AP Chemistry. Le-Chatlier's Principle. Slide 1 / 27 Slide 2 / 27. Slide 4 / 27. Slide 3 / 27. Slide 5 / 27. Slide 6 / 27

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1 Slide 1 / 27 Slide 2 / 27 AP Chemistry Equilibrium Part B: Le-Chatelier's Principle, Q, and Calculating K values Slide 3 / 27 Slide 4 / 27 Table of Contents click on the topic to go to that section Le-Chatlier's Principle Equilibrium Constant Le-Chatlier's Principle Return to Table of Contents Slide 5 / 27 Equilibrium Slide 6 / 27 Q and K The reaction quotient, "Q", is the ratio of products to reactants at any stage in a reaction. Q = [Products] x [Reactants] y The value of Q and it's relation to K provides information as to which way a reaction will shift to reach The formation of rust (Fe 2O 3) is favorable and has a large K value. The reverse process ("unrusting") is therefore highly unfavorable. Q > K Q < K Too many products Too many reactants Reaction shifts left to reach equilibrium Reaction shifts right to reach equilibrium Q = K At Equilibrium No shift

2 Slide 7 / 27 Q and K Given initial amounts, the direction the reaction will shift to reach equilibrium can be determined by comparing Q to K. Given the following reaction, which way will the reaction shift to reach equilibrium given the following initial concentrations? N 2(g) + 3H 2(g) --> 2NH 3(g) K = [NH 3] 0 = [H 2] 0 = [N 2] 0 = 0.41 M Q = (0.41) 2 / (0.41)*(0.41) 3 = 5.94 Q > K = too many products Reaction will shift LEFT to reach equilibrium Slide 8 / 27 If a reaction at equilibrium is disturbed, it will shift to re-establish Effect of changing amount of reactant or product. Increasing the amount of B increases the value of "Q" making the reverse reaction happen more readily shifting the reaction left until equilibrium is re-established. Decreasing the amount of B makes the reverse reaction less likely thereby shifting the reaction towards the products until equilibrium is reestablished. Note: Changing the concentrations does NOT affect the favorability of a reaction, so the equilibrium constant is not changed. Slide 9 / 27 If a reaction at equilibrium is disturbed, it will shift to re-establish Effect of changing volume. Increasing the volume will shift the reaction to the side with more moles of gas, so in this case, to the right. Changes in volume can be created by changing the pressure. In this case, increasing the pressure will lower the volume and shift the reaction to the side with fewer gaseous moles. Note: Since volume changes amount to changing concentrations, changing the concentrations does NOT affect the favorability of a reaction, so the equilibrium constant is not changed. Slide 11 / 27 1 Given initial partial pressures of hydrogen and chlorine gas of 2.5 atm each, and an initial partial pressure of HCl of 1.5 atm, which of the following would occur? H 2(g) + Cl 2(g) --> 2HCl(g) K = 2.3 Slide 10 / 27 If a reaction at equilibrium is disturbed, it will shift to reestablish Effect of changing temperature The effect of increasing or decreasing the temperature depends on whether the reaction is exo or endo thermic Exothermic Reaction + E Increasing the temperature increases the thermal energy and will shift the reaction left. Endothermic Reaction E + Increasing the temperature increases the thermal energy and will shift the reaction right. Unlike a change in concentration, V, or P, a change in the temperature DOES change the value of the equilibrium constant. Slide 12 / 27 2 Will a precipitate form if 100 ml of M AgNO 3 is mixed with 100 ml of M NaCl? AgCl(s) --> Ag + + Cl - K = 1.77 x A The concentration of HCl will increase due to Q > K B The concentration of HCl will increase due to Q < K C The concentration of H 2 will decrease due to Q > K D The concentration of Cl 2 will decrease due to Q < K Yes No

