C h a p t e r 13 Chemical Equilibrium

The Equilibrium State01 Chemical Equilibrium: A state achieved when the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant. 1. What is the relationship between the concentrations of reactants and products in an equilibrium mixture? 2. Determine equilibrium concentrations from initial concentrations. 3. How to alter the composition of an equilibrium mixture (concentration, pressure, temp, catalyst).

The Equilibrium State02 Thus far, we have assumed complete conversion from reactants to products. Many reactions do not go to completion Concentrations do not reach constant values because the reaction stops, but because the rates of the forward and reverse reactions become equal.

Reversible reactions (arrows in both directions) “reactants” “products” The Equilibrium State03

The Equilibrium State04 Graphs of reactant and product concentrations change with time as shown below.

The Equilibrium State05 Rate of the forward reaction (N 2 O 4  2NO 2 ) decreases as concentration of N 2 O 4 decreases, while the rate of the reverse reaction (N 2 O 4  2NO 2 ) increases as the concentration of NO 2 increases

Equilibrium Constant K c 06 What is the relationship between the concentrations of reactants and products in an equilibrium mixture? aA + bB  cC + dD Equilibrium Constant:K c = [C] c [D] d  products [A] a [B] b  reactants K c is independent of concentration changes, but dependent on the temperature.

Equilibrium Constant K c 07 K c values are reported without units Thermodynamic state: [ ] = 1M N 2 O 4  2NO 2 Equilibrium Constant:K c = [NO 2 ] 2 = (0.0125M/1M) 2 [N 2 O 4 ](0.0337M/1M) = 4.64 x at 25˚C

Equilibrium Constant K c ’ 08 cC + dD  aA + bB Equilibrium Constant:Kc’ = [A] a [B] b  products [C] c [D] d  reactants K c = 1/ K c ’ * important to specify the form of the balanced equation

Example09 What is the equilibrium equation? (a) N 2 (g) + 3H 2 (g)  2NH 3 (g) (b) 2NH 3 (g)  N 2 (g) + 3H 2 (g)

Equilibrium Constant K p 10 Equilibrium equations for gas-phase reactions (partial pressures) aA(g) + bB(g)  cC(g) + dD(g) Equilibrium Constant:K p = (P C ) c (P D ) d  products (P A ) a (P B ) b  reactants

Equilibrium Constant K p 11 Convert between K c and K p using PV = nRT P A V = n A RT P A = n A RT = [A]RT V K p = K c (RT) ∆n ∆n = moles gas products – moles of gas reactants

Equilibrium Constant12 Homogeneous Equilibrium: When all reacting species are in the same phase, all reactants and products are included in the expression. Amounts of components are given as molarity or partial pressure of a gas. K c  NO 2  2 N 2 O 4  K p  P 2 2 P N 2 O 4

Example13 The following pictures represent mixtures of A molecules (red) and B molecules (blue), which interconvert according to the equation A  B. If Mixture (1) is at equilibrium, which of the other mixtures is also at equilibrium?

Examples14 1. Write the K p and K c expressions for: 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) 2. The equilibrium concentrations for the reaction between CO and Cl 2 to form carbonyl chloride (phosgene gas) CO(g) + Cl 2 (g)  COCl 2 (g) at 74°C are: [CO] = 1.2 x 10 –2 M, [Cl 2 ] = M, and [COCl 2 ] = 0.14 M. Calculate K c and K p.

Example15 Methane (CH 4 ) reacts with hydrogen sulfide to yield H 2 and carbon disulfide, a solvent used in manufacturing. What is the value of K p at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm of CH 4, 0.25 atm of H 2 S, 0.52 atm of CS 2, and 0.10 atm of H 2 ?

Equilibrium Constant16 Heterogeneous Equilibrium: When reacting species are in different phases, solid and liquid phases are excluded from the expression because their concentrations “do not change.” CaCO 3 (s)  CaO(s) + CO 2 (g) K c = [CO 2 ] because CaCO 3 and CaO are solids.

Equilibrium Constant17

Examples18 Write the equilibrium equation for each of the following reactions: (a)CO 2 (g) + C(s)  2 CO(g) (b)Hg(l) + Hg 2+ (aq)  Hg 2 2+ (aq) (c)2 Fe(s) + 3 H 2 O(g)  Fe 2 O 3 (s) + 3 H 2 (g) (d)2 H 2 O(l)  2 H 2 (g) + O 2 (g)

Using Equilibrium Constants19 We can make the following generalizations concerning the composition of equilibrium mixtures: If K c > 10 3, products predominate over reactants. If K c is very large, the reaction is said to proceed to completion. If K c is in the range 10 –3 to 10 3, appreciable concentrations of both reactants and products are present. If K c < 10 –3, reactants predominate over products. If K c is very small, the reaction proceeds hardly at all.

Predicting Reaction direction20 The reaction quotient (Q c ) is obtained by substituting initial concentrations into the equilibrium constant. H 2 (g) + I 2 (g)  2HI(g) Q c = [HI] t 2 [H 2 ] t [I 2 ] t Q c > K c System proceeds to form reactants. Q c = K c System is at equilibrium. Q c < K c System proceeds to form products.

Predicting Reaction direction 21

Example22 The equilibrium constant (K c ) for the formation of nitrosyl chloride, from nitric oxide and chlorine gas: 2 NO(g) + Cl 2 (g)  2 NOCl(g) is 6.5 x 10 4 at 35°C. In an experiment, 2.0 x 10 –2 moles of NO, 8.3 x 10 –3 moles of Cl 2, and 6.8 moles of NOCl are mixed in a 2.0-L flask. In which direction will the system proceed to reach equilibrium?