Ch_07_summary

# Ch_07_summary - CHAPTER 7 EQUILIBRIUM(IB TOPICS 7 AND 17...

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Unformatted text preview: CHAPTER 7 EQUILIBRIUM (IB TOPICS 7 AND 17) SUMMARY Equilibrium Dynamic Equilibrium is recognized by a constant macroscopic property in a closed system due to the concentration of all the reactants and products remaining constant at the same temperature. It is explained in terms of the rate of the forward reaction equal to the rate of the reverse reaction. For the equilibrium aA + bB cC + dD where a, b, c, and d are coefficients in the balanced equation, the equilibrium expression at a particular temperature (where square brackets [ ]eq represent equilibrium concentration in mol dm−3) is: Kc= [C ] c eq [D ] d eq [A ] a eq [B ] b eq ; K c is the equilibrium constant at a particular temperature. In the Kc expression, the right side of the chemical equation is written as the numerator, the left side of the chemical equation is written as the denominator, and the coefficients become exponents. Square brackets represent equilibrium concentrations. Units of Kc depend on the powers to the concentration in the equilibrium expression. Significance of Kc • When KC > 1, concentration of products generally exceeds that of reactants at equilibrium. • When KC >> 1, reaction goes almost to completion. • When KC < 1, concentration of reactants generally exceeds that of products at equilibrium. • When KC << 1, reaction hardly proceeds. Law of Multiple Equilibria • When equations are added, multiply KC values. • When equations are subtracted, divide KC values. • When an equation is doubled, square KC value. • When an equation is halved, take square root of KC value (a new equilibrium constant value). • When an equation is reversed, inverse KC value. Le Chatelier’s Principle This principle helps predict the direction of the effect: If a stress or change is applied to a system at equilibrium, the processes that take place counteract (minimize) the effect of that change and establish a new equilibrium. Thus, concentrations of the species change in such a way as to minimize the effect of the stress (the stress can not be totally eliminated). Conditions that alter the position of equilibrium: Change in concentration changes the position of equilibrium (favors forward or reverse reaction), but does not change the value of Kc. A change in temperature changes both the position of equilibrium and the value of Kc. Effect of adding a Catalyst: Catalysts increase rates of reactions. Any catalyst has the same effect on the rates of the forward and reverse reactions. The addition of a catalyst to a system at equilibrium does not alter the position of equilibrium. The system reaches the same equilibrium more quickly but with no net change in equilibrium concentrations; thus Kc does not change. Concentration Change and Le Chatelier’s Principle: If the concentration of one of the reactants or products in a system at equilibrium is increased, the effect of the change may be minimized by the establishment of a new equilibrium so that the concentration of the added substance is reduced. KC does not change. © IBID Press 2007 1 CHAPTER 7 EQUILIBRIUM (IB TOPICS 7 AND 17) SUMMARY A+B C+D A+B C+D 1. Remove any reactant: KC remains 1. Remove any constant. constant. product: KC remains 2. Equilibrium shifts to the left, to replace 2. Equilibrium shifts to the right, to replace some of the reactant removed, to some of the product removed, to minimize minimize change. change. Pressure Change and Le Chatelier’s Principle: If pressure of the system is increased, concentration increases, and the effect of this change may be counteracted by the establishment of a new equilibrium so that the volume is decreased. The equilibrium constant, KC, does not change as only concentration is changed: (1) If ΔVgases = VP – VR = +, for example: H2O (g) H2 (g) + ½ O2 (g). If pressure is increased, the system now occupies less volume, i.e., the increase in pressure is counteracted if hydrogen and oxygen combine to form gaseous water which occupies less volume. Thus the reverse reaction is favoured until a new equilibrium is established. Kc does not change. (2) If ΔVgases = −, for example: 3 H2 (g) + N2 (g) 2 NH3 (g). There is a decrease in volume of gases. So increase in pressure will favour formation of more ammonia, i.e., the forward reaction is favoured to relieve the stress of pressure until a new equilibrium is established. KC doesn’t change. (3) If ΔVgases = VP – VR = 0, for example : CO (g) + NO2 (g) CO2 (g) + NO (g). There is no change in volume of gases (ΔV = 0). Thus, no change in equilibrium position when the pressure is altered in such cases. The rates of the forward and reverse reactions will increase equally as concentrations of reactants and products increase. Kc remains constant. Temperature Change and Le Chatelier’s Principle: Changing the temperature of a system results in the establishment of a new equilibrium constant. Increase in temperature favours the change that takes place with absorption of thermal energy: Endothermic reactions: e.g.: A + B + heat C+D Exothermic reactions: e.g.: E + F G + H + heat Increase in temperature Favors forward reaction to use up Favors reverse reaction to use up some of the heat supplied some of the heat supplied • Effect on Concentration [C] and [D] increase and [A] and [G] and [H] decrease and [E] and [F] [B] decrease increase • Effect on KC KC increases Decrease in temperature KC decreases Favours the reverse reaction to Favours the forward reaction to produce some of the heat taken produce some of the heat taken way. away. • Effect on Concentration [C] and [D] decrease and [A] and [G] and [H] increase and [E] and [F] [B] increase decrease • Effect on KC KC decreases KC increases. Thus, the equilibrium shifts in such a way as to absorb heat when the temperature is increased and to evolve heat when the temperature is decreased. Do not confuse effect of change in temperature on equilibrium with effect on rate of reaction. It helps if you think of the change in heat as part of the reactants or products. Recall that KC varies only with temperature. © IBID Press 2007 2 CHAPTER 7 EQUILIBRIUM (IB TOPICS 7 AND 17) SUMMARY Solving Equilibrium Problems • Write the balanced equation and the Kc expression for the reaction • List the initial concentrations of all reactants and products • Determine the concentration changes to reach equilibrium, and the equilibrium concentrations • Substitute into the equilibrium expression to solve for the unknown concentration, or determine Kc value. The Haber Process: Summary: N2(g) + 3H2 (g) 2NH3(g); ΔH = − • All the species are gases • There are 4 moles (volumes) of gases in the reactants and only 2 in the product • The forward reaction is exothermic. Change in concentration: Increased concentration of H2 (g) and/or N2 (g) increases the number of collisions of reactants per unit time and the rate of the forward reaction increases. This also increases the pressure on the system; in order to relieve the pressure, the forward reaction is favored with more ammonia produced; Kc remains constant. Change in pressure: Since there are 4 mol gases on the reactant side, and 2 in the product side, an increase in pressure forces the system to occupy less volume so as to relieve some of the pressure applied. Thus the forward reaction is favored. Rate also increases as concentration of gases increases. Kc remains constant. Change in temperature: An increase in temperature increases the number of particles with E ≥ Ea, and reaction rate increases. Since the forward reaction is exothermic, increased temperature favors the reverse reaction in order to use up some of the heat supplied to relieve the stress. Kc decreases with increase in temperature (Kc is temperature dependent). Addition of iron as a catalyst speeds up both the forward and reverse reactions equally; the same equilibrium is reached faster; there is no change in equilibrium concentrations and the value of Kc remains the same. The Contact Process Summary: 2 SO2 (g) + O2 (g) 2 SO3 (g); ΔH = – • All the species are gases • There are 3 moles (volumes) of gases in the reactants and only 2 in the product • The forward reaction is exothermic. Change in concentration: Increased concentration of SO2 (g) and O2 (g) increases the number of collisions of reactants per unit time and the rate of the forward reaction increases. This also increases the pressure on the system; in order to relieve the pressure, the forward reaction is favored with more sulfur trioxide produced; Kc remains a constant. Change in pressure: Since there are 3 mol gases on the reactant side, and 2 on the product side, an increase in pressure forces the system to occupy less volume so as to relieve some of the pressure applied. Thus the forward reaction is favored. Rate also increases as concentration increases. Kc remains constant. Change in temperature: An increase in temperature increases the number of particles with E ≥ Ea and the reaction rate increases. Since the forward reaction is exothermic, a temperature increase favors the reverse reaction in order to to use up some of the heat supplied to relieve the stress. Kc decreases with increase in temperature (Kc is temperature dependent). Addition of V2O5 or Pt as a catalyst speeds up both the forward and reverse reactions equally; the same equilibrium is reached faster; there is no change in equilibrium concentrations and the value of Kc remains the same. © IBID Press 2007 3 CHAPTER 7 EQUILIBRIUM (IB TOPICS 7 AND 17) SUMMARY Phase Equilibrium (Between Liquid and Its Own Vapor) When the rate of condensation equals the rate of evaporation, saturated vapor pressure is present and a dynamic equilibrium is established: The saturated vapor pressure of a liquid at a given temperature is independent of the amount of liquid or vapor present and of the volume of the container. Enthalpy of Vaporization, Boiling Point & Intermolecular Forces: The molar enthalpy of vaporization, ΔHvap, is the heat required to convert one mole of a substance at its boiling point from a liquid to a gas If intermolecular attraction is strong (as in H2O), then molecules in a liquid cannot easily escape into the vapor phase. Thus higher the ΔHvap, the lower the volatility of the liquid. Consequently, the liquid will have a low vapor pressure (and high molar enthalpy of vaporization) because more energy is needed to overcome inter-particle forces to form the vapor. • Weaker the inter-particle forces, higher the vapor pressure, lower the boiling point, lower the ΔHvap. • Stronger the inter-particle forces, lower the vapor pressure, higher the boiling point, higher the ΔHvap. (N.B. shading indicates Topic 17 (AHL) material) © IBID Press 2007 4 ...
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