aq)}}) \\tag{36}\\]\r\n\r\nPure H2<\/sub>S is a gas and has a different \u0394G\u00b0f<\/sub> than does H2<\/sub>S formed in solution. Thus,\r\n\r\n\\[ \\ce{H2S_{(g)} <=> H2S_{_{(aq)}}} \\tag{37}\\]\r\n\r\nis a reaction with,\r\n\r\n\\[ \\Delta G^\\circ = \\Delta G^\\circ_f(\\ce{H2S_{(aq)}}) - \\Delta G^\\circ_f(\\ce{H2S_{(g)}}) = -27.83 - (-33.56)\\ \\text{kJ\/mol} = 5.73\\ \\text{kJ\/mol} \\tag{38}\\]\r\n\r\nfrom which K can be found:\r\n\r\n\\[ K = e^{-\\Delta G^\\circ\/RT} = e^{-5730\\ \\text{J\/mol} \/ (8.314\\ \\text{J\/mol\u00b7K} \\times 298.15\\ \\text{K})} = 0.099 \\tag{39}\\]\r\n\r\nH2<\/sub>S is modestly soluble in water.\r\n\r\nSome other example formation reactions are:\r\n\r\n\\[ \\ce{2Fe_{(s)} + 3\/2O2_{(g)} <=> Fe2O3_{(s)}} \\tag{40}\\]\r\n\r\n\\[ \\ce{2Cu_{(s)} + C_{(s)} + 5\/2O2_{(g)} + H2_{(g)} <=> Cu2CO3(OH)2_{(s)}} \\tag{41}\\]\r\n\r\n\\[ \\ce{Fe_{(s)} + 1\/8S8_{(s)} + 5O2_{(g)} + 6H2_{(g)} <=> FeSO4\u00b76H2O_{(s)}} \\tag{42}\\]\r\n\r\nNote that Fe(SO4<\/sub>)\u00b76H2<\/sub>O is an ionic solid, but as a neutral compound we can write its formation reaction as above.\r\n\r\nFor ions in solution we can also write formation reactions and obtain their \u0394G\u00b0f<\/sub>\u00a0 values. Ions, however, have charge and we need to eliminate that. (We cannot have a charge imbalance, as noted in the section on balancing chemical reactions.) The \u0394G\u00b0f<\/sub> of H+<\/sup> is also assigned a value of 0 kJ\/mol. This allows us to write formation reactions for ions, for which the measured \u0394G\u00b0 = \u0394G\u00b0f<\/sub>. For example, we can write the following formation reactions by the procedures shown below:\r\n\r\n\\[ \\ce{Cu_{(s)} <=> Cu^2+_{(aq)} + 2e^-} \\tag{43}\\]\r\n\r\n\\[ \\ce{2H^+_{(aq)} + 2e^- <=> H2_{(g)}} \\tag{44}\\]\r\n\r\n\\[ \\ce{Cu_{(s)} + 2H^+_{(aq)} <=> Cu^2+_{(aq)} + H2_{(g)}} \\tag{45}\\]\r\n\r\n\\[ \\ce{1\/8S8_{(s)} + 2O2_{(g)} + 2e^- <=> SO4^2-_{(aq)}} \\tag{46}\\]\r\n\r\n\\[ \\ce{H2_{(g)} <=> 2H^+_{(aq)} + 2e^-} \\tag{47}\\]\r\n\r\n\\[ \\ce{1\/8S8_{(s)} + 2O2_{(g)} + H2_{(g)} <=> SO4^2-_{(aq)} + 2H^+_{(aq)}} \\tag{48}\\]\r\n\r\nFormation reactions for any ions can be written as illustrated above. The \u0394G\u00b0f<\/sub> for all reactants in formation reactions are zero. Then \u0394G\u00b0 for a formation reaction then is \u0394G\u00b0f<\/sub> for the one product of the reaction. However, a certain amount of care is needed. It will be noted that the products on the right of reaction (48) are equal to those for dissociated sulfuric acid. However, what is intended here is the formation reaction for sulfate ion in water. If we want the \u0394G\u00b0f<\/sub> for pure H2<\/sub>SO4<\/sub> we have to write:\r\n\r\n\\[ \\ce{H2_{(g)} + 1\/8S8_{(s)} + 2O2_{(g)} <=> H2SO4_{(l)}} \\tag{49}\\]\r\n\r\nAnd for aqueous sulfuric acid we write:\r\n\r\n\\[ \\ce{H2_{(g)} + 1\/8S8_{(s)} + 2O2_{(g)} <=> H2SO4_{(aq)}} \\tag{50}\\]\r\n\r\nThe Reaction Quotient. When we have some combination of reactants and products for a reaction, under some set of conditions (concentration, pressure and temperature), there are three possibilities:\r\n
    \r\n \t
  1. Reaction proceeds from left to right.<\/li>\r\n \t
  2. Reaction proceeds from right to left.<\/li>\r\n \t
  3. No reaction occurs; the system is at equilibrium.<\/li>\r\n<\/ol>\r\nHow can we know what will occur? How big is the driving force for it to occur? First, we can write the reaction quotient:\r\n\r\n\\[Q = \\frac{[C]^c [D]^d}{[A]^a [B]^b}\\tag{51}\\]\r\n\r\nwhere the concentrations are not necessarily those at equilibrium! (Properly, we should write activities.) If Q < K then [C] and [D] are less than they would be at equilibrium and [A] and [B] are greater than they would be at equilibrium. The reaction will proceed to the right. Conversely, if Q > K the reaction will proceed to the left. And if Q = K then the system is at equilibrium.\r\n\r\nSince \u0394G\u00b0 is related to K, it tells us the magnitude of the driving force for the reaction under standard conditions. Likewise \u0394G tells us the magnitude of the driving force for the reaction under more general, non-standard conditions. If \u0394G is positive then this says that the reaction as written is unfavourable and will proceed from right to left (\u0394G\u00b0 \u00b5 -lnK). Furthermore, the bigger the value of \u0394G the farther we are from equilibrium. Likewise if \u0394G < 0 the reaction as written is favourable; it proceeds from left to right. The bigger the magnitude of \u0394G, the stronger the driving force and the farther we are from equilibrium. Hence when \u0394G = 0 we are at equilibrium; there is no driving force left for any change in composition. By this point it should be evident that \u0394G and Q for a reaction should be related somehow. The relationship is:\r\n\r\n\\[ \\Delta G = \\Delta G^\\circ + RT \\ln Q \\tag{52}\\]\r\n\r\nThis simply reverts to \u0394G\u00b0 = -RT lnK when Q = K, because at equilibrium \u0394G = 0. These considerations give quantitative expression to Le Chatelier\u2019s qualitative deductions. Since,\r\n\r\n\\[ \\Delta G = \\Delta H - T\\Delta S \\tag{53}\\]\r\n\r\nwe can see why increasing temperature (the effect is to increase heat input to the reaction system) causes the effects rationalized by Le Chatelier\u2019s principle. However, it can be a bit more complex than the effects of enthalpy change alone.\u00a0 Often near room temperature \u0394H > |T\u0394S| and the \u0394H term dominates. Where \u0394S has the same sign as \u0394H the effect is amplified. If \u0394S has the opposite sign and the temperature is high, then the T\u0394S term may dominate. This happens for reactions where there is more gas produced than consumed, since gases generally have much higher entropies than condensed phases.\r\n\r\n<\/div>","rendered":"
    \n

