Applications of the Equilibrium Constant

The position of the equilibrium CaC03(s) CaO(s) + C02(g) does not depend on the amounts of CaC03(s) and CaO(s) present. 

Thus the equilibrium expressions for the decomposition of solid CaC03 are 

In summary, we can make the following general statement The activity of a pure solid or liquid is always 1. 

Note that the net effect of inserting an activity of 1 into the equilibrium expression for each pure solid or liquid in the reaction has the same effect as simply disregarding them. If pure solids or pure liquids are involved in a chemical reaction, their concentrations are not included in the equilibrium expression for the reaction. This simplification occurs only with pure solids or liquids, not with solutions or gases, because in these last two cases the activity cannot be assumed to be 1. 

For example, in the decomposition of liquid water to gaseous hydrogen and oxygen, 

Knowing the equilibrium constant for a reaction allows us to predict several important features of the reaction the tendency of the reaction to occur (but not the speed of the reaction), whether a given set of concentrations represents an equilibrium condition, and the equilibrium position that will be achieved from a given set of initial concentrations. 

To introduce some of these ideas, we will first consider the reaction 

In a given experiment, the two types of molecules are mixed together in the following amounts  

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After the system reacts and comes to equilibrium, what wil that at equilibrium the ratio 

The direction of the reaction may change, but the expression for the equilibrium constant does not 

Conversely, starting with only C and D, the value of Q is infinite as the values of PA and PB approach zero  

If all four substances are present, Q can have any value between zero and infinity. 

The form of the expression for Q, known as the reaction quotient, is the same as that for the equilibrium constant, K. The difference is that the partial pressures that appear in Q are those that apply at a particular moment, not necessarily when the system is at equilibrium. By comparing the numerical value of Q with that of K, it is possible to decide in which direction the system will move to achieve equilibrium. 

If Q Kthe reaction proceeds from left to right  

Sometimes, knowing only the magnitude of the equilibrium constant, it is possible to decide on the feasibility of a reaction. Consider, for example, a possible method for fixing atmospheric nitrogen—converting it to a compound—by reaction with oxygen  

Because K is so small, the partial pressure of NO in equilibrium with N2 and O2, and hence the amount of NO, must be extremely small, approaching zero. Clearly, this would not be a suitable way to fix nitrogen, at least at 25°C. 

An alternative approach to nitrogen fixation involves reacting it with hydrogen  

In this case, the equilibrium system must contain mostly ammonia. A mixture of N2 and H2 should be almost completely converted to NH3 at equilibrium. 

In general, if JC is a very small number, the equilibrium mixture will contain mostly unreacted starting materials for all practical purposes, the forward reaction does not go. Conversely, a large K implies a reaction that, at least in principle, is feasible products should be formed in high yield. Frequently, K has an intermediate value, in which case you must make quantitative calculations concerning the direction or extent of reaction. 

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Application of the equilibrium constant equation along with the value for [Fe+ ] and then the Nernst equation containing values for E and E° and the concentration changes (0.10 M for all other soluble species) will give accurate pH values for comparison with the estimates from the diagram. The estimated values from the diagram are again presented in brackets. 

Another important application of the equilibrium constant K in general, [Equation (11)], is in the coagulation treatment of water using alum, Al2(S04)3-14 H2O. (The 14 actually varies from 13 to 18.) In coagulating a raw water using alum, a number of complex reactions are formed by the Af ion. These reactions are as follows ... 

The association rate coefficient is obtained from equation (2.61) by application of the equilibrium constant and gives... 

These discrepancies have raised questions about the validity and linearity of the rate law and the applicability of the equilibrium constant equation. Many of these points are highlighted in the rate law expression for reactions (1)—(3), viz. 

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