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Week 8.3 thermodynamics and equilibria

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  • 1. Chapter 4 Thermodynamic and Equilibria First Law of Thermodynamics Second Law of Thermodynamics Physical Transformation of Pure Substances Prepared by: Mrs Faraziehan Senusi PA-A11-7C Simple Mixtures Chemical Equilibrium Reference: Chemistry: the Molecular Nature of Matter and Change, 6th ed, 2011, Martin S. Silberberg, McGraw-Hill
  • 2. Chemical Equilibrium
  • 3. Kinetics addresses the speed of a reaction, the change in product concentration (or reactant concentration) per unit time. Equilibrium addresses the extent of a reaction, the concentration of product that results given unlimited reaction time. At equilibrium: rateforward = ratereverse A system at equilibrium is dynamic on the molecular level; no further net change in reactant and product concentrations is observed because changes in one direction are balanced by changes in the opposite direction.
  • 4. The Concept of Equilibrium • Consider colorless frozen N2O4. At room temperature, it decomposes to brown NO2: N2O4(g) 2NO2(g) • At some time, the color stops changing and we have a mixture of N2O4 and NO2. • Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate.
  • 5. Reaching Equilibrium on the Macroscopic and Molecular Levels When the experiment begins, the reaction mixture consists mostly of colorless N2O4. As N2O4 decomposes to reddish brown NO2, the color of the mixture becomes pale brown N2O4(g) When equilibrium is reached, the concentrations of NO2 and N2O4 are constant, and the color reaches its final intensity. 2NO2(g) N2O4(g) is colorless and NO2(g) is brown Copyright ©The McGraw-Hill Companies, Inc. Because the reaction continues in the forward and reverse directions at equal rates, the concentrations (and color) remain constant.
  • 6. • Using the collision model: – as the amount of NO2 builds up, there is a chance that two NO2 molecules will collide to form NO2. – At the beginning of the reaction, there is no NO2 so the reverse reaction (2NO2(g) N2O4(g)) does not occur. , N2O4 , NO2
  • 7. • The point at which the rate of decomposition: N2O4(g) 2NO2(g) equals the rate of dimerization (a compound formed by combination of two identical molecules) : 2NO2(g) N2O4(g). is dynamic equilibrium. • The equilibrium is dynamic because the reaction has not stopped: the opposing rates are equal.
  • 8. • • Consider frozen N2O4: only white solid is present. On the microscopic level, only N2O4 molecules are present. As the substance warms it begins to decompose: N2O4(g) • • A mixture of N2O4 (initially present) and NO2 (initially formed) appears light brown. When enough NO2 is formed, it can react to form N2O4: 2NO2(g) • N2O4(g) At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4: N2O4(g) • 2NO2(g) 2NO2(g) The double arrow implies the process is dynamic.
  • 9. B • For an equilibrium we write A • Consider Forward reaction: A B Rate = kf [A] Reverse reaction: B A Rate = kr [B] • At equilibrium kf [A] = kr [B]. • As the reaction progresses Reaction rate constant – [A] decreases to a constant, – [B] increases from zero to a constant. – When [A] and [B] are constant, equilibrium is achieved. • Alternatively: – kf [A] decreases to a constant, – kr [B] increases from zero to a constant. – When kf [A] = kr [B] equilibrium is achieved.
  • 10. • • • • Consider a case in which the coefficients in the equation for a reaction are all 1. When substances A and B react, the rate of the forward reaction decreases as time passes because the concentration of A and B decrease. A+ B  C + D (1) As the concentrations of C and D build up, they start to re-form A and B. C+DA+B (2) As more C and D molecules are formed, more can react, and so the rate of reaction between C and D increases with time. Eventually, the two reactions occur at the same rate, and the system is at equilibrium If a reaction begins with only C and D present, the rate of reaction (2) decreases with time, and the rate of reaction (1) increases with time until the two rates are equal.
  • 11. • Consider the Haber Process: N2(g) + 3H2(g) 2NH3(g) • If we start with a mixture of nitrogen and hydrogen (in any proportions), the reaction will reach equilibrium with a constant concentration of nitrogen, hydrogen and ammonia. • However, if we start with just ammonia and no nitrogen or hydrogen, the reaction will proceed and N2 and H2 will be produced until equilibrium is achieved.
  • 12. • No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium.
  • 13. The Equilibrium Constant A reversible reaction occurs by a one-step mechanism : • • • • • The rate of the forward reaction is Ratef = kf[A]2[B] , kf = specific rate constants of the forward reaction; The rate of the reverse reaction is Rater = kr[A2B] , kr =specific rate constants of the reverse reactions. The two rates are equal at equilibrium (Ratef = Rater). So, At any specific temperature, both kf and kr are constants, so kf/kr is also a constant.
  • 14. • This ratio is given a special name and symbol, the equilibrium constant, Kc or simply K. • Equilibrium constant is the product and ratio of the rate constants for each step of the mechanism.