    1.1 The Concept of Equilibrium.<\/h2>\n

    All chemical reactions have a natural spontaneous tendency to move toward equilibrium, however fast or slow the reaction may be. Chemical equilibrium is the state where no net change in composition occurs under specified conditions. More specifically, at equilibrium, all species have the same thermodynamic chemical potential. The hypothetical reaction,<\/p>\n

    \\[\\ce{A + B <–> C + D}\\tag{1}\\]<\/p>\n

    is at equilibrium when the composition of a mixture containing A, B, C and D cannot change. (Of course if conditions change, such as temperature or addition of one of A or B etc, then change will occur.) The state of equilibrium does not imply that all chemical processes have stopped. Rather, the rates of the forward and the reverse reactions at equilibrium are equal. Hence there is no net change in composition with time. Equilibrium is a dynamic balance of forward and reverse reactions rates. Phenomenologically, this is how equilibrium is defined.<\/p>\n

    For reaction (1), the reactants A and B react to form C and D at a rate that can be stated as the change in concentration of A or B with time, i.e. rate = -d[A]\/dt = -d[B]\/dt\u00a0 (where [A] is concentration of A, etc. and t is time; negative because the reactants are decreasing in concentration). But C and D also react to form A and B in a reverse reaction, at a rate = -d[C]\/dt = -d[D]\/dt, as indicated by the double-headed arrow. If the rate of the forward reaction is fast relative to that of the back reaction, C and D will predominate when a stoichiommetric mixture of A and B are combined. If the forward reaction is slow relative to the back reaction, little of C and D will form. At equilibrium the forward and backward reaction rates are equal. (Not equal rate constants, but equal actual rates in concentration per unit time.) Some reactions will go virtually to completion. For example, when an equimolar mixture of H2<\/sub>\u00a0and O2 <\/sub>gas\u00a0is ignited, the reaction is,<\/p>\n

    \\[\\ce{H2_{(g)} + O2_{(g)} = H2O_{(l)}} \\tag{2}\\]<\/p>\n

    In the case of reactions that go to completion, as in the H2<\/sub> + O2<\/sub> case, the forward reaction is so fast compared to the reverse reaction that virtually no net reverse reaction occurs. Then the equilibrium constant is immense; e.g. 3.5 x 1041<\/sup> for reaction (2).<\/p>\n

    Note:<\/strong> That a mixture of A, B, C and D is not undergoing any compositional change is not necessarily indicative of equilibrium having been attained. For instance, a mixture of H2<\/sub> and O2 <\/sub>in a container will persist without change for a long time. But, we know that this is not the equilibrium situation: provide a spark or a suitable catalytic surface, and water will form until one or the other of H2<\/sub> or O2<\/sub> is consumed. The equilibrium state is water plus left over excess O2 <\/sub>or H2<\/sub>, whichever was present in greater amount to start with.<\/p><\/blockquote>\n

    This illustrates a key point, which is the distinction between kinetics and thermodynamics. In the limit of infinite time, equilibrium will be attained. Thermodynamics will prevail. It\u2019s just that thermodynamics cannot tell us how fast the reactions will be. That is the province of kinetics. Many hydrometallurgical processes are too slow to attain equilibrium within the residence time provided because the temperatures are typically quite low.<\/p>\n

    1.2 The Equilibrium Constant<\/h2>\n

    Equilibrium constants are defined in terms of the activity of each species involved in the reaction. Activity is a modified concentration. It is observed that as concentration of a solute increases, the solution\u2019s apparent concentration is not equivalent to its actual concentration. For example, the rate of reaction (1) was expressed as -d[A]\/dt, where [A] is the actual concentration of reactant A. Over a limited range of concentration this holds true, especially at low concentration. But, as concentration increases it becomes apparent that there are deviations from this simple rule. The rate may appear to be slower than would be expected; reactant A is behaving as if its concentration is less than its true concentration. We refer to the apparent concentration as the activity. The effect is much more pronounced with ionic solutes than with neutral ones.<\/p>\n

    The origins of the activity have to do with interactions between the solvent and the solute, and between the solutes themselves. When the solution is very dilute the solvent molecules keep the solutes ions quite far apart, and the solutes interact only weakly. But as the solution gets more concentrated, the solutes get closer together, and the interaction starts to become strong. This makes them less readily able to participate in chemical reactions; they behave as if their concentration is actually less than what it actually is. Their activity is lower than their concentration. In cases where very high concentration solutions can be obtained the activity may become significantly greater than the concentration.<\/p>\n

    The activity takes into account the fact that many solutions are non-ideal. For ideal solutions the solute and the solvent do not interact significantly. There are numerous cases where solutions of molecular solutes do behave ideally over a wide range of concentrations. But aqueous solutions, particularly of ionic solutes are highly non-ideal.<\/p>\n

    For the general reaction,<\/p>\n

    \\[\\ce{aA + bB = cC + dD}\\tag{3}\\]<\/p>\n

    where a, b, c and d are the stoichiommetric coefficients, the equilibrium constant is defined as,<\/p>\n

    \\[
    \nK = \\frac{a_C^c\\, a_D^d}{a_A^a\\, a_B^b}\\tag{4}
    \n\\]<\/p>\n

    where aC<\/sub>c<\/sup> etc. are the activities of the various species raised to the powers of their stoichiommetric coefficients. For a solution that behaves ideally the activities are equal to the concentrations. Aqueous solutions of ionic solutes will behave ideally at quite low concentrations, i.e. <10-3<\/sup> M for singly charged anions; lower still for more highly charged ions. Generally though, solutions of practical interest in hydrometallurgy are quite concentrated and hence far from ideal. Unfortunately activities are usually quite difficult to measure.<\/p>\n

    For the gas phase we can often assume the ideal gas law. Pressure is a proxy for concentration when dealing with gases. For an ideal gas,<\/p>\n

