1 General Concepts of Chemical Equilibrium

2 In this chapter you will be introduced to basic equilibrium concepts and related calculations. The type of calculations you will learn to perform in this chapter will be essential for solving equilibrium problems in later chapters. The ease of solving equilibrium problems will depend on the skills you will master in this chapter.

3 The Equilibrium Constant For chemical reactions that do not proceed to completion, an equilibrium constant can be written as the quotient of multiplication of molar concentrations of products divided by that of reactants, each raised to a power equal to its number of moles. aA + bB  cC + dD Where, the small letters represent the number of moles of substances A, B, C, and D. The equilibrium constant is written as: K eq = ([C] c [D] d )/([A] a [B] b )

4 In a chemical reaction, as reactants start to react products start to form. Therefore, reactants continuously decrease and products continuously increase till a point is reached where eventually no change in concentrations can be detected. This is the point of equilibrium which is a point where the rate of the forward reaction (product formation) equals the rate of backward reaction (reactants formation). In fact equilibrium implies continuous transformation between infinitesimally small amounts of products and reactants.

5 Progress of a chemical reaction. The rate of the forward reaction diminishes with time, while that of the backward reaction increases, until they are equal. A large K means the reaction lies far to the right at equilibrium.

6 Equilibrium constants may be written for dissociations, associations, reactions, or distributions.

7 Calculation of Equilibrium Constants We know from thermodynamics that a reaction occurs spontaneously if it has a negative  G (Gibbs free energy) where:  G =  H - T  S Where,  H is the change in enthalpy of the reaction and  S is the change in entropy. At standard conditions of temperature and pressure (standard state) we have the standard free energy  G o where:  G o =  H o - T  S o

8 The standard free energy is related to equilibrium constant by the relation:  G o = - RT ln K K = e -  Go/RT R is the gas constant (8.314 deg K -1 mol -1 ) It should be clear that  G o gives us good information about the spontaneity of the reaction but it offers no clue on the rate at which the reaction may occur.

9 Le Chatelier's Principle The equilibrium concentrations of reactants and products can be changed by applying an external stress to the system, e.g. increasing or decreasing the concentration of a reactant or product, changing temperature or pressure. The change occurs in a direction which tends to counterpart the applied stress. For example:

10 a. Increasing the temperature of an exothermic reaction will shift the reaction to left (more reactants and less products). The opposite is observed if the reaction is exothermic or if heat is removed from an exothermic reaction. b. In a reaction where the number of gaseous molecules produced is more than the number of reacting gaseous molecules, increasing the pressure of the system will shift the reaction toward the reactants in an attempt to decrease the number of moles and vice versa. Reactions in solutions are usually insensitive to changes in pressure.

11 c. Increasing the concentration of a reactant or removing a product will result in a shift of reaction towards more products and vice versa. d. Some reactions can be facilitated by addition of a catalyst (a substance that is not part of reactants or products but its presence makes the reaction faster). The catalyst does not change the position of equilibrium but makes the time required to reach this equilibrium point shorter. Le Chatelier's principle can be advantageously used to force reactions that are close to completion to proceed to completion. This is usually done by addition of more reagent in a gravimetric procedure, allowing a gas to escape if one of the products is a gas, etc..

12 Stepwise Equilibrium Constants Sometimes a chemical equilibrium can be written as a multi step equilibrium. The overall equilibrium constant for the reaction is the multiple of stepwise equilibrium constants. Look at the stepwise equilibrium below: A 2 B  A + AB K 1 = [A][AB]/[A 2 B] AB = A + B K 2 = [A][B]/[AB] K 1 K 2 = [A] 2 [B]/[A 2 B]

13 Now look at the equilibrium A 2 B  2 A + B K = [A] 2 [B]/[A 2 B] From K and K 1 K 2 we see that they are equal. This is valid for any multi step equilibrium reaction as will be seen later.

14 Calculations Using Equilibrium Constants Here, we are faced with three situations which should be considered separately: 1. When the equilibrium constant is very small. Most reactants do not undergo a reaction and very little products are produced. Usually the concentration of products can be neglected as compared to reactants concentrations.

15 2. When the equilibrium constant is very high. In this case, most reactants disappear and the reaction container contains mainly the products. In calculations, one can neglect the concentration of remaining reactants as compared to concentrations of products. 3. A situation where the equilibrium constant is moderate. Appreciable amounts of reactants are left and appreciable amounts of products are also formed in the reaction mixture. One can not neglect the reactants or the products concentration. Following are examples on each case.