3 Slide 13 / 27 3 Which of the following would shift the reaction towards the products? Slide 14 / 27 4 Which of the following changes would dissolve more precipitate? CO(g) + H 2O(g) --> CO 2(g) + H 2(g) H = -45 kj AgCl(s) --> Ag + + Cl - H = +12 kj A Increasing the temperature B Decreasing the volume C Decreasing the pressure D Increasing the concentration of CO(g) E Adding an inert gas such as helium A Removing chloride ions by precipitating them with Cu+ ions B Increasing amount of solid AgCl C Lowering the temperature D Increasing the pressure E None of these Slide 15 / 27 5 What would be true if the pressure was increased in the reaction vessel containing the reaction below? 2KClO 3(s) --> 2KCl(s) + 3O 2(g) H = +35 kj Slide 16 / 27 6 Which of the following would decrease the value of K for the reaction below? Ca(OH) 2(s) --> Ca OH - H = -32 kj A The partial pressure of O 2(g) will diminish B No change will occur C The temperature will decrease D The amount of solid KCl will increase E None of these A Increasing the temperature B Decreasing the temperature C Adding H 2O D Decreasing solid Ca(OH) 2 E Adding a strong acid Slide 17 / 27 Application Slide 18 / 27 Shower scale is typically composed partly of metal carbonate precipitates which do not readily dissolve in tap water. Ksp = 6 x 10-9 Equilibrium Constant Using an acidic cleaning agent however can shift the equilibrium to the right via the depletion of carbonate ions through the reaction of them with H+ ions. Return to Table of Contents

4 Slide 19 / 27 The equilibrium constant for a reaction is specific to how the reaction is written. If a reaction is reversed, the new K value will be the inverse of the original K value. Ksp = 6 x 10-9 Ca 2+ + CO 3 --> CaCO 3(s) Ksp = 2 x 10 8 This makes intuitive sense. If a reaction is favored in one direction, it will be unfavorable in the reverse direction. Slide 20 / 27 The equilibrium constant for a reaction is specific to how the reaction is written. If the coefficients of a reaction are changed, the exponents of each substance will change in the expression thereby exponentially changing K. Ksp = 6 x /2CaCO 3(s) --> 1/2Ca /2CO 3 Ksp = (6 x 10-9 ) 1/2 Ksp = 8 x 10-5 Slide 21 / 27 If reactions are added together, the equilibrium constants are multiplied. Slide 22 / 27 7 At 300 K, the production of 2 moles of ammonia from nitrogen and hydrogen gas has a K = 4.3 x What is the value of K for the production of 3 moles of ammonia? Many reactions occur as a series of steps. Consider the production of Ag(NH 3) 2 + Step 1: AgCl(s) --> Ag + + Cl - K sp = 1.8 x Step 2: Ag + + 2NH 3 --> Ag(NH 3) 2+ K f = 1.6 x 10 8 Overall AgCl(s) + 2NH 3 --> Ag(NH 3) Cl - K sp*k f = K overall =0.028 Slide 23 / 27 8 At 300 K, the production of 2 moles of ammonia from nitrogen and hydrogen gas has a K = 4.3 x What is the value of K for the decomposition of 1 mole of ammonia? Slide 24 / 27 9 Given the reactions below, what would be the value of K for the reaction: Mn 2+ + H 2S --> CaS(s) + 2H + MnS(s) --> Mn 2+ + S Ksp = 3 x H 2S --> H + + HS - Ka = 1.0 x 10-7 HS - --> H + + S Ka = 1.3 x 10-13

5 Slide 25 / Given the following data below, what would be the value of K for the reaction: Ca 2+ + Ag 2CO 3(s) --> 2Ag + + CaCO 3(s) Ksp = 6 x 10-9 Ag 2CO 3(s) --> 2Ag + + CO 3 Ksp = 8 x Slide 26 / Given the following data, which of the following would be MOST favorable? Ksp = 6.0 x H 2(g) + N 2(g) --> 2NH 3(g) K = 4.3 x 10-3 A Formation of 4 moles of ammonia B Formation of 2 moles of Ca 2+ ions C Decomposition of 6 moles of ammonia D Formation of 0.1 moles of CaCO 3(s) from the ions E All are equally favorable Slide 27 / 27 Application Coupling a non-favorable reaction with a small K value with one that is highly favorable with a large K value is a common way to drive a process towards product. For example, Co(OH) 3(s) is rather insoluble... Co(OH) 3 --> Co + 3OH - Ksp = 1.6 x But can be made more soluble by coupling it with a reaction that forms the complex Co(NH 3) 6 Co + 6NH 3 --> Co(NH 3) 6 Kf = 4.5 x Coupled Reaction Co(OH) 3 + 6NH 3 --> Co(NH 3) 6 + 3OH - K overall = 7.2 x 10-13