    \\[ P = \\frac{nRT}{V} = \\frac{n(RT)}{V} \\tag{5}\\]<\/p>\n

    where n\/V (moles\/L) is equal to concentration. Again, however, interactions between gas molecules lead to deviations from ideality, and then in general we need an activity for gases in order to take attractive and repulsive interactions into account. For gases we call this activity the fugacity. It can be thought of as an effective pressure. There are good methods for estimating or measuring fugacities. At room temperature, typically 20\u00b0C, and 1 atm pressure the ideal gas law is a moderately good approximation for many gases. As pressure increases and\/or temperature decreases the ideal gas law becomes a poorer approximation.<\/p>\n

    Where a gas is involved in a reaction, we include it in the equilibrium constant expression by using its fugacity, e.g. for:<\/p>\n

    \\[\\ce{aA + bB = cC + eE_{(g)}}\\tag{6}\\]<\/p>\n

    \\[ K = \\frac{a_C^c \\, f_E^e}{a_A^a \\, a_B^b}\\tag{7}\\]<\/p>\n

    Solution concentration is commonly expressed in terms of molarity (M), molality (m) or mole fraction (X). Molality has units of moles of solute per kg of solvent (not kg of solution!). It is commonly used in thermodynamics because it is independent of solution density, whereas molarity varies with density; the volume of the solution changes with temperature. Mole fraction is less commonly used. It is equal to moles of a species divided by the total moles of all species. To illustrate, a 1.032 M NaCl solution at 20\u00b0C has a molality of 1.054 and a mole fraction of 0.01863. For dilute aqueous solutions with densities close to 1 g\/mL, molality and molarity do not differ greatly. For concentrated solutions, with densities significantly different from that of pure water, molality and molarity diverge.<\/p>\n

    The activity is defined so as to be unitless, which is helpful because it is used heavily in logarithmic functions. To make this work the activity is defined as the ratio:<\/p>\n

    \\[ a = \\frac{\\text{absolute activity}}{\\text{standard state activity}}\\tag{8}\\]<\/p>\n

    Absolute activity is the effective concentration; it has concentration units. The standard state is a reference point which we define so as to have unit activity. The common standard state for solutions is 1 molal and behaving as if ideal. It is a hypothetical state; most real solutions at unit molality are far from ideal. But, the construct is helpful. The activity of a solute in a solution as defined in equation [4] is numerically equal to its absolute activity. The standard state for gases is 1 bar pressure and acting as if ideal. (1 bar = 0.987 atm.) For liquids and gases the standard state is the pure liquid or solid. (A fuller discussion of these ideas is usually presented in thermodynamics courses and texts.)<\/p>\n

    Where pure solids, for example, are involved in a reaction the activity as defined by equation (8) is,<\/p>\n

    \\[\\frac{\\text(activity of pure solid)}{\\text(activity of pure solid)} =\u00a0 a\u00a0 = 1\\tag{9}\\]<\/p>\n

    In other words, the pure solid is in its standard state. If the solid is not pure, i.e. it is a solid solution, or a mixture, or it is in a chemical form that is less stable than the standard state form, then the activity is not equal to 1. (Many compounds can exist in a number of different crystalline forms or compositions; elemental sulfur might be formed not as the stable S8<\/sub>, but as polymeric sulfur, for example.)<\/p>\n

    To a first approximation the equilibrium constant is often expressed in terms of the concentrations and pressures of solutes and gases, respectively. The approximation can be rather poor at high concentrations, but for the purposes of this course the important principles can be presented without needing to develop the concept of activity further than this. For accurate work, activities and fugacities are required.)<\/p>\n

    We will take the equilibrium constant to involve concentrations in molal or molar and gas pressures in atm. (Strictly speaking we should use pressure in bars, but much older and still useful data is referenced to pressure in atm.) As above, the activities of pure solids and liquids are taken to be 1. Where water is involved in the chemical reaction, we will treat it as if it is a pure liquid. (Properly one would include its activity; in dilute solution this is very close to 1, but in concentrated solutions it may actually be considerably lower than 1.) The concentration of water can be found from its density (0.998 g\/mL at 20\u00b0C) and molecular weight:<\/p>\n

    \\[
    \n\\frac{998\\ \\text{g H2O}}{1\\ \\text{L}}\\times
    \n\\frac{1\\ \\text{mol H2O}}{18.015\\ \\text{g H2O}}
    \n= 55.4\\ \\text{mol\/L}
    \n\\tag{10}\\]<\/p>\n

