Dr. Saleha Shamsudin

CHEMICAL EQUILIBRIUM Discuss the concept of chemical reaction: the rate concept, type of equilibria, Le Chatelier’s principle. Effects of temperature, pressure and concentration on equilibria. ACID-BASE EQUILIBRIUM Repeat acids and bases theories, acid base equilibria in water, pH scale. DISCUSS weak acids and bases, calculate the pH of salt of weak acids and bases. DISCUSS buffers. Introduce strong acids and strong bases, APPLY detection of end point, indicators. DISCUSS weak base versus strong acid.

What is chemical equilibrium? A dynamic state in which the rates of forward (f) and reverse (r) reaction are identical.

Chemical Equilibrium – A Dynamic Equilibrium Upon addition of reactants and/or products, they react until a constant amount of reactants and products are present = equilibrium. Equilibrium is dynamic since product is constantly made (forward reaction), but at the same rate it is consumed (reverse reaction).

1. Chemical Reactions 1.1 Introduction - Chemical reactions Involves chemical change. Chemical change: the atoms rearrange themselves in such a way that new substances are formed. Components decrease in quantity are called reactants and those increase in quantity are called products. The chemical reaction is: aA + bB ↔ cC + dD

Forward reaction: k f = rate constant [A] and [B]= molar concentrations of A and B Forward and reverse rates are equal k f [A] a [B] b = k b [C] c [D] d The rate of reaction = constant x concentration each species raised to the power of the number of molecules participating in the reaction.

Molar equilibrium constant, K, k b [C] c [D] d = kf = K [A] a [B] b k b Rearranging these equation gives molar equilibrium constant (K) : [C] c [D] d = k f = K [A] a [B] b k b When the two rates become equal, the system is in a state of equilibrium. Forward and reverse rates are equal k f [A] a [B] b = k b [C] c [D] d

Type of equilibria We can write equilibrium constants for many types of chemical processes. The equilibria may represent : Dissociation (acid/base, solubility) Formation of products (complexes) Reaction (redox) Distribution between 2 phases (water and non-aqueous solvent)

Types of equilibria

Equilibrium constant of reverse reaction: aA + bB  cC + dD cC + dD  aA + bB

For weak electrolyte Slightly soluble substances Example: 1) AB  A + B Equilibrium constant written as: [A][B] = Keq [AB] Equilibrium constants for dissociating and associating species

2) Stepwise dissociation A 2 B  A + AB K 1 = [A][AB] [A 2 B] AB  A + B K 2 = [A][B] [AB] Overall dissociation: A 2 B  2A + B Keq = [A] 2 [B] [A 2 B]

3) Heterogeneous equilibria - Involving more than one physial phase, example, solids, water and pure liquids. - Concentrations of any solid reactants and products are omitted from the equi. cons. expression. - Molar concentrations of water (or of any liquid reactant or product) is omitted from the equi. cons. expression.

E.g. Write out the equilibrium expression for K using the reaction below: N 2 (g) + 3H 2 (g)  2NH 3 (g) K c = ?

Try this example: An aqueous solution of ethanol and acetic acid, each with a concentration of M, is heated to 100 o C. At equilibrium, the acetic acid concentration is M. Calculate K at 100 o C for the reaction C 2 H 5 OH(aq) + CH 3 CO 2 H(aq)  CH 3 CO 2 C 2 H 3 (aq) + H 2 O(l) ethanol acetic acid ethyl acetate 1.Write the equilibrium constant expressions in terms of concentrations: 2.Calculate K at 100 o C for the reaction: Calculating Equilibrium Constants

Equation C 2 H 5 OH(aq)CH 3 CO 2 H(aq)CH 3 CO 2 C 2 H 3 Initial (M) Change (M) Equilibrium (M) The equilibrium constant expressions is: K = [CH 3 CO 2 C 2 H 3 ] notice that liquid [C 2 H 5 OH][CH 3 CO 2 H] water does not 2. The calculation of K is: appear in equi. = = 0.11 expressions. (0.748)(0.748)

Problem: K= for H 2 (g) + I 2 (g)  2HI(g) Has been determined at 425 o C. If 1.00 mol Each of H 2 and I 2 are placed in 0.5 L flask at 425 o C, what are the concentrations of H 2, I 2 and HI when equilibrium has been achieved? Calculating an Equilibrium from an Equilibrium Constant

Solution: 1. Write the equi. Cons. Expression. 2. Set up an ICE table Equation H 2 (g)I 2 (g)2HI(g) Initial (M) Change (M) Equilibrium(M) 1.00 mol 0.5 L = 2.0 M - x 2.0-x 1.00 mol 0.5 L 2.0 M - x 2.0-x 0 +2x 2x -Assume x is equal to the quantity of H 2 or I 2 consumed in the reaction

- Substitute the equi. Con. Into K expression = (2x) 2 = (2x) 2 (2.0 – x)(2.0 – x) (2.0 – x)2  = = 2x 2.0 – x x = 1.58 H = I = 2.0 – x = 0.42 M HI = 2x = 3.16 M

Calculating Equilibrium Concentrations Using initial concentrations, stoichiometry and K c, equilibrium concentrations of all components can be determined. E.g.2 [H 2 ] o = [I 2 ] o = 24 mM, were mixed and heated to 490 o C in a container. Calculate equil. composition. Given that K c = 46. Solution: set up an equilibrium table and solve for unknown after substitution into equilibrium expression. H 2 (g) +I 2 (g)  2HI(g)

Equilibrium concentrations of each component are in last row. Substitute into equilibrium expression and solve for x. It may be necessary to rearrange so that quadratic equation can be used.. Rearrange to ax 2 + bx + c = 0; determine a, b, c and substitute into quadratic equation: