AP Chemistry Rapid Learning Series - 20

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Chemistry :: Biology :: Physics :: Math

Rapid Learning Cen...
AP Chemistry Rapid Learning Series - 20

Learning Objectives
By completing this tutorial you will learn about…
Dynamic equ...
AP Chemistry Rapid Learning Series - 20

Equilibrium

5/66

Definition - Equilibrium
Reversible Reaction – A
c e ca eact o...
AP Chemistry Rapid Learning Series - 20

Establishing Equilibrium
It takes time to establish equilibrium.
At first, there ...
AP Chemistry Rapid Learning Series - 20

Equilibrium
Constants

9/66

Definition - Equilibrium Constant
Equilibrium Consta...
AP Chemistry Rapid Learning Series - 20

Writing Equilibrium Constant Expressions
To write an equilibrium constant express...
AP Chemistry Rapid Learning Series - 20

Concentrations of Solids and Liquids
Pure solids and pure liquids have constant “...
AP Chemistry Rapid Learning Series - 20

Calculating “K” Example
Example: Solve for equilibrium constant for
Fe2O3 (s) + 3...
AP Chemistry Rapid Learning Series - 20

Using “K” to Find Equilibrium Concentration
Example: Find the equilibrium concent...
AP Chemistry Rapid Learning Series - 20

What is the Reaction Quotient?
Reaction Quotient is “Q”
K

Q

Equilibrium Constan...
AP Chemistry Rapid Learning Series - 20

Using Reaction Quotient
Reaction Quotient is used to determine if a system
is at ...
AP Chemistry Rapid Learning Series - 20

Solubility
Equilibrium
Constants

23/66

Definition - Solubility Product
Solubili...
AP Chemistry Rapid Learning Series - 20

Breaking Compounds into Electrolytes
How do you break up a compound when forming
...
AP Chemistry Rapid Learning Series - 20

Solubility Product Example
Example: Find the solubility product of Cd(OH)2 if at ...
AP Chemistry Rapid Learning Series - 20

Using Reaction Quotient with Solubility
Reaction Quotient can be used to determin...
AP Chemistry Rapid Learning Series - 20

Solving Equilibrium
Problems

31/66

ICE Charts in Equilibrium Problems
Equilibri...
AP Chemistry Rapid Learning Series - 20

How to Use an ICE Chart
The ICE chart is an efficient way to organize the
informa...
AP Chemistry Rapid Learning Series - 20

Finding Equilibrium Constant - 1
Example: H2 (g) + I2 (g) 2 HI (g) A mixture of 1...
AP Chemistry Rapid Learning Series - 20

Definition: Quadratic Formula
Quadratic formula –
formula used to find
“x”
“ ” in...
AP Chemistry Rapid Learning Series - 20

Finding Equilibrium Concentrations #1
Example: H2 (g) + I2 (g) 2 HI (g) A flask i...
AP Chemistry Rapid Learning Series - 20

Finding Equilibrium Concentrations #3
Example: H2 (g) + I2 (g) 2 HI (g) A flask i...
AP Chemistry Rapid Learning Series - 20

Finding Equilibrium Concentrations #4
Keq = 1.6 × 10-5. If the
Example: 2NOCl (g)...
AP Chemistry Rapid Learning Series - 20

Finding Solubility
Example: Find the solubility of CaF2 if the Ksp is 4.0×10-11
C...
AP Chemistry Rapid Learning Series - 20

Definition—Le Chatelier’s Principle
Le Chatelier’s Principle – If a system
at equ...
AP Chemistry Rapid Learning Series - 20

Decreasing Concentrations
How does removing a reactant or product affect a
system...
AP Chemistry Rapid Learning Series - 20

Definition - Endo & Exothermic Reactions
Endothermic Reaction – The reaction
gy
p...
AP Chemistry Rapid Learning Series - 20

Temperature and Exothermic
For exothermic, think of temperature (or energy) as
a ...
AP Chemistry Rapid Learning Series - 20

Le Chatelier’s Examples
Example: Which way will the reaction shift for each of th...
AP Chemistry Rapid Learning Series - 20

Haber Process
The reaction to produce ammonia, NH3, is very
important to manufact...
AP Chemistry Rapid Learning Series - 20

Equilibrium in the Exam
Common equilibrium problems:
The first free response ques...
AP Chemistry Rapid Learning Series - 20