    Hence even a 1 M solution is still roughly 98% water.<\/p>\n

    1.3 Understanding and Using Equilibrium Constants.<\/h2>\n

    1. Numerical Significance.<\/h3>\n

    Where equilibrium constants are concerned, size matters. A large equilibrium constant indicates that a reaction is very favourable and will proceed to the right to a very high degree. A very small equilibrium constant indicates a very unfavourable reaction, which will proceed to the right only to a very small extent. When the equilibrium constant is near one, similar proportions of reactants and products will result. For example, water dissociates weakly into H+<\/sup> and OH–<\/sup>:<\/p>\n

    \\[\\ce{H2O_{(l)} = H+_{(aq)} + OH-_{(aq)}}\\tag{11}\\]<\/p>\n

    \\[\\ce{K_{w25\u00b0C} = [H+][OH-] = 1 x 10^{-14}} \\tag{12}\\]<\/p>\n

    Only about 2 H2<\/sub>O molecules per billion dissociate on average. On the other hand, if we added solid NaOH (strong base) to a solution of HCl (strong acid), the reaction would be:<\/p>\n

    \\[\\ce{NaOH_{(s)} = Na+_{(aq)} + OH-_{(aq)}}\\tag{13}\\]<\/p>\n

    \\[\\ce{H+_{(aq)} + OH-_{(aq)} = H2O_{(l)}}\\tag{14}\\]<\/p>\n

    Clearly, if reaction (11) is unfavourable, reaction (12) is very favourable. It is important to remember that in principle reactions can proceed in either direction. This is the basis for equilibrium.<\/p>\n

    2. Manipulating Equilibrium Constants.<\/h3>\n

    (a). Reversing the reaction.<\/h5>\n

    What is the equilibrium constant for reaction (14)?<\/p>\n

    \\[K = \\frac{1}{[H^+][OH^-]}
    \n= \\frac{1}{1 \\times 10^{-14}}
    \n= 1 \\times 10^{14}\\tag{15}\\]<\/p>\n

    In other words, when the reaction is reversed, the equilibrium constant is inverted. For,<\/p>\n

    \\[aA + bB = cC
    \n\\qquad
    \nK = \\frac{[C]^c}{[A]^a\\, [B]^b}
    \n\\tag{16}\\]<\/p>\n

    and for the reverse reaction,<\/p>\n

    \\[
    \ncC = bB + aA
    \n\\qquad
    \nK’ = \\frac{[A]^a [B]^b}{[C]^c} = \\frac{1}{K}\\tag{17}\\]<\/p>\n

    (b). Multiplying the Reaction by a Factor.<\/h5>\n

    The only constraint on the stoichiommetric coefficients is that the reaction must be balanced. Other than that they can be any numbers. When a reaction is written with different stoichiommetric coefficients the equilibrium constant changes. The same is true if a reaction is multiplied by some factor. Consider the reaction,<\/p>\n

    \\[ \\ce{NH3 + H2O = NH4+ + OH-} \\tag{18}\\]<\/p>\n

    \\[ K = \\frac{\\ce{[NH4+][OH-]}}{\\ce{[NH3]^2}} = 1.8 \\times 10^{-5} \\tag{19}\\]<\/p>\n

    We can also write the reaction as,<\/p>\n

    \\[ \\ce{2NH3 + 2H2O = 2NH4+ + 2OH-} \\tag{20}\\]<\/p>\n

    \\[ K = \\frac{\\ce{[NH4+]^2[OH-]^2}}{\\ce{[NH3]^2}} = (1.8 \\times 10^{-5})^2 = 3.2 \\times 10^{-10} \\tag{21}\\]<\/p>\n

    In general, multiply a reaction by n, raise its equilibrium constant to the power n.<\/p>\n

    (c). Adding Reactions.<\/h5>\n

    Balanced reactions are statements of mass balance. Just like any arithmetic equation, they can be added together in the same way. When two or more reactions are added together, their equilibrium constants are multiplied. An example is the series of reactions listed below. Mercury oxide will react readily with thiosulfate ion to form a very stable complex. This could afford an important way to control mercury in some hydrometallurgical processes. This reaction can be written as:<\/p>\n

    \\[ \\ce{HgO_{(s)} + 2S2O3^2- + H2O_{(l)} = [Hg(S2O3)2]^2- + 2OH-} \\tag{22}\\]<\/p>\n

    This reaction can be split into two, the sum of which is the preceding reaction:<\/p>\n

    First Half<\/p>\n

    \\[ \\ce{HgO_{(s)} + H2O_{(l)} = Hg^2+ + 2OH-} \\tag{23}\\]<\/p>\n

    \\[K_{\\mathrm{sp}} = [\\mathrm{Hg}^{2+}][\\mathrm{OH}^{-}]^{2} = 3.6 \\times 10^{-26}\\tag{24}\\]<\/p>\n

