Presentation is loading. Please wait.

Presentation is loading. Please wait.

7. Equilibria Learning Outcomes Candidates should be able to: (a)Explain in terms of rates of the forward and reverse reactions, what is meant by a reversible.

Similar presentations


Presentation on theme: "7. Equilibria Learning Outcomes Candidates should be able to: (a)Explain in terms of rates of the forward and reverse reactions, what is meant by a reversible."— Presentation transcript:

1

2 7. Equilibria Learning Outcomes Candidates should be able to: (a)Explain in terms of rates of the forward and reverse reactions, what is meant by a reversible reaction and dynamic equilibrium (b)State Le Chatelier’s Principle and apply it to deduce qualitatively (from appropriate information) the effects of changes in temperature, concentration or pressure, on a system at equilibrium (c)Deduce whether changes in concentration, pressure or temperature or the presence of a catalyst affect the value of the equilibrium constant for the reaction (d)Deduce expressions for equilibrium constants in terms of concentrations, Kc, and partial pressures, Kp (e)Calculate the values of equilibrium constants in terms of concentrations or partial pressures from appropriate data (f)Calculate the quantities present at equilibrium, given appropriate data (g)Describe 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) (h)Show understanding of, and use the Brønsted-Lowry theory of acids and bases (i)Explain qualitatively the differences in behaviour between strong and weak acids and bases in terms of the extent of dissociation

3 The Concept of Equilibrium A, B C, D Although 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 processes take 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.

4 The Equilibrium Constant For the reversible reaction we saw that chemical equilibrium is attained when The 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).

5 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 constant Applies irrespective of whether the reaction is elementary or complex.

6 The Equilibrium Constant (contd.) Examples Write down expressions for the concentration equilibrium constants of the following reactions Balanced chemical equation

7 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 mixture It follows that at a constant temperature

8 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.

9 The Equilibrium Constant (contd.) Examples Write down expressions for the concentration equilibrium constants of the following reactions Balanced chemical equation

10 The Equilibrium Constant (contd.) For the reversible reaction the concentration equilibrium constant is given by The magnitude of equilibrium constants The 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.

11 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.

12 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 reactant than product. The position of equilibrium is said to be far to the left.

13 To summarise:

14 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 H 2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of K c ? CO(g) + 2H 2 (g)  CH 3 OH(g) I. C. E. 0.1500 0.30000 -x-x - 2x +x 0.1500 - x 0.3000 - 2x x Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187 Therefore x = 0.0313 We can now solve for each of the other Eq. terms.

15 Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H 2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of K c ? CO(g) + 2H 2 (g)  CH 3 OH(g) I. C. E. 0.1500 0.30000 -x-x - 2x +x 0.1500 - x 0.3000 - 2x x Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187 Therefore x = 0.0313 We can now solve for each of the other Eq. terms. H 2 : 0.3000 - 2x = 0.2374 moles CH 3 OH: x = 0.0313 moles Therefore E. 0.1187 0.2374 0.0313 CO(g) + 2H 2 (g)  CH 3 OH(g)

16 Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H 2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of K c ? CO(g) + 2H 2 (g)  CH 3 OH(g) I. C. E. 0.1500 0.30000 -x-x - 2x +x 0.1500 - x 0.3000 - 2x x Therefore E. 0.1187 0.2374 0.0313 CO(g) + 2H 2 (g)  CH 3 OH(g) Now find K c : K c = 10.52

17 Effect of temperature and pressure on reversible reactions Experimentally it has been found that the composition of equilibrium mixtures obtained from reversible reactions may be affected by: Temperature Addition or removal of reactants or products, with consequential changes in concentrations or partial pressures Changes in total pressure

18 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 reactions An example:

19 Haber Process

20 Le Châtelier’s 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 Châtelier’s Principle Could these effects have been predicted?

21 Change in Reactant or Product Concentrations Example Consider the Haber process reaction If H 2 is added while the system is at equilibrium, the system must respond to counteract the added H 2 (by Le Châtelier’s Principle). The system must consume the H 2 and produce more of the products until a new equilibrium is established. So, most of the additional H 2 will be consumed, and [N 2 ] will decrease, while [NH 3 ] will increase. Le Châtelier’s Principle (contd.)

22

23 Change in Reactant or Product Concentrations Adding 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 Châtelier’s Principle). We illustrate the application of these principles with the industrial preparation of ammonia. Le Châtelier’s Principle (contd.)

24 Haber Process Iron catalyst

25 Change in Reactant or Product Concentrations N 2 and H 2 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 N 2, H 2 and NH 3 ) is passed over a cooler to a refrigeration unit. In the refrigeration unit, ammonia liquefies, but not N 2 or H 2. Le Châtelier’s Principle (contd.)

26 Change in Reactant or Product Concentrations The unreacted nitrogen and hydrogen are recycled with the new N 2 and H 2 feed gas. The equilibrium amount of ammonia is optimized because the product (NH 3 ) is continually removed and the reactants (N 2 and H 2 ) are continually being added. Le Châtelier’s Principle (contd.)

27 Effects of Volume and Pressure Changes As volume is decreased pressure increases. Le Châtelier’s 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 Châtelier’s Principle (contd.)


Download ppt "7. Equilibria Learning Outcomes Candidates should be able to: (a)Explain in terms of rates of the forward and reverse reactions, what is meant by a reversible."

Similar presentations


Ads by Google