16 Example Calculate the equilibrium concentration of A and B in a 0.10 M solution of weak electrolyte AB if the equilibrium constant of the reaction is 3.0x Solution AB  A + B First we should look at the value of the equilibrium constant to have an appreciation of what is going on. It is clear that we have a small equilibrium constant which suggests that very little products may have been formed and thus we build our solution on the assumption that AB will mainly remain unreacted except for a very little concentration x.

17 Therefore, AB concentration will decrease by x and A and B will be formed in a concentration equals x for each. The following table represents what is happening.

18 K = [A][B]/[AB] Substitution of equilibrium concentration gives K = (x)(x)/(0.1-x) = 3.0 x In order to solve this equation, we should make a justified assumption which we have discussed above; that is x is very small as compared to 0.10 M and later we will check whether our assumption is valid or not.

19 assume 0.10 >>x. We have 3.0 x = x 2 /0.10, thus we have x = 5.5 x M Now we should check the validity of our assumption that 0.10 >>x by calculating the relative error Relative error = (5.5x10 -4 /0.10) x 100 = 0.55%

20 In this course, we will consider our assumption valid if the relative error is below 5%. Therefore, our assumption is valid and we have: [A] =5.5x10 -4 M, [B] = 5.5x10 -4 M, and [AB] = (0.10 – 5.5x10 -4 ) ~ 0.10 M

21 Example In the reaction A + B  C + D If 0.2 mol of A is mixed with 0.5 mol of B in 1.0 L, find the concentration of A, B, C, and D. The equilibrium constant of the reaction is 4.0x Solution It is clear that we have a small equilibrium constant which suggests that very little products may have been formed and thus we build our solution on the assumption that A and B will mainly remain unreacted except for a very little concentration x.

22 Therefore, A and B concentration will decrease by x and C and D will be formed in a concentration equals x for each. The following table represents what is happening.

23 K = [C][D]/[A][B] Substitution using equilibrium concentrations results in K = (x)(x)/(0.20 – x)(0.50 – x) Now from the value of the equilibrium constant we can assume that 0.20 >> x. Then we have 4.0x10 -9 = x 2 /(0.20x0.50) x = 2.0x10 -5 M

24 Now check the validity of the assumption by calculating the relative error Relative Error = (2.0x10 -5 /0.20) x 100 = 0.01% The assumption is valid and the concentrations of the different species are [A] = (0.20 – 2.0x10 -5 ) ~ 0.20 M [B] = 0.50 M [C] = [D] = 2.0x10 -5 M

25 Example In the reaction A + B  C + D If 0.2 mol of A is mixed with 0.5 mol of B in 1.0 L, find the concentration of A, B, C, and D. The equilibrium constant of the reaction is 4.0x10 8. Solution It is clear that we have a high equilibrium constant which suggests that very little reactants may have been left and thus we build our solution on the assumption that A will mainly be of a very little concentration x as a result of the chemical equilibrium. B is already present in excess and an equivalent amount to A reacts while the rest of B remains in unreacted.

26 K = [C][D]/[A][B] Substitution using equilibrium concentrations results in K = (0.20 – x) (0.20 – x)/(x( x)) Now from the value of the equilibrium constant we can assume that 0.20 >> x. Then we have 4.0x10 8 = 0.040/0.3x x = 3.3x M

27 Now check the validity of the assumption by calculating the relative error Relative Error = (3.3x /0.20) x 100 = 0.005% The assumption is valid and the concentrations of the different species are [A] = 3.3x M [B] = ( x ) = 0.30 M [C] = [D] = ( x ) ~ 0.20 M

28 Example A and B react as follows A + 2 B  2 C Assume 0.10 mol of A reacts with 0.20 mol of B in 1 L. If K = 1.0x10 10, find the equilibrium concentrations of A, B, and C. Solution We have a high equilibrium constant which suggests that most reactants are changed into products except for a tiny equilibrium concentration. Therefore, at equilibrium we mainly have C.

29 K = [C] 2 /[A][B] 2 K = (0.20 – 2x) 2 /x(2x) 2 Again, we are justified to assume that 0.20 >> 2x since the equilibrium constant is high. We then have: 1.0x10 10 = (0.2) 2 /4x 3 x = 1.0x10 -4

30 Now let us check for our assumption by determining the relative error Relative error = (2 x 1.0x10 -4 /0.20) x 100 = 0.10% The assumption is valid and the equilibrium concentrations are [A] = x =1.0x10 -4 M [B] = 2 x = 2.0x10 -4 M [C] = (0.20 – 2.0x10 -4 ) ~ 0.20 M