Chemical equilibrium – 2 opposing reactions occur simultaneously at the same rate ⇌ D E E D when the rate D E is equal to rate E D,

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Presentation transcript:

chemical equilibrium – 2 opposing reactions occur simultaneously at the same rate ⇌ D E E D when the rate D E is equal to rate E D, the condition of equilibrium has been established D E ⇌

⇌ 2 NO (g) + O2 (g) 2 NO2 (g) 2 NO2 (g) 2 NO (g) + O2 (g)

equilibrium constant, K when a process is at equilibrium, it has been found experimentally, for all processes, that the ratio of products to reactants is constant ! equilibrium constant, K K = [Products] [Reactants] Figure: 15-03 Kc = for [ ] in molar, M Kp = for [ ] in partial pressure in atm for gases

2 NO (g) + O2 (g) 2 NO2 (g) ⇌ K = [NO2]2 [NO]2 [O2]

⇌ [NO2]2 [N2O4] N2O4 (g) 2 NO2 (g) Kc = Figure: 15-T01 When equilibrium is established, the concentration of reactants and products vary….but the RATIO of P/R remains CONSTANT !

⇌ [NO2]2 [NO]2 [O2] 2 NO (g) + O2 (g) 2 NO2 (g) Kc = At equilibrium, the following [ ]’s are found: [NO2] = 0.896 M [NO] = 0.0126 M [O2] = 0.00413 M Determine the equilibrium constant, Kc

By convention, equilibrium constants, Kc are always UNITLESS !!

Determine the equilibrium constant, Kc for the reverse process 2 NO2 (g) 2 NO (g) + O2 (g) ⇌ This equilibrium expression is the reciprocal of the previous example Kc = [NO]2[O2] [NO2]2 Kc = 1 1.22 x 106 = 8.17 x 10-7 The value of the equilibrium constant, K, for a reaction in one direction is the reciprocal of the equilibrium reaction written in the reverse direction

⇌ ⇌ 2 NO (g) + O2 (g) 2 NO2 (g) Kc = 1.22 x 106 reverse reaction reciprocal 2 NO2 (g) 2 NO (g) + O2 (g) ⇌ Kc = 8.17 x 10-7

P R P R when K >> 1 (bigger than 1000) K = [products] > [reactants] K = P R “Equilibrium lies to the right” “Equilibrium lies to the left” when K << 1 (smaller than 0.001) [reactants] > [products] K = P R

If K  1 (0.001 – 1000) The equilibrium mixture will have similar (or comparable) amounts of reactants and products K is constant and does NOT vary with [ ] K does vary with temperature

Kc and Kp are related Kp = Kc (RT) Δn R = 0.08206 L·atm/mole·K T = temperature in Kelvin n = (sum of coefficients of gaseous products) − (sum of coefficients of gaseous reactants)

A container is initially charged with 2 A container is initially charged with 2.00 M phosgene, COCl2 (g) at 395 °C. An equilibrium with carbon monoxide and chlorine gas is established. The equilibrium concentration of chlorine gas was found to be 0.0398 M. What is Kc for this reaction ? 1. Write the balanced chemical reaction 2. Build a chart under the reaction 3. Fill in the appropriate information (this is hard !!!) 4. Solve the problem that is presented

A container is initially charged with 0. 260 atm Cl2 (g) and 0 A container is initially charged with 0.260 atm Cl2 (g) and 0.520 atm Br2 (g) at 75 °C. The reactants combine to produce BrCl (g). Kp = 56.9 at this temperature. What are the partial pressures of all species at equilibrium ?

A container is initially charged with 0. 18 M CH4 (g) and 0 A container is initially charged with 0.18 M CH4 (g) and 0.18 M CCl4 (g) at 455 °C. The reactants combine to produce CH2Cl2 (g). Kc = 0.559 at this temperature. What are the molarities of all species at equilibrium ?

reaction quotient, Q – an equilibrium expression for a reaction NOT necessarily at equilibrium When Q < K, the reaction will proceed to the right (toward products) to establish equilibrium When Q > K, the reaction will proceed to the left (toward reactants) to establish equilibrium When Q = K, the reaction is at equilibrium

Q < K rxn forms products Q > K rxn forms reactants Q = K rxn at equilibrium Q > K rxn forms reactants

homogeneous equilibrium – all species are in the same phase heterogeneous equilibrium – all species are NOT in the same phase * ALL pure liquids and solids are left out of the equilibrium expression aqueous solutions, ex. NaCl (aq), are always included in the equilibrium expression

Fact: When a system is at equilibrium, it will remain at equilibrium forever, unless disturbed by some outside force. outside force: - change in concentration - change in temperature - change in pressure (gas phase) Le Châtelier’s Principle – when a “stress” is applied to a system at equilibrium, the reaction “shifts” to relieve the stress, and re-establish the condition of equilibrium Henry Le Chatelier 1850 – 1936

N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g) Figure: 15-06

Figure: 15-12

Le Châtelier states, “if a system is at equilibrium, and you… 1. add reactant, the reaction shifts to the Right, to consume the excess reactant and restore equilibrium” 2. add product, the reaction shifts to the Left, to consume the excess product and restore equilibrium” 3. remove reactant, the reaction shifts to the Left, to replace the lost reactant and restore equilibrium” 4. remove product, the reaction shifts to the Right, to replace the lost product and restore equilibrium”

Changes in applied pressure affects gas phase reactions N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g) increase in pressure results in the reaction shifting toward the side with the fewest number of gas particles decrease in pressure results in the reaction shifting toward the side with the largest number of gas particles

N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g) Figure: 15-13

Changes in Temperature affects equilibrium Changes in concentration or pressure results in reactions shifting but does NOT affect the numerical value of the equilibrium constant, K Changes in temperature also results in reactions shifting and DOES affect the numerical value of the equilibrium constant, K Changes in Temperature affects equilibrium predicting how temperature affects equilibria requires thermodynamic knowledge of the equilibrium reaction heat is a reactant for an endothermic reaction and heat is a product for an exothermic reaction