  • 15. • For a general reaction in the gas phase aA + bB cC + dD • The equilibrium constant expression is K eq KP c d PC PD a b PA PB c d PC PD a b PA PB where Keq is the equilibrium constant.
  • 16. • The thermodynamic definition of the equilibrium constant involves activities rather than concentrations. • The activity of a component of an ideal mixture is the ratio of its concentration or partial pressure to a standard concentration (1 M) or pressure (1 atm). • Ex : A concentration of 1.20 M becomes 1.20 M/1 M� = 1.20 ; • Ex: A pressure of 0.53 atm becomes 0.53 atm/1 atm = 0.53.
  • 17. • For now, we can consider the activity of each species to be a dimensionless quantity whose numerical value can be determined as follows: 1. For any pure liquid or pure solid, the activity is taken as 1. 2. For components of ideal solutions, the activity of each component is taken to be the ratio of its molar concentration to a standard concentration of 1 M, so the units cancel. 3. For gases in an ideal mixture, the activity of each component is taken to be the ratio of its partial pressure to a standard pressure of 1 atm, so again the units cancel. • Because of the use of activities, the equilibrium constant has no units; the values we put into Kc are numerically equal to molar concentrations, but are dimensionless.
  • 18. The Magnitude of Equilibrium Constants • The equilibrium constant, K, is the ratio of products to reactants. • Therefore, the larger K, the more products are present at equilibrium. • Conversely, the smaller K, the more reactants are present at equilibrium.
  • 19. • If K >> 1, then products dominate at equilibrium and equilibrium lies to the right. • If K << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.
  • 20. Calculating Equilibrium Constants • Proceed as follows: – Tabulate initial and equilibrium concentrations (or partial pressures) given. – If an initial and equilibrium concentration is given for a species, calculate the change in concentration. – Use stoichiometry on the change in concentration line only to calculate the changes in concentration of all species. – Deduce the equilibrium concentrations of all species. • Usually, the initial concentration of products is zero. (This is not always the case.)
  • 21. Example 1 Some nitrogen and hydrogen are placed in an empty 5.00-liter container at 500°C. When equilibrium is established, 3.01 mol of N2, 2.10 mol of H2, and 0.565 mol of NH3 are present. Evaluate Kc for the following reaction at 500°C. Solution The equilibrium concentrations are obtained by dividing the number of moles of each reactant and product by the volume, 5.00 liters. Then we substitute these equilibrium concentrations into the equilibrium constant expression.
  • 22. Example 2 The equation for the following reaction and the value of Kc at a given temperature are given. An equilibrium mixture in a 1.00 liter container contains 0.25 mol of PCl5 and 0.16 mol of PCl3. What equilibrium concentration of Cl2 must be present? Solution Because the volume of the container is 1.00 liter, the molar concentration (mol/L) of each substance is numerically equal to the number of moles. The equilibrium constant expression and its numeric value are
  • 23. Example 3 In an equilibrium mixture at 500°C, we find PNH3 = 0.147atm, PN2 = 6.00atm, and PH2 = 3.70atm. Evaluate KP at 500°C for the following reaction. Solution
  • 24. Example 4 Consider the reversible reaction of sulfur dioxide with oxygen to form sulfur trioxide at 1500 K. Suppose 0.400 mole of SO2 and 0.200 mole of O2 are injected into a closed 1.00 liter container. When equilibrium is established, we find that 0.056 mole of SO3 has formed and that 0.344 mole of SO2 and 0.172 mole of O2 remain unreacted. Calculate the value of the equilibrium constant for this reaction at 1500 K.
  • 25. Solution Substituting the numerical values (without units) into the equilibrium expression gives the value of the equilibrium constant. * If [SO3]eq is only given, use coefficients of the balanced equation or vice versa.
  • 26. RELATIONSHIP BETWEEN KP AND Kc If the ideal gas equation is rearranged, the molar concentration of a gas is Substituting P/RT for n/V in the Kc expression for the N2–H2–NH3 equilibrium gives the relationship between Kc and KP for this reaction.
  • 27. • In general the relationship between Kc and KP is • For reactions in which equal numbers of moles of gases appear on both sides of the equation, Δn = 0 and KP = Kc. From example 1, for the ammonia reaction at 500°C (or 773 K), Kc = 0.286 •
  • 28. THE REACTION QUOTIENT • The reaction quotient has the same form as the equilibrium constant, but it involves specific values that are not necessarily equilibrium concentrations. • Reaction quotient, Q, as a measure of the progress of the reaction.
  • 29. • If Q > K then the reverse reaction must occur to reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K). • If Q < K then the forward reaction must occur to reach equilibrium.
  • 30. Heterogeneous Equilibria • When all reactants and products are in one phase, the equilibrium is homogeneous. • If one or more reactants or products are in a different phase, the equilibrium is heterogeneous. • Consider: CaCO3(s) CaO(s) + CO2(g) – experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?
