James Richard Fromm
A state of chemical equilibrium is a dynamic state in which chemical reaction is occurring in both directions, as in the example of H2(g) + I2(g) 2HI(g). That is, the reaction is reversible. In this reaction, one is continuously forming HI, and continuously destroying it (at same rate, when at equilibrium) so that there is no net change in the quantity of hydrogen, iodine, or hydrogen iodide present. Therefore, for any reaction at chemical equilibrium, RATE FORWARD = RATE BACKWARD. In this case, for the reaction written above, the rate of the forward reaction is
kfp(H2)p(I2) and the rate of the reverse reaction is kb p(HI)p(HI). Then at equilibrium
kf p(H2)p(I2) = kb p(HI)p(HI), so that we can now define the equilibrium constant K as the ratio of the rate constants kf /kb:
kf /kb = p(HI)2/p(H2)p(I2) = K
In general, this can be written as:
K = p(products)POWERS/p(reactants)POWERS
For gases, at low pressures, we can use the partial pressure of the gas as the measure of the active mass of the gas. By active mass we mean here mean the reactive amount of substance per unit volume, or per space. We have also referred to this as the chemical activity, for which the usual symbol is a. For gases, the common measure of amount of substance/volume is pressure; a(gas) is taken as approximately equal to p(gas).
For solutes, a more common measure of amount of substance per unit volume is the molar concentration, usually measured in moles/dm3, for which the usual symbol is c; a(solute) is taken as approximately equal to c(solute).
For pure elements or compounds, the amount of substance per unit volume is a constant, because the molar mass of a compound is a constant and so is its density; a(pure solid or pure liquid) is therefore taken as a constant which wherever possible is given the value of exactly one.
Example. The density of water is 1000 g/dm3; the molar mass of water is 18.0152 g/mol. The molar concentration of water in pure water is therefore 1000/18.0152 = 55.51 mol/litre.
Example. The density of solid AgCl is 5.6 g/mL or 5600 g/L; the molar mass of AgCl is 143.32 g/mol. The molar concentration of AgCl in pure solid AgCl is therefore 5600/143.32 = 39.07 mol/litre.
When a reaction occurs entirely in the gas phase, the activities of the gaseous reactants and products are replaced by their partial pressures. These partial pressures are the partial pressures which actually prevail at equilibrium, not the partial pressures of the gases which may have been mixed initially.
Example. The equilibrium constant for the reaction H2(g) + Cl2(g) 2HCl(g) is written as:
K = a(HCl)a(HCl)/a(H2)a(Cl2) = a2(HCl)/a(H2)a(Cl2)
K = p2(HCl)/p(H2)p(Cl2)
The effect of adding reactants to a system at chemical equilibrium is to increase the concentration or partial pressures of the products; the equilibrium concentrations of the reactants will be less than the sum of the original concentrations and those added due to the equilibrium reaction.
Example. The equilibrium for the formation of hydrogen iodide, H2(g) + I2(g) 2HI(g), is governed by the equilibrium constant K = p2(HI)/p(H2)p(I2). If the equilibrium partial pressures of hydrogen and iodine are taken to be, or made to be, exactly 1.0 on some pressure scale (atmospheres, bars, torr, pascals, or kilopascals) then the equilibrium partial pressure m of hydrogen iodide is given by K = x2/(1.0)2. If the value of the equilibrium constant is known, the value of x can be calculated.
If an additional partial pressure z of the reactant hydrogen is added, then the partial pressure of hydrogen would become 1.0 + z. The partial pressure of hydrogen having increased, there is a stress on the equilibrium which is relieved by removal of some partial pressure of hydrogen, y, through the chemical reaction. The stoichiometry of the reaction and the equilibrium constant would then give
K = (x + 2y)2/(1.0 + z - y)(1.0 - y)
because x + 2y is the new equilibrium partial pressure of hydrogen iodide, 1.0 + z - y is the new equilibrium partial pressure of hydrogen, and 1.0 - y is the new equilibrium partial pressure of iodine. The new equilibrium partial pressure of hydrogen is m + 2y because two moles of HI are produced for every mole of H2 which reacts. If the value of x is calculated as above, and the value of the added partial pressure z is known, then the only remaining unknown is y. One equation in one unknown can be solved for y and then the values of the actual partial pressures at the new position of chemical equilibrium can be calculated. We will carry out these calculations in another section.
The effect of adding a reactant to a reaction at equilibrium is to decrease the concentrations of all of the reactants (as well as the concentration of the reactant added, though only to the extent of part of the original increase) and to increase the concentrations of all products. A chemical equilibrium shifts to the right when a component is added on the left, or reactant, side.
The effect of adding products to a system at equilibrium is the reverse of adding reactants. The equilibrium will shift to the left, increasing the pressures or concentrations of the reactants.
