Consider the following reaction: Fe3+(aq) + SCN-(aq) ↔︎ [Fe(SCN)]2-(aq)

1)Explain what is meant by the term dynamic equilibrium.

2) Write the equilibrium expression for the reaction.

3) The numerical value of Kc at 298K is about 1000 for the reaction. What does this tell you about the position of the equilibrium at 298K?

In 1860s, Norwegian scientists C. M. Guldberg and P. Waage noted a peculiar relationship between the amounts of reactants and products in an equilibrium. Today, we call this observation the law of mass action. It relates the amounts of reactants and products at equilibrium for a chemical reaction. For a general chemical reaction occurring in solution, aA + bB ⇄ cC + dD the equilibrium constant, also known as Keq, is defined by the following expression: Keq = [C]c/[D]d where [A] is the molar concentration of species A at equilibrium, and so forth. The coefficients a, b, c, and d in the chemical equation become exponents in the expression for Keq. The Keq is a characteristic numerical value for a given reaction at a given temperature. That is, each chemical reaction has its own characteristic Keq. The concentration of each reactant and product in a chemical reaction at equilibrium is related; the concentrations cannot be random values, but they depend on each other. The numerator of the expression for Keq has the concentrations of every product (however many products there are), while the denominator of the expression for Keq has the concentrations of every reactant, leading to the common products over reactants definition for the Keq. Let us consider a simple example. Suppose we have this equilibrium: A ⇄ B .There is one reactant, one product, and the coefficients on each are just 1. The Keq expression for this equilibrium is Keq = [B]/[A]. Exponents of 1 on each concentration are understood. Suppose the numerical value of Keq for this chemical reaction is 2.0. If [B] = 4.0 M, then [A] must equal 2.0 M so that the value of the fraction equals 2.0: Keq = [B]/[A] = 4.0/2.0 =2.0 .By convention, the units are understood to be M and are omitted from the Keq expression. Suppose [B] were 6.0 M. For the Keq value to remain constant (it is, after all, called the equilibrium constant), then [A] would have to be 3.0 M at equilibrium: Keq = [B]/[A] = 60/3.0= 2.0 .If [A] were not equal to 3.0 M, the reaction would not be at equilibrium, and a net reaction would occur until that ratio was indeed 2.0. At that point, the reaction is at equilibrium, and any net change would cease. However, that the forward and reverse reactions do not stop because chemical equilibrium is dynamic.

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