Free Response Questions
The equilibrium free response question is often about
sol...
AP Chemistry Rapid Learning Series - 20

Answering Free Response Questions
At 25°C, the value of Ksp for PbCl2 is 1.6 × 10...
AP Chemistry Rapid Learning Series - 20

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  1. 1. AP Chemistry Rapid Learning Series - 20 Rapid Learning Center Chemistry :: Biology :: Physics :: Math Rapid Learning Center Presents … p g Teach Yourself AP Chemistry Visually in 24 Hours 1/66 http://www.RapidLearningCenter.com Equilibrium E ilib i AP Ch i t R id Learning Series Chemistry Rapid L i S i Wayne Huang, PhD Kelly Deters, PhD Russell Dahl, PhD Elizabeth James, PhD Debbie Bilyen, M.A. © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com Rapid Learning Center www.RapidLearningCenter.com/ © Rapid Learning Inc. All rights reserved. 1
  2. 2. AP Chemistry Rapid Learning Series - 20 Learning Objectives By completing this tutorial you will learn about… Dynamic equilibrium Equilibrium constants E ilib i t t Reaction Quotients Solubility equilibrium How to solve equilibrium problems Le Chatelier’s Principle p 3/66 Concept Map Previous content Chemistry New content Equilibrium Studies When Forward & reverse are equal Reaction Rates Dissolution Reaction Equation with Ratio of products : reactants Equilibrium Constant Expression Solubility Product Can reach Matter Undergo When it’s disturbed, follow Chemical Reactions With values plugged in Equilibrium Constant Le Chatelier’s Principle 4/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 2
  3. 3. AP Chemistry Rapid Learning Series - 20 Equilibrium 5/66 Definition - Equilibrium Reversible Reaction – A c e ca eact o that can chemical reaction t at ca proceed in both directions (represented by a “ ”). Equilibrium – When the rate of the forward reaction equals the rate of the reverse l th t f th reaction. 6/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 3
  4. 4. AP Chemistry Rapid Learning Series - 20 Establishing Equilibrium It takes time to establish equilibrium. At first, there are only reactants present. Only the forward reaction is p possible. Reactants 7/66 Products But once there are products as well, they can begin to reform p , y g reactants. The reverse reaction becomes possible. Forward rate slows and reverse rate increases until they are the same. Once the rate of the forward and reverse process are equal, it is at equilibrium. When equilibrium is established, the number of products and reactants doesn’t change…but the reaction keeps going. Definition - Dynamic Equilibrium Dynamic Equilibrium – The reaction continues to proceed in both directions, but at the same rate. The number of products and reactants no longer change, it may look as thought the reaction has stopped… But the reaction continues! 8/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 4
  5. 5. AP Chemistry Rapid Learning Series - 20 Equilibrium Constants 9/66 Definition - Equilibrium Constant Equilibrium Constant Expression – Equation showing the ratio of the concentrations of products to reactants at equilibrium. Concentration is symbolized with brackets “[A]”. Equilibrium Constant (K) – The number calculated from the equilibrium constant expression. “K” is different for every reaction at every temperature! 10/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 5
  6. 6. AP Chemistry Rapid Learning Series - 20 Writing Equilibrium Constant Expressions To write an equilibrium constant expression: 1 Write the concentration of products on the top— take each one to a power of the coefficient in the balanced equation. b l d ti 2 Write the concentration of reactants on the bottom—also take each to the power of the balanced equation coefficient. Example: Write the equilibrium constant expression for the following: 2 H2 (g) + O2 (g) 2 H2O (g) K= [H2O] [H2] [O2] 11/66 Definition: Homo- and Heterogeneous Equilibrium Homogeneous Equilibrium – All of the species are the same state of matter. 2 H2 (g) + O2 (g) 2 H2O (g) Heterogeneous Equilibrium – There are at least 2 states of matter. 2 H2 (g) + O2 (g) 2 H2O (l) 12/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 6