    Second Half:<\/p>\n

    \\[ \\ce{Hg^2+ + 2S2O3^2- = [Hg(S2O3)2]^2-} \\tag{25}\\]<\/p>\n

    \\[K = \\frac{[\\text{Hg(S2O3)2}^{2-}]}{[\\text{Hg}^{2+}]\\,[\\text{S2O3}^{2-}]^{2}}= 1 \\times 10^{32}\\tag{26}\\]<\/p>\n

    Combine:<\/p>\n

    \\[ \\ce{HgO_{(s)} + H2O_{(l)} = Hg^2+_{(aq)} + 2OH-_{(aq)}} \\tag{27}\\]<\/p>\n

    \\[ \\ce{Hg^2+_{(aq)} + 2S2O3^2-_{(aq)} = [Hg(S2O3)2]^2-_{(aq)}} \\tag{28}\\]<\/p>\n

    \\[ \\ce{HgO_{(s)} + 2S2O3^2-_{(aq)} + H2O_{(l)} = [Hg(S2O3)2]^2-_{(aq)} + 2OH-_{(aq)}} \\tag{29}\\]<\/p>\n

    The equilibrium constant for reaction (22) is,<\/p>\n

    \\[K’ = \\frac{[\\mathrm{Hg(S_2O_3)_2}^{2-}][\\mathrm{OH^-}]^{2}}
    \n{[\\mathrm{S_2O_3}^{2-}]^{2}}
    \n= \\frac{[\\mathrm{Hg(S_2O_3)_2}^{2-}][\\mathrm{Hg^{2+}}][\\mathrm{OH^-}]^{2}}
    \n{[\\mathrm{S_2O_3}^{2-}]^{2}[\\mathrm{Hg^{2+}}]}\\tag{30}\\]<\/p>\n

    The latter part of equation [15] can be clearly seen to be equal to K Ksp:<\/p>\n

    \\[ K’ = K \\cdot K_{sp} = 3.6 \\times 10^6 \\tag{31}\\]<\/p>\n

    (d) Le Chatelier\u2019s Principle.<\/h5>\n

    Henri Louis Le Chatelier (born 1850) was an influential chemist and engineer. His landmark chemical principle states that for any change made to a chemical system at equilibrium, the system will respond to minimize the change. In other words, add a reagent to a reaction at equilibrium and the reaction will tend to be driven to the right, i.e. to form products, consuming some of the reactant, and lessening the added concentration of it. Likewise, add a product of a reaction and the reaction will tend to be driven toward the left, i.e. to form reactants. Similarly for heat, if a reaction is endothermic (\u0394H > 0) then raising the temperature (adding heat) will usually drive the reaction to the right. And if the reaction is exothermic (\u0394H < 0), raising the temperature will usually drive the reaction toward the left, or lessen its tendency to proceed to the right. The extent of these effects depends on the concentration of substance added, \u0394H, the equilibrium constant and the composition of the reaction mixture. The value of this simple principle is that it allows one to make qualitative (and quantitative) predictions of the effects of changes on a reaction system.<\/p>\n

    (e) Relationship to Thermodynamics.<\/h5>\n

    The equilibrium constant for a reaction is related to the Gibbs free energy change for the reaction as follows:<\/p>\n

    \\[ \\Delta G^\\circ = -RT \\ln K = -2.303RT \\log K \\tag{32}\\]<\/p>\n

    where R is the ideal gas constant (8.31441 J\/mol K), T is the temperature in Kelvin and \u0394G\u00b0 is the standard Gibbs free energy change for the reaction. The term standard refers to standard state, i.e. 1 atm pressure, pure compounds in their most stable states (for gases, liquids and solids) and 1 molal solutions behaving as if ideal. If one knows the standard free energy change for a reaction then the equilibrium constant can be calculated, or vice versa. The procedure in part (c), involving adding reactions and calculating new equilibrium constants may also be used to determine the equilibrium constant of a new reaction. For the general reaction.<\/p>\n

    \\[ \\ce{aA + bB <=> cC + dD} \\tag{33}\\]<\/p>\n

    the \u0394G\u00b0 for the reaction is given by the appropriate combination of \u0394G\u00b0f values for the species involved:<\/p>\n

    \\[ \\Delta G^\\circ = c\\Delta G^\\circ_f(C) + d\\Delta G^\\circ_f(D) – a\\Delta G^\\circ_f(A) – b\\Delta G^\\circ_f(B) \\tag{34}\\]<\/p>\n

