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Slide 1 / 27 Slide 2 / 27 AP Chemistry Equilibrium Part B: - - PowerPoint PPT Presentation
Slide 1 / 27 Slide 2 / 27 AP Chemistry Equilibrium Part B: - - PowerPoint PPT Presentation
Slide 1 / 27 Slide 2 / 27 AP Chemistry Equilibrium Part B: Le-Chatelier's Principle, Q, and Calculating K values . 2014-10-29 www.njctl.org Slide 3 / 27 Table of Contents click on the topic to go to that section Le-Chatlier's Principle
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Table of Contents
click on the topic to go to that section
Le-Chatlier's Principle Equilibrium Constant
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Le-Chatlier's Principle
Return to Table of Contents
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Equilibrium
The formation of rust (Fe2O3) is favorable and has a large K value. The reverse process ("unrusting") is therefore highly unfavorable.
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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 equilibrium. Q > K Too many products Reaction shifts left to reach equilibrium Q < K Too many reactants Reaction shifts right to reach equilibrium Q = K At Equilibrium No shift
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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? N2(g) + 3H2(g) --> 2NH3(g) K = 0.037 [NH3]0 = [H2]0 = [N2]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
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Le-Chateliers' Principle
If a reaction at equilibrium is disturbed, it will shift to re-establish equilibrium. Effect of changing amount of reactant or product. A(g) --> 2B(g) 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. A(g) --> 2B(g) 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
- f a reaction, so the equilibrium constant is not changed.
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Le-Chateliers' Principle
If a reaction at equilibrium is disturbed, it will shift to re-establish equilibrium. Effect of changing volume. Increasing the volume will shift the reaction to the side with more moles
- f gas, so in this case, to the right.
A(g) --> 2B(g) A(g) --> 2B(g) 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.
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Le-Chateliers' Principle
If a reaction at equilibrium is disturbed, it will shift to reestablish equilibrium. Effect of changing temperature The effect of increasing or decreasing the temperature depends on whether the reaction is exo or endo thermic Endothermic Reaction E + A(g) --> 2B(g) Exothermic Reaction A(g) --> 2B(g) + E Increasing the temperature increases the thermal energy and will shift the reaction left. 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.
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1 Given initial partial pressures of hydrogen and chlorine gas of 2.5 atm each, and an initial partial pressure of HCl
- f 1.5 atm, which of the following would occur?
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 H2 will decrease due to Q > K D The concentration of Cl2 will decrease due to Q < K
H2(g) + Cl2(g) --> 2HCl(g) K = 2.3
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1 Given initial partial pressures of hydrogen and chlorine gas of 2.5 atm each, and an initial partial pressure of HCl
- f 1.5 atm, which of the following would occur?
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 H2 will decrease due to Q > K D The concentration of Cl2 will decrease due to Q < K
H2(g) + Cl2(g) --> 2HCl(g) K = 2.3
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Answer B or D
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2 Will a precipitate form if 100 mL of 0.003 M AgNO
3 is
mixed with 100 mL of 0.005 M NaCl? Yes No
AgCl(s) --> Ag+(aq) + Cl-(aq) K = 1.77 x 10-10
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2 Will a precipitate form if 100 mL of 0.003 M AgNO
3 is
mixed with 100 mL of 0.005 M NaCl? Yes No
AgCl(s) --> Ag+(aq) + Cl-(aq) K = 1.77 x 10-10
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Answer
Yes – Q > K
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3 Which of the following would shift the reaction towards the products? 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
CO(g) + H2O(g) --> CO2(g) + H2(g) H = -45 kJ
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3 Which of the following would shift the reaction towards the products? 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
CO(g) + H2O(g) --> CO2(g) + H2(g) H = -45 kJ
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Answer
D
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4 Which of the following changes would dissolve more precipitate? 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
AgCl(s) --> Ag+(aq) + Cl-(aq) H = +12 kJ
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4 Which of the following changes would dissolve more precipitate? 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
AgCl(s) --> Ag+(aq) + Cl-(aq) H = +12 kJ
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Answer
A
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5 What would be true if the pressure was increased in the reaction vessel containing the reaction below? A The partial pressure of O2(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
2KClO3(s) --> 2KCl(s) + 3O2(g) H = +35 kJ
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5 What would be true if the pressure was increased in the reaction vessel containing the reaction below? A The partial pressure of O2(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
2KClO3(s) --> 2KCl(s) + 3O2(g) H = +35 kJ
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Answer
A
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6 Which of the following would decrease the value of K for the reaction below? A Increasing the temperature B Decreasing the temperature C Adding H2O D Decreasing solid Ca(OH)2 E Adding a strong acid
Ca(OH)2(s) --> Ca2+(aq) + 2OH-(aq) H = -32 kJ
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6 Which of the following would decrease the value of K for the reaction below? A Increasing the temperature B Decreasing the temperature C Adding H2O D Decreasing solid Ca(OH)2 E Adding a strong acid
Ca(OH)2(s) --> Ca2+(aq) + 2OH-(aq) H = -32 kJ
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Answer
A
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Application
Shower scale is typically composed partly of metal carbonate precipitates which do not readily dissolve in tap water. CaCO3(s) --> Ca2+(aq) + CO32-(aq) Ksp = 6 x 10-9 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. CaCO3(s) --> Ca2+(aq) + CO32-(aq)
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Equilibrium Constant
Return to Table of Contents
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The Equilibrium Constant
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. CaCO3(s) --> Ca2+(aq) + CO32-(aq) Ksp = 6 x 10-9 Ca2+(aq) + CO32-(aq) --> CaCO3(s) Ksp = 2 x 108 This makes intuitive sense. If a reaction is favored in one direction, it will be unfavorable in the reverse direction.
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The Equilibrium Constant
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. CaCO3(s) --> Ca2+(aq) + CO32-(aq) Ksp = 6 x 10-9 1/2CaCO3(s) --> 1/2Ca2+(aq) + 1/2CO32-(aq) Ksp = (6 x 10-9)1/2 Ksp = 8 x 10-5
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The Equilibrium Constant
If reactions are added together, the equilibrium constants are multiplied. Many reactions occur as a series of steps. Consider the production of Ag(NH3)2+ Step 1: AgCl(s) --> Ag
+(aq) + Cl-(aq) Ksp = 1.8 x 10-10
Step 2: Ag+(aq) + 2NH3(aq) --> Ag(NH3)2+(aq) Kf = 1.6 x 108 Overall AgCl(s) + 2NH3(aq) --> Ag(NH3)2+ + Cl-(aq) Ksp*Kf = Koverall =0.028
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7 At 300 K, the production of 2 moles of ammonia from nitrogen and hydrogen gas has a K = 4.3 x 10-3. What is the value of K for the production of 3 moles of ammonia?
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7 At 300 K, the production of 2 moles of ammonia from nitrogen and hydrogen gas has a K = 4.3 x 10-3. What is the value of K for the production of 3 moles of ammonia?
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Answer
2.82 x 10-4
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8 At 300 K, the production of 2 moles of ammonia from nitrogen and hydrogen gas has a K = 4.3 x 10-3. What is the value of K for the decomposition of 1 mole of ammonia?
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8 At 300 K, the production of 2 moles of ammonia from nitrogen and hydrogen gas has a K = 4.3 x 10-3. What is the value of K for the decomposition of 1 mole of ammonia?
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Answer
15.2
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9 Given the reactions below, what would be the value of K for the reaction: Mn2+(aq) + H2S(aq) --> CaS(s) + 2H+(aq)
MnS(s) --> Mn2+(aq) + S2-(aq) Ksp = 3 x 10-14 H2S(aq) --> H+(aq) + HS-(aq) Ka = 1.0 x 10-7 HS-(aq) --> H+(aq) + S2-(aq) Ka = 1.3 x 10-13
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9 Given the reactions below, what would be the value of K for the reaction: Mn2+(aq) + H2S(aq) --> CaS(s) + 2H+(aq)
MnS(s) --> Mn2+(aq) + S2-(aq) Ksp = 3 x 10-14 H2S(aq) --> H+(aq) + HS-(aq) Ka = 1.0 x 10-7 HS-(aq) --> H+(aq) + S2-(aq) Ka = 1.3 x 10-13
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Answer
4.33 X10-7
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10 Given the following data below, what would be the value
- f K for the reaction:
Ca2+(aq) + Ag2CO3(s) --> 2Ag+(aq) + CaCO3(s)
CaCO3(s) --> Ca2+(aq) + CO32-(aq) Ksp = 6 x 10-9 Ag2CO3(s) --> 2Ag+(aq) + CO32-(aq) Ksp = 8 x 10-12
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10 Given the following data below, what would be the value
- f K for the reaction:
Ca2+(aq) + Ag2CO3(s) --> 2Ag+(aq) + CaCO3(s)
CaCO3(s) --> Ca2+(aq) + CO32-(aq) Ksp = 6 x 10-9 Ag2CO3(s) --> 2Ag+(aq) + CO32-(aq) Ksp = 8 x 10-12
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Answer
1.33 X10-3
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11 Given the following data, which of the following would be MOST favorable? A Formation of 4 moles of ammonia B Formation of 2 moles of Ca2+ ions C Decomposition of 6 moles of ammonia D Formation of 0.1 moles of CaCO3(s) from the ions E All are equally favorable
CaCO3(s) --> Ca2+(aq) + CO32-(aq) Ksp = 6.0 x 10-9 3H2(g) + N2(g) --> 2NH3(g) K = 4.3 x 10-3
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11 Given the following data, which of the following would be MOST favorable? A Formation of 4 moles of ammonia B Formation of 2 moles of Ca2+ ions C Decomposition of 6 moles of ammonia D Formation of 0.1 moles of CaCO3(s) from the ions E All are equally favorable
CaCO3(s) --> Ca2+(aq) + CO32-(aq) Ksp = 6.0 x 10-9 3H2(g) + N2(g) --> 2NH3(g) K = 4.3 x 10-3
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Answer
C
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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 --> Co3+(aq) + 3OH-(aq) Ksp = 1.6 x 10-44 But can be made more soluble by coupling it with a reaction that forms the complex Co(NH3)63+(aq) Co3+(aq) + 6NH3(aq) --> Co(NH3)63+(aq) Kf = 4.5 x 1031 Coupled Reaction Co(OH)3 + 6NH3(aq) --> Co(NH3)63+ + 3OH- Koverall = 7.2 x 10-13