Example. For the example already used above, let z now represent the added partial pressure of the product HI. Then K = p2(HI)/p(H2)p(I2) with the equilibrium state before the addition being, as before, K = x2/(1.0)2. The partial pressure of hydrogen iodide after addition and re-equilibration is then x + z - 2y, where y is again the change in partial pressure of hydrogen or iodine due to reaction. These partial pressures are both originally taken as 1.0 (atmospheres, bars, torr, pascals, or kilopascals), so
K = (x + z - 2y)2/(1.0 + y)2
The effect of addition of product is to shift the equilibrium to the left, back toward the reactants. The final concentrations of hydrogen iodide, iodine, and hydrogen will all be higher after addition and re-equilibration than before the addition of hydrogen iodide.
All chemical equilibria are affected to some degree by pressure, but in most cases the equilibrium constant varies only slightly with pressure and this is the only effect of pressure on the equilibrium. When gases are involved in the equilibrium, however, the effect of pressure can be much more significant. This greater effect can be understood in terms of the principle of Le Chatelier.
Example. The Haber process for ammonia synthesis is based on the equilibrium
N2(g) + 3H2(g) 2NH3(g). In this equilibrium, there are four moles of gasesous reactants on the left yielding two moles of gaseous product on the right. Since by Avogadro's hypothesis equal volumes of gases contain equal numbers of molecules, four volumes of gas yield two volumes of gas. An increase in total pressure will be a force tending to decrease the volume of the system described by this equilibrium reaction and constant, and the equilibrium will respond by reacting so as to reduce the total volume - thereby increasing the proportion of ammonia. Conversely, a lower total pressure will tend to increase the proportion of reactants. For this reason, the synthesis of ammonia is normally carried out at high pressure. The same equilibrium, in reverse, is used to prepare reducing atmospheres for heat-treating furnaces. When this is done, the ammonia is dissociated at atmospheric pressure in order to increase the extent of dissociation to hydrogen and nitrogen.
Example. The carbonated beverages so popular in our world are obtained by dissolving carbon dioxide in water. The equilibrium can be written as CO2(g) CO2(aq). Since carbon dioxide in gaseous form is significantly affected by pressure, increasing the CO2 pressure in the bottling plant causes dissolution; when the pressure cap is removed, effervescence occurs as the equilibrium shifts to produce carbon dioxide gas from solution.
Example. The heating of limestone to produce lime involves the equilibrium
CaCO3(s) CaO(s) + CO2(g).
Production will be adversely affected unless the carbon dioxide gas is removed as formed, so provision for ventilation must be made.
Gas phase reactions are not significantly affected by pressure when equal volumes, or moles, of gas are found on the reactant and product sides. Thus there will be no significant shift in the position of the equilibrium
H2(g) + Cl2(g) 2HCl(g)
when the pressure is changed while the position of the equilibrium
2H2(g) + O2(g) 2H2O(g)
will be affected by changes in pressure.
All chemical equilibria are affected by temperature and for most equilibria the effect of temperature is significant. Equilibrium constant values are therefore always given at a particular temperature, often 25oC. Alternatively, they can be given as a function of temperature explicitly in the form of an equation such as K = ((2900/T ) + 4623); a numeric value can be obtained by calculation for the desired temperature. As will be shown in other sections, values of these constants can also be calculated from tabulated values of certain thermodynamic properties of pure substances and thermodynamic properties of aqueous solutes. Such calculations are necessary for quantitative treatment.
Le Chatelier's principle, however, permits a simple qualitative explanation of the effect of temperature on chemical equilibria. An exothermic reaction is one in which heat is produced, as a product, while an endothermic reaction is one in which heat is absorbed, or consumed, as a reactant. Heat, considered as a product or a reactant, acts as a stress just as would the addition of moles of a substance which is a reactant or a product. When heat is a product, the addition of heat which must occur as the temperature is raised drives the equilibrium backward, while if heat is a reactant then the addition of heat drives the equilibrium forward.
Example. The standard molar enthalpy of formation of HI(g) is +135.1 kJ/mole. The enthalpy change of the formation reaction
H2(g) + I2(g) 2HI(g)
is therefore +270.2 kJ and the reaction is endothermic. The effect of adding heat to the equilibrium is
heat + H2(g) + I2(g) 2HI(g)
and therefore the addition of heat will drive the equilibrium to the right. Increasing the temperature will therefore form additional hydrogen iodide.
Example. The standard molar enthalpy of formation of ammonia is -46.11 kJ/mole. The heat of the formation reaction
N2(g) + 3H2(g) 2NH3(g)
is -92.22 kJ/mole and the reaction is exothermic. The effect of adding heat to the equilibrium is
N2(g) + 3H2(g) 2NH3(g) + heat
and therefore the addition of heat will drive the equilibrium to the left. Increasing the temperature will therefore increase the extent of dissociation of ammonia to nitrogen and hydrogen.