    The \u0394G\u00b0f<\/sub> is called the free energy of formation. These refer to \u0394G\u00b0 for reactions which form a species from its elements, in the case of neutral compounds and elements plus H+<\/sup> and H2<\/sub> in the case of ions. Reactants and products are understood to be in their standard states (as above). The usual standard state for a solution is 1 atm pressure, unit molality and treated as if ideal. Finally, the elements in their standard states are assigned \u0394G\u00b0f<\/sub> = 0. (A detailed explanation of these considerations can be found in any physical chemistry or thermodynamics text.) The following are some examples:<\/p>\n

    \\[ \\ce{H2_{(g)} + 1\/8S8_{(s)} <=> H2S_{(g)}} \\quad \\Delta G^\\circ = \\Delta G^\\circ_f(\\ce{H2S_{(g)}}) \\tag{35}\\]<\/p>\n

    \\[ \\ce{H2_{(g)} + 1\/8S8_{(s)} <=> H2S_{(aq)}} \\quad \\Delta G^\\circ = \\Delta G^\\circ_f(\\ce{H2S_{(aq)}}) \\tag{36}\\]<\/p>\n

    Pure H2<\/sub>S is a gas and has a different \u0394G\u00b0f<\/sub> than does H2<\/sub>S formed in solution. Thus,<\/p>\n

    \\[ \\ce{H2S_{(g)} <=> H2S_{_{(aq)}}} \\tag{37}\\]<\/p>\n

    is a reaction with,<\/p>\n

    \\[ \\Delta G^\\circ = \\Delta G^\\circ_f(\\ce{H2S_{(aq)}}) – \\Delta G^\\circ_f(\\ce{H2S_{(g)}}) = -27.83 – (-33.56)\\ \\text{kJ\/mol} = 5.73\\ \\text{kJ\/mol} \\tag{38}\\]<\/p>\n

    from which K can be found:<\/p>\n

    \\[ K = e^{-\\Delta G^\\circ\/RT} = e^{-5730\\ \\text{J\/mol} \/ (8.314\\ \\text{J\/mol\u00b7K} \\times 298.15\\ \\text{K})} = 0.099 \\tag{39}\\]<\/p>\n

    H2<\/sub>S is modestly soluble in water.<\/p>\n

    Some other example formation reactions are:<\/p>\n

    \\[ \\ce{2Fe_{(s)} + 3\/2O2_{(g)} <=> Fe2O3_{(s)}} \\tag{40}\\]<\/p>\n

    \\[ \\ce{2Cu_{(s)} + C_{(s)} + 5\/2O2_{(g)} + H2_{(g)} <=> Cu2CO3(OH)2_{(s)}} \\tag{41}\\]<\/p>\n

    \\[ \\ce{Fe_{(s)} + 1\/8S8_{(s)} + 5O2_{(g)} + 6H2_{(g)} <=> FeSO4\u00b76H2O_{(s)}} \\tag{42}\\]<\/p>\n

    Note that Fe(SO4<\/sub>)\u00b76H2<\/sub>O is an ionic solid, but as a neutral compound we can write its formation reaction as above.<\/p>\n

    For ions in solution we can also write formation reactions and obtain their \u0394G\u00b0f<\/sub>\u00a0 values. Ions, however, have charge and we need to eliminate that. (We cannot have a charge imbalance, as noted in the section on balancing chemical reactions.) The \u0394G\u00b0f<\/sub> of H+<\/sup> is also assigned a value of 0 kJ\/mol. This allows us to write formation reactions for ions, for which the measured \u0394G\u00b0 = \u0394G\u00b0f<\/sub>. For example, we can write the following formation reactions by the procedures shown below:<\/p>\n

    \\[ \\ce{Cu_{(s)} <=> Cu^2+_{(aq)} + 2e^-} \\tag{43}\\]<\/p>\n

    \\[ \\ce{2H^+_{(aq)} + 2e^- <=> H2_{(g)}} \\tag{44}\\]<\/p>\n

    \\[ \\ce{Cu_{(s)} + 2H^+_{(aq)} <=> Cu^2+_{(aq)} + H2_{(g)}} \\tag{45}\\]<\/p>\n

    \\[ \\ce{1\/8S8_{(s)} + 2O2_{(g)} + 2e^- <=> SO4^2-_{(aq)}} \\tag{46}\\]<\/p>\n

    \\[ \\ce{H2_{(g)} <=> 2H^+_{(aq)} + 2e^-} \\tag{47}\\]<\/p>\n

    \\[ \\ce{1\/8S8_{(s)} + 2O2_{(g)} + H2_{(g)} <=> SO4^2-_{(aq)} + 2H^+_{(aq)}} \\tag{48}\\]<\/p>\n