  7. 7. AP Chemistry Rapid Learning Series - 20 Concentrations of Solids and Liquids Pure solids and pure liquids have constant “concentrations”. If concentration (Molarity) = mole liters And Density = grams liters And Molar Mass = grams mole Then f Th for a pure solid or liquid, Molarity = grams / liters lid li id M l it grams / mole Or, Molarity = 13/66 Density . Molar Mass Both Density and Molar Mass are constants—they don’t change. Therefore, “concentration” of a pure solid or liquid is a constant. “K” Expressions with Solids or Liquids How does this affect the writing of Equilibrium Constant Expressions? If the “concentration” of a pure solid or liquid is constant, then it will not change during equilibrium and it is not written in the “K” expression. 2 H2 (g) + O2 (g) 2 H2O (g) K= [ H 2 O ]2 [ H 2]2 [O2 ] 2 H2 (g) + O2 (g) 2 H2O (l) K= 1 [ H 2]2 [O2 ] H2O is not included in this “K” expression because it’s a liquid. Only gases and solutions are included in “K” expressions! 14/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 7
  8. 8. AP Chemistry Rapid Learning Series - 20 Calculating “K” Example Example: Solve for equilibrium constant for Fe2O3 (s) + 3 H2 (g) 2 Fe (s) + 3 H2O (g) if the following are concentrations at equilibrium: [H2] = 0.45 M and [H2O] = 0.18 M [H2]eq = 0.45 M K= [ H 2O ]3 [ H 2 ]3 K= [H2O]eq = 0.18 M [0.18]3 [0.45]3 K=? Note that Fe2O3 and Fe were not included in the K expression as they are solids! K = 0.064 Most instructors and textbooks do not require units for “K” as each one would be different. 15/66 Meaning of Equilibrium Constant What general meaning can you get from the magnitude of the equilibrium constant? If K is very large… [Products] [Reactants] There is a much larger ratio of products to reactants at equilibrium. The reaction is said to “lie to the right” (products are on the right). If K is very small small… [Products] [Reactants] There is a much smaller ratio of products to reactants at equilibrium. The reaction is said to “lie to the left”. 16/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 8
  9. 9. AP Chemistry Rapid Learning Series - 20 Using “K” to Find Equilibrium Concentration Example: Find the equilibrium concentration for NO if the equilibrium constant for N2 (g) + O2 (g) 2 NO (g) is 1.24×10-4, and the other equilibrium concentrations are [N2] = 0.166 M and [O2] = 0.145 M [N2]eq = 0.166 M K= [O2]eq = 0.145 M K = 1.24×10-4 [NO]eq = ? M 1.24 ×10 −4 = [ NO ]2 [ N 2 ][O2 ] [ NO ]2 (0.166 M )(0.145M ) (1.24 ×10 )(0.166M )(0.145M ) = [ NO] −4 [NO]eq = 0.00173 M 17/66 Reaction Quotient 18/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 9
  10. 10. AP Chemistry Rapid Learning Series - 20 What is the Reaction Quotient? Reaction Quotient is “Q” K Q Equilibrium Constant Reaction Quotient Expression is ratio of products to reactants with balanced equation coefficients as powers Expression is ratio of products to reactants with balanced equation coefficients as powers Only includes gases and solutions Only includes gases and solutions To solve for K, plug in concentrations at equilibrium To solve for Q, plug in concentrations at any time 19/66 The Difference between K and Q What exactly is the difference? 2 H2 (g) + O2 (g) 2 H2O (g) K= [ H 2 O ]2 [ H 2]2 [O2 ] 2 H2 (g) + O2 (g) 2 H2O (g) Q= [ H 2 O ]2 [ H 2]2 [O2 ] The expressions for K and Q are the same. To solve for “K”, plug in concentrations at equilibrium only. To solve for “Q”, plug in concentrations at any time. 20/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 10
  11. 11. AP Chemistry Rapid Learning Series - 20 Using Reaction Quotient Reaction Quotient is used to determine if a system is at equilibrium…and if it’s not, which way does it need to go to get there. [products now] = Q K = [reactants at equilibrium] [reactants now] Q=K [products at equilibrium] [now] = [equilibrium] [Products now] too large Q>K [Reactants now] too small [Products now] too small Q<K [Reactants now] too large 21/66 System is at equilibrium System will make more S t ill k reactants to reach equilibrium System will make more products to reach equilibrium Reaction Quotient Example Example: For N2 (g) + O2 (g) 2 NO (g), if [N2] = 0.81 M, [O2] = 0.75 M and [NO] = 0.030 M, is the reaction at equilibrium if K = 0.0025? If not, which way will it go to reach equilibrium? [N2] = 0.81 M [O2] = 0.75 M Q= [ NO ]2 [ N 2 ][O2 ] Q= (0.030M ) 2 (0.81M )(0.75M ) [NO] = 0.030 M K = 0.0025 At equilibrium = ? Q = 0.0015 Q<K Reaction is not at equilibrium More products will need to be made (and also thereby reducing reactants) to have Q = K Reaction will go to the right (products) to reach equilibrium 22/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 11
