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7. Equilibria

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7. Equilibria. Learning Outcomes Candidates should be able to: Explain in terms of rates of the forward and reverse reactions, what is meant by a reversible reaction and dynamic equilibrium - PowerPoint PPT Presentation

Text of 7. Equilibria

  • 7. EquilibriaLearning OutcomesCandidates should be able to:Explain in terms of rates of the forward and reverse reactions, what is meant by a reversible reaction and dynamic equilibriumState Le Chateliers Principle and apply it to deduce qualitatively (from appropriate information) the effects of changes in temperature, concentration or pressure, on a system at equilibriumDeduce whether changes in concentration, pressure or temperature or the presence of a catalyst affect the value of the equilibrium constant for the reactionDeduce expressions for equilibrium constants in terms of concentrations, Kc, and partial pressures, KpCalculate the values of equilibrium constants in terms of concentrations or partial pressures from appropriate dataCalculate the quantities present at equilibrium, given appropriate dataDescribe and explain the conditions used in the Haber process and the Contact process, as examples of the importance of an understanding of chemical equilibrium in the chemical industry (see also section 9.6)Show understanding of, and use the Brnsted-Lowry theory of acids and basesExplain qualitatively the differences in behaviour between strong and weak acids and bases in terms of the extent of dissociation

  • The Concept of EquilibriumA , B

    C , DAlthough it would appear that no further change takes place once the system has reached equilibrium, be aware that at the molecular level the following two reaction processestake place continuously. There does not appear to be any change taking place because the two processes take place at exactly the same rates.This is an example of dynamic equilibrium in chemistry.

  • The Equilibrium ConstantFor the reversible reaction we saw that chemical equilibrium is attained whenThe constant is called the equilibrium constant of the reaction. Note: in deriving this expression we have assumed that the reaction of interest is an elementary reaction in both directions. However, it turns out that the above expression holds true for ANY chemical reaction (whether elementary or complex).

  • The Equilibrium Constant (contd.)In 1864 GULDBERG and WAAGE postulated the now famous LAW OF MASS ACTION. According to the law of mass action:

    Any reversible chemical reaction will have associated with it an equilibrium constantApplies irrespective of whether the reaction is elementary or complex.

  • The Equilibrium Constant (contd.)ExamplesWrite down expressions for the concentration equilibrium constants of the following reactions

    Balanced chemical equation

  • The Equilibrium Constant (contd.)When dealing with reactions taking place in the gas phase, it is often more convenient to deal with partial pressures of reactants and products than with concentrations.From the ideal gas equation for a component of a gas mixtureIt follows that at a constant temperature

  • The Equilibrium Constant (contd.)Accordingly, equilibrium constants for reactions involving gaseous reactants and products can be expressed in terms of partial pressures in the place of concentrations.

  • The Equilibrium Constant (contd.)ExamplesWrite down expressions for the concentration equilibrium constants of the following reactions

    Balanced chemical equation

  • The Equilibrium Constant (contd.)For the reversible reaction the concentration equilibrium constant is given by The magnitude of equilibrium constantsThe magnitude of the equilibrium constant gives information about what we could expect the reaction mixture to look like at equilibrium. If then we would expect that, when the system has reached equilibrium [C] , [D] [A] , [B]. The equilibrium mixture will contain significant amounts of all the reactants and products.

  • The Equilibrium Constant (contd.)The magnitude of equilibrium constants If then we would expect that, when the system has reached equilibrium [C] , [D] >> [A] , [B]. The equilibrium mixture will contain much more product than reactant. The position of equilibrium is said to be far to the right.

  • The Equilibrium Constant (contd.)The magnitude of equilibrium constants If then we would expect that, when the system has reached equilibrium [C] , [D]
  • To summarise:

  • Using I.C.E. (Initial, Change, Equilibrium)Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc? CO(g) + 2H2(g) CH3OH(g)I.C.E.0.15000.30000-x-2x+x0.1500 - x0.3000 - 2xxSince @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187Therefore x = 0.0313We can now solve for each of the other Eq. terms.

  • Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc? CO(g) + 2H2(g) CH3OH(g)I.C.E.0.15000.30000-x-2x+x0.1500 - x0.3000 - 2xxSince @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187Therefore x = 0.0313We can now solve for each of the other Eq. terms.H2: 0.3000 - 2x = 0.2374 molesCH3OH: x = 0.0313 molesTherefore E. 0.1187 0.2374 0.0313CO(g) + 2H2(g) CH3OH(g)

  • Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc? CO(g) + 2H2(g) CH3OH(g)I.C.E.0.15000.30000-x-2x+x0.1500 - x0.3000 - 2xxTherefore E. 0.1187 0.2374 0.0313CO(g) + 2H2(g) CH3OH(g)Now find Kc: Kc = 10.52

  • Effect of temperature and pressure on reversible reactionsExperimentally it has been found that the composition of equilibrium mixtures obtained from reversible reactions may be affected by:TemperatureAddition or removal of reactants or products, with consequential changes in concentrations or partial pressuresChanges in total pressure

  • In the Haber Process for the production of ammonia, based on the reversible reaction:

    it is observed that:As the total pressure increases, the amount of ammonia present at equilibrium increases.As the temperature decreases, the amount of ammonia at equilibrium increases.

    Effect of temperature and pressure on reversible reactionsAn example:

  • Haber Process

  • Le Chteliers Principle states that if a system at equilibrium is disturbed, the position of equilibrium will shift in such a way as to counteract the disturbance.Le Chteliers PrincipleCould these effects have been predicted?

  • Change in Reactant or Product ConcentrationsExampleConsider the Haber process reaction

    If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 (by Le Chteliers Principle).The system must consume the H2 and produce more of the products until a new equilibrium is established.So, most of the additional H2 will be consumed, and [N2] will decrease, while [NH3] will increase.Le Chteliers Principle (contd.)

  • Change in Reactant or Product ConcentrationsAdding a reactant or product shifts the position of 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 (by Le Chteliers Principle).We illustrate the application of these principles with the industrial preparation of ammonia.Le Chteliers Principle (contd.)

  • Haber ProcessIron catalyst

  • Change in Reactant or Product ConcentrationsN2 and H2 are pumped into a chamber.The pre-heated gases are passed through a heating coil to the catalyst bed.The catalyst bed is kept at 460 - 550 C under high pressure.The product gas stream (containing N2, H2 and NH3) is passed over a cooler to a refrigeration unit.In the refrigeration unit, ammonia liquefies, but not N2 or H2.Le Chteliers Principle (contd.)

  • Change in Reactant or Product ConcentrationsThe 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.

    Le Chteliers Principle (contd.)

  • Effects of Volume and Pressure ChangesAs volume is decreased pressure increases. Le Chteliers Principle: if pressure is increased the position of equilibrium will shift to counteract the increase.That is, the position of equilibrium shifts to remove gases and decrease pressure.An increase in pressure favors the direction that has fewer moles of gas.In a reaction with the same number of product and reactant moles of gas, pressure has no effect.Le Chteliers Principle (contd.)

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