Chapter 15 Chemical Equilibrium BLB 11th

2 NO2(g) → 2 NO(g) + O2(g)

15.1 The Concept of Equilibrium N2O4(g) 2 NO2(g) ⇌ Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate.

A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant. N2O4(g) 2 NO2(g) ⇌

The Concept of Equilibrium The reaction system is closed. Opposing reactions occur at equal rates, i.e. the rates of forward and reverse reactions are equal. Dynamic process, i.e. never stops. Concentrations of reactants and products are constant. Equilibrium can be reached from either direction, reactants or products.

15.2 The Equilibrium Constant aA + bB ⇌ cC + dD K is constant for a reaction, regardless of the [A]0. K depends on stoichiometry, not on the mechanism. Pure solid or liquid reactant & products are not included for heterogeneous equilibria. (15.4) Water (or other pure solvent) is not included if the reactant and product concentrations are low. (p. 642) Kc – concentrations (M) of solutions Kp – partial pressures of gases; p. 636 to convert between Kp & Kc

Write the expression for K. N2(g) + O2(g) ⇌ 2 NO(g) 2 SO2(g) + O2(g) ⇌ 2 SO3(g)

15.3 Interpreting and Working with Equilibrium Constants The Magnitude of K K >> 1 Essentially all products; lies to right K > 1 product-favored K < 1 reactant-favored K << 1 Essentially all reactants; lies to left

Manipulating K (pp. 638-39) Reverse reaction: Adding reactions: Multiplying reaction by some factor: #1: C(s) + ½ O2(g) ⇌ CO(g) #2: 2 C(s) + O2(g) ⇌ 2 CO(g)

15.4 Heterogeneous Equilibria Reactant components are in different phases. When a pure solid or liquid is involved in a heterogeneous equilibrium, its concentration is not included in the K expression. Write the K expression: SnO2(s) + 2 CO(g) ⇌ Sn(s) + 2 CO2(g)

15.5 Calculating Equilibrium Constants Must plug in the equilibrium concentrations of reactants and products A concentration table with initial, change, and equilibrium concentrations is set up. See Sample Exercise 15.9, pp. 643. More in lab and in Chapter 16 with weak acids and bases.

Calculating K N2(g) + 3 H2(g) ⇌ 2 NH3(g) p. 642 @ 472°C 2.46 7.38 0.166 ← eq. partial pressures (atm)

15.6 Applications of Equilibrium Constants Predicting the direction of a reaction The reaction quotient, Q (p. 644) Q has the same form as K, but for non-equilibrium conditions. Comparing Q and K: Q < K achieves equilibrium by shifting to right Q = K @ equilibrium Q > K achieves equilibrium by shifting to left

Sample problem 38a N2(g) + 3 H2(g) ⇌ 2 NH3(g) KP = 4.51 x 10-5 at 450°C 45 atm 55 atm 98 atm

15.7 Le Châtelier’s Principle If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its position to counteract the effect of the disturbance. (p. 648)

Le Châtelier’s Principle Disturbance Effect More reactant or product Shifts to consume Volume decrease (total pressure increase) Shifts to reduce pressure (fewer moles) Inert substance added None Temperature increase If exothermic, shifts left and K decreases. If endothermic, shifts right and K increases.

2 SO2(g) + O2(g) ⇌ 2 SO3(g) ΔH < 0 Sample problem 51 2 SO2(g) + O2(g) ⇌ 2 SO3(g) ΔH < 0 Effect on equilibrium mixture when… (a) O2(g) is added? (b) the reaction is heated? (c) the volume of container is doubled? (d) a catalyst is added? (e) the total P is increased by adding a noble gas? (f) SO3(g) is removed?

6 CO2(g) + 6 H2O(l) ⇌ C6H12O6(s) + 6 O2(g) ΔH = 2816 kJ Write the K expression. Effect on equilibrium mixture when… (a) PCO2 is increased? (b) the reaction is heated? (c) some CO2(g) is removed? (d) the total P is decreased? (e) part of C6H12O6(s) is removed? (f) a catalyst is added?