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RETSD Adult Ed Program CHEM 40S Module 3, Lesson 4 LESSON 4 -- LE CHATELIER’S PRINCIPLE Definition: Le Chatelier’s Principle: If a system is in equilibrium and a condition is changed, then the equilibrium will shift toward restoring the original conditions. The conditions affecting equilibrium are temperature, pressure, and concentration of either reactants or products. If stress is put on a reversible reaction at equilibrium, the equilibrium will shift in such a way as to relieve the stress. Example: The reaction for the preparation of ammonia by the Haber process is: N2(g) + 3H2(g) 2NH3(g) + heat Consider the effects of concentration, pressure, and temperature on the equilibrium. If the concentration of either of the reactants is increased, the number of collisions between reactant particles will increase. The result is an increase in the reaction rate toward the right, or in other words, an increase in the rate of the forward reaction. Then, as the amount of NH3 increases, the rate of the reverse reaction also increases. However, the net result to the system as a whole is to shift the equilibrium toward the right. That is, more product is produced. For example, if the [H2] is increased, the [N2] is decreased and the [NH3] increases. The value of the equilibrium constant remains the same. The removal of a product will have the effect of driving the equilibrium to the right to replace the lost product. Consider the example of an organic acid reacting with an alcohol to produce an ester and water. An ester is a compound with a pleasant odor that can be synthesized in the laboratory by reacting an alcohol and an organic acid (chemistry 30S students do this as part of their course at the CLC). Esters account for the distinctive odors of many fruits such as bananas, pineapples, and oranges. Organic acid + alcohol ester + water The reactions are very slow; therefore the reactants must be heated, and a catalyst must be added in order to make the reaction occur with any speed. In esterification reactions, the products are in chemical equilibrium with the reactants. They will seldom go to completion unless the water produced in the reaction is removed as it forms. The catalyst added is concentrated sulfuric acid, which also acts as a dehydrating agent (it removes the water from the equilibrium). The removal of water helps to force the equilibrium to the right, producing more product (ester). 2010 22 RETSD Adult Ed Program CHEM 40S Module 3, Lesson 4 If a reactant or a product is added to a system at equilibrium, the system will shift away from the added component. If a reactant or product is removed, the system will shift toward the removed component. Returning to the Haber process as an example, if the pressure is increased, more product is formed. What if the pressure was doubled? Doubling the pressure exerted on a gas will decrease its volume to half the original volume. Therefore, the concentrations of nitrogen, hydrogen, and ammonia are all doubled. N2(g) + 3H2(g) 2NH3(g) Kr = [NH3]2 The reverse reaction must then speed up by a factor of 4, as indicated by the rate law for the reverse reaction. On the other hand, the concentration of hydrogen is cubed. Further, it is multiplied by the concentration of nitrogen: Kf = [N2] [H2]3 Doubling the pressure, then, should increase the rate of the forward reaction by 23 x 2, or 16 times. The net result is clearly an increase in product. Increased pressure on a reaction system with a gas phase has the same effect as increased concentration. Pressure increases concentration, not Keq. Temperature changes cause a change in the value Keq. An increase in pressure on a system at equilibrium favors the side with fewer gas molecules -- the equilibrium will shift to that side. Using the Haber process as an example, consider the application of pressure on the whole system at equilibrium. By shifting to produce more product, NH3, the system reduces the stress because it reduces the number of gas molecules and thus the pressure in the container is reduced. For every four reactant molecules, two product molecules are produced. To reduce the added pressure, the system will shift to the side with fewer gas molecules. 