    Formation reactions for any ions can be written as illustrated above. The \u0394G\u00b0f<\/sub> for all reactants in formation reactions are zero. Then \u0394G\u00b0 for a formation reaction then is \u0394G\u00b0f<\/sub> for the one product of the reaction. However, a certain amount of care is needed. It will be noted that the products on the right of reaction (48) are equal to those for dissociated sulfuric acid. However, what is intended here is the formation reaction for sulfate ion in water. If we want the \u0394G\u00b0f<\/sub> for pure H2<\/sub>SO4<\/sub> we have to write:<\/p>\n

    \\[ \\ce{H2_{(g)} + 1\/8S8_{(s)} + 2O2_{(g)} <=> H2SO4_{(l)}} \\tag{49}\\]<\/p>\n

    And for aqueous sulfuric acid we write:<\/p>\n

    \\[ \\ce{H2_{(g)} + 1\/8S8_{(s)} + 2O2_{(g)} <=> H2SO4_{(aq)}} \\tag{50}\\]<\/p>\n

    The Reaction Quotient. When we have some combination of reactants and products for a reaction, under some set of conditions (concentration, pressure and temperature), there are three possibilities:<\/p>\n

      \n
    1. Reaction proceeds from left to right.<\/li>\n
    2. Reaction proceeds from right to left.<\/li>\n
    3. No reaction occurs; the system is at equilibrium.<\/li>\n<\/ol>\n

      How can we know what will occur? How big is the driving force for it to occur? First, we can write the reaction quotient:<\/p>\n

      \\[Q = \\frac{[C]^c [D]^d}{[A]^a [B]^b}\\tag{51}\\]<\/p>\n

      where the concentrations are not necessarily those at equilibrium! (Properly, we should write activities.) If Q < K then [C] and [D] are less than they would be at equilibrium and [A] and [B] are greater than they would be at equilibrium. The reaction will proceed to the right. Conversely, if Q > K the reaction will proceed to the left. And if Q = K then the system is at equilibrium.<\/p>\n

      Since \u0394G\u00b0 is related to K, it tells us the magnitude of the driving force for the reaction under standard conditions. Likewise \u0394G tells us the magnitude of the driving force for the reaction under more general, non-standard conditions. If \u0394G is positive then this says that the reaction as written is unfavourable and will proceed from right to left (\u0394G\u00b0 \u00b5 -lnK). Furthermore, the bigger the value of \u0394G the farther we are from equilibrium. Likewise if \u0394G < 0 the reaction as written is favourable; it proceeds from left to right. The bigger the magnitude of \u0394G, the stronger the driving force and the farther we are from equilibrium. Hence when \u0394G = 0 we are at equilibrium; there is no driving force left for any change in composition. By this point it should be evident that \u0394G and Q for a reaction should be related somehow. The relationship is:<\/p>\n

      \\[ \\Delta G = \\Delta G^\\circ + RT \\ln Q \\tag{52}\\]<\/p>\n

      This simply reverts to \u0394G\u00b0 = -RT lnK when Q = K, because at equilibrium \u0394G = 0. These considerations give quantitative expression to Le Chatelier\u2019s qualitative deductions. Since,<\/p>\n

      \\[ \\Delta G = \\Delta H – T\\Delta S \\tag{53}\\]<\/p>\n

      we can see why increasing temperature (the effect is to increase heat input to the reaction system) causes the effects rationalized by Le Chatelier\u2019s principle. However, it can be a bit more complex than the effects of enthalpy change alone.\u00a0 Often near room temperature \u0394H > |T\u0394S| and the \u0394H term dominates. Where \u0394S has the same sign as \u0394H the effect is amplified. If \u0394S has the opposite sign and the temperature is high, then the T\u0394S term may dominate. This happens for reactions where there is more gas produced than consumed, since gases generally have much higher entropies than condensed phases.<\/p>\n<\/div>\n","protected":false},"author":2529,"menu_order":1,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1203","chapter","type-chapter","status-publish","hentry"],"part":1188,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/1203","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/users\/2529"}],"version-history":[{"count":26,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/1203\/revisions"}],"predecessor-version":[{"id":3858,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/1203\/revisions\/3858"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/parts\/1188"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/1203\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/media?parent=1203"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapter-type?post=1203"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/contributor?post=1203"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/license?post=1203"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}