  • 31. • The concentration of a solid or pure liquid is its density divided by molar mass. • Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant. • For the decomposition of CaCO3: K eq [CaO] [CO 2 ] constant [CO 2 ] [CaCO3 ] • We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions. • The amount of CO2 formed will not depend greatly on the amounts of CaO and CaCO3 present.
  • 32. Le Châtelier’s Principle • Consider the production of ammonia N2(g) + 3H2(g) 2NH3(g) • As the pressure increases, the amount of ammonia present at equilibrium increases. • As the temperature decreases, the amount of ammonia at equilibrium increases. • Can this be predicted? Le Châtelier’s Principle: If a change of conditions (stress) is applied to a system at equilibrium, the system shifts in the direction that reduces the stress to move toward a new state of equilibrium.
  • 33. • The reaction quotient, Q, helps us predict the direction of this shift. • Three types of changes can disturb the equilibrium of a reaction. 1. Changes in concentration 2. Changes in pressure or volume (for reactions that involve gases) 3. Changes in temperature
  • 34. Change in Reactant or Product Concentrations • Consider the Haber process N2(g) + 3H2(g) 2NH3(g) • If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 (by Le Châtelier). • The system must consume the H2 and produce products until a new equilibrium is established. • So, [H2] and [N2] will decrease and [NH3] increases.
  • 35. • Adding a reactant or product shifts the equilibrium away from the increase. • Removing a reactant or product shifts the equilibrium towards the decrease. • To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier). • We illustrate the concept with the industrial preparation of ammonia.
  • 36. 4. The product gas stream (containing N2, H2 and NH3) is passed over a cooler to a refrigeration unit. 1. N2 and H2 are pumped into a chamber. 3. 6. 7. The unreacted nitrogen and hydrogen are recycled with the new N2 and H2 feed gas. The equilibrium amount of ammonia is optimized because the product (NH3) is continually removed and the reactants (N2 and H2) are continually being added. 5. The catalyst bed is kept at 460 550 C under high pressure. In the refrigeration unit, ammonia liquefies not N2 or H2. 2. The pre-heated gases are passed through a heating coil to the catalyst bed.
  • 37. Consider the following system starting at equilibrium:
  • 38. Effects of Volume and Pressure Changes • As volume is decreased pressure increases. • Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase. • In a reaction with the same number of product and reactant moles of gas, pressure has no effect.
  • 39. For an ideal gas • The term (n/V) represents concentration, that is, mol/L. • At constant temperature, n, R, and T are constants. • Thus, if the volume occupied by a gas decreases, its partial pressure increases and its concentration (n/V) increases. • If the volume of a gas increases, both its partial pressure and its concentration decrease.
  • 40. • Consider the following gaseous system at equilibrium • At constant temperature, a decrease in volume (increase in pressure) increases the concentrations of both A and D. • In the expression for Q, the concentration of D is squared and the concentration of A is raised to the first power. • As a result, the numerator of Q increases more than the denominator as pressure increases. Thus, Q > Kc, and this equilibrium shifts to the left. • Conversely, an increase in volume (decrease in pressure) shifts this reaction to the right until equilibrium is reestablished, because Q < Kc.
  • 41. In general, for reactions that involve gaseous reactants or products, LeChatelier’s Principle allows us to predict the following results:
  • 42. Effect of Temperature Changes • The equilibrium constant is temperature dependent. • For an endothermic reaction, H > 0 and heat can be considered as a reactant. • For an exothermic reaction, H < 0 and heat can be considered as a product.
  • 43. • Adding heat (i.e. heating the vessel) favors away from the increase: – if H > 0, adding heat favors the forward reaction, – if H < 0, adding heat favors the reverse reaction. • Removing heat (i.e. cooling the vessel), favors towards the decrease: – if H > 0, cooling favors the reverse reaction, – if H < 0, cooling favors the forward reaction.
  • 44. • Consider Cr(H2O)62+(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l) for which H > 0. – Co(H2O)62+ is pale pink and CoCl42- is blue. – If a light purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue. – Since H > 0 (endothermic), adding heat favors the forward reaction, i.e. the formation of blue CoCl42-. • Consider Cr(H2O)62+(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l) – If the room temperature equilibrium mixture is placed in a beaker of ice water, the mixture turns bright pink. – Since H > 0, removing heat favors the reverse reaction which is the formation of pink Co(H2O)62+.
  • 45. The Effect of Catalysis • A catalyst lowers the activation energy barrier for the reaction. • Therefore, a catalyst will decrease the time taken to reach equilibrium. • Adding a catalyst to a system changes the rate of the reaction, but this cannot shift the equilibrium in favor of either products or reactants. – it affects the activation energy of both forward and reverse reactions equally • A catalyst does not effect the composition of the equilibrium mixture ~ Kc, does not change.
  • 46. Relationship Between ΔG°rxn and the Equilibrium Constant The free energy change for any other concentrations or pressures is ΔGrxn As reaction occurs, the free energy of the mixture and the concentrations change until at equilibrium ΔGrxn = 0, and the concentrations of reactants and products satisfy the equilibrium constant. At that point, Q becomes equal to K Rearranging gives

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