2010 23 RETSD Adult Ed Program CHEM 40S Module 3, Lesson 4 When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume. This is done by decreasing the total number of gaseous molecules in the system. Example: In the reaction H2(g) + Cl2(g) 2HCl(g), all substances are gases. Pressure would not shift the equilibrium, as the rate in each direction would be affected the same way. The number of gas molecules is the same on both sides. Pressure, of course, has an effect only on the gases in the reaction. A reaction that takes place in solution would be unaffected by pressure. TEMPERATURE It is important to realize that although the changes we have just discussed may alter the equilibrium position, they do not alter the equilibrium constant, Keq. For example, the addition of a reactant shifts the equilibrium position to the right but has no effect on the value of the equilibrium constant; the new equilibrium concentrations still work to calculate the same value of Keq in the equilibrium expression. The effect of temperature on equilibrium is different, however, because the value of Keq changes with temperature. Both the forward and reverse reactions of any system at equilibrium are sped up by an increase in temperature. However, their rates are increased by different amounts. Also, the value of the equilibrium constant itself is changed by a change in temperature. One easy way to predict the shift in an equilibrium subjected to a temperature change is to consider heat as a reactant or product. In the Haber process: N2(g) + 3H2(g) 2NH3(g) + heat heat is produced when hydrogen and nitrogen react. If heat is considered to be a product, the addition of heat (a product) will increase the concentration of the product. The equilibrium will shift to the left. In the Haber process, the reverse reaction (the decomposition of ammonia) is favored by the addition of heat. Optimum conditions are those which produce the highest yield of product. In industry, a desired chemical can sometimes be obtained by a reversible reaction. However, equilibrium is often attained before enough product is produced to make the process economical. In such circumstances, the equilibrium can be shifted to get a higher yield of product. The chemist determines what conditions will tend to produce the highest 2010 24 RETSD Adult Ed Program CHEM 40S Module 3, Lesson 4 yield. The conditions which produce the highest yield are called the optimum conditions. For example, the optimum conditions for the Haber process are: High concentration of H2 and N2 must be maintained removal of NH3 as it is formed precise temperature control - high enough to maintain a reasonable rate but low enough not to favor the reverse reaction use of a contact catalyst to lower the required activation energy a high pressure should be maintained through the reaction Each condition increases the yield by shifting the equilibrium to favor the product. LESSON 4, ASSIGNMENT 1 Answer the following questions. For each situation, predict the shift in equilibrium using Le Chatelier’s Principle and explain the shift using collision - rate theory. 1. When a few drops of a 0.20 mol/L Fe(NO3)3(aq) solution are added to 50 mL of a 1.0 mol/L KSCN(aq) solution, the following equilibrium is established: Fe3+(aq) + pale yellow SCN(aq) FeSCN2+(aq) colorless red a. A crystal of KSCN(s) is added to the equilibrium mixture. Hint: KSCN is an ionic compound – when added to the solution, it will dissociate into its component ions. b. A few drops of dilute NaOH(aq) are added to the equilibrium mixture. NaOH is also an ionic compound which will dissociate into its ions when added. 2. For each of the following systems at equilibrium, predict how they will shift as temperature is increased: a. N2(g) + O2(g) + energy 2NO (g) b. 2SO2(g) + O2(g) 2SO3(g) (forward reaction is exothermic) 3. For the exothermic reaction 2SO2(g) + O2(g) 2SO3(g) predict the equilibrium shift caused by each of the following changes: 2010 25 RETSD Adult Ed Program CHEM 40S a. b. c. d. Module 3, Lesson 4 SO2 is added SO3 is removed The volume is decreased The temperature is decreased 4. Complete the following table: Reaction: Energy + N2O4(g) 2NO2(g) Change Addition of N2O4(g) Addition of NO2(g) Removal of N2O4(g) Removal of NO2(g) Decrease in container volume Increase in container volume Increase in temperature Decrease in temperature Shift (left/right) Reading Assignment Chemistry: The Study of Matter Sections 18-4 to 18-8: Pages 518 to 532 Problem Assignment 2 Chemistry: The Study of Matter Practice Problems No. 9 to 11: Page 523 Practice Problems No. 14 & 15: Page 529 End of Module 3 – Read over and do all questions a second time in order to be well prepared for the test. Following is a list of important terms you should know: Reaction rate collision theory activated complex activation energy REACTION MECHANISM rate determining step Concentration rate law expression specific rate constant heterogeneous reactions 2010 26 RETSD Adult Ed Program CHEM 40S Module 3, Lesson 4 activated complex catalyst homogeneous catalyst coefficient equilibrium solubility solvent solute saturated spontaneous reaction nonspontaneous reaction CHEMICAL EQUILIBRIUM equilibrium constant Keq law of mass action incomplete reaction complete reaction homogeneous equilibria heterogeneous equilibria Le Chatelier’s Principle the Haber process factors necessary for equilibrium to exist 2010 27