  12. 12. AP Chemistry Rapid Learning Series - 20 Solubility Equilibrium Constants 23/66 Definition - Solubility Product Solubility Product (Ksp) – Equilibrium Constant for a dissolution equation. Equation showing a solid dissolving and producing ions NaCl (s) Na+ (aq) + Cl- (aq) Ksp describes the equilibrium between the solid forming dissociated ions and the dissociated ions joining back together to reform the solid. 24/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 12
  13. 13. AP Chemistry Rapid Learning Series - 20 Breaking Compounds into Electrolytes How do you break up a compound when forming electrolytes? 1 Do not break up polyatomic ions 2 Use subscripts that are not a part of a polyatomic ion as coefficients e.g. CaCl2 doesn’t have “Cl2” ions, it has 2 “Cl” ions Example: Break up the following strong electrolytes: Na3PO4 (NH4)2CO3 3 Na+ + PO432 NH4+ + CO32- 25/66 Writing Solubility Product Expressions Writing solubility products expressions are just like writing equilibrium constant expressions 1 Break the solid into it’s electrolytes. 2 Write the concentration of products on the top— take each one to a power of the coefficient in the balanced equation. 3 Write the concentration of reactants on the bottom—Except for dissolution equations, it’s always a solid so it’s “1”. solid…so it s 1 Example: Write the equilibrium constant expression for the following: CaCl2 (s) Ca2+ (aq) + 2 Cl- (aq) K sp = [Ca 2+ ][Cl − ]2 1 or K sp = [Ca 2+ ][Cl − ]2 26/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 13
  14. 14. AP Chemistry Rapid Learning Series - 20 Solubility Product Example Example: Find the solubility product of Cd(OH)2 if at equilibrium the solution is 1.7×10-5 M Cd2+ and 3.4×10-5 M OHCd(OH)2 (s) Cd2+ (aq) + 2 OH- (aq) [Cd2+]eq = 1.7×10-5 M [OH-]eq = 3.4×10-5 M Ksp = ? K sp = [Cd 2+ ][OH − ]2 K sp = (1.7 ×10 −5 M )(3.4 ×10 −5 M ) 2 Ksp = 2.0×10-14 27/66 Definition - Saturated Solution Saturated Solution – Th S t t d S l ti The solution has reached the equilibrium between the solid and the dissociated particles. 28/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 14
  15. 15. AP Chemistry Rapid Learning Series - 20 Using Reaction Quotient with Solubility Reaction Quotient can be used to determine if a solution is saturated or not. [dissolved ions now] = Qsp Ksp = [dissolved ions at saturation] Qsp = Ksp [now] = [equilibrium] Equilibrium (saturated solution) Q>K Too many ions dissolved Solution will be saturated. Extra ions will precipitate back out (go back to the solid form) Q<K Too few ions dissolved System is not saturated (it could hold more) 29/66 Using Reaction Quotient with Ksp Example: Is a solution of Mg(OH)2 saturated if 1.8×10-4 M Mg2+ and 3.0×10-4 M OH- is present? Ksp is 8.9×10-12 Mg(OH)2 (s) [Mg2+] = 1.8×10-4 M [OH-] = 3.0×10-4 M Ksp = 8.9×10-12 At equilibrium = ? Mg2+ (aq) + 2 OH- (aq) Qsp = [ Mg 2+ ][OH − ]2 Qsp = (1.8 ×10 −4 )(3.0 ×10 −4 M ) 2 Qsp = 1.62×10-11 Qsp > Ksp Solution is saturated There are extra ions that will reform a solid So there is a saturated solution with a precipitate 30/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 15
  16. 16. AP Chemistry Rapid Learning Series - 20 Solving Equilibrium Problems 31/66 ICE Charts in Equilibrium Problems Equilibrium problems are often difficult for people to solve. Using “ICE” charts is an excellent technique. I nitial C hange ilib i E quilibrium 32/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 16
  17. 17. AP Chemistry Rapid Learning Series - 20 How to Use an ICE Chart The ICE chart is an efficient way to organize the information in an equilibrium problem. “ICE” Place it down the side Reactants & Products Place each reactant and then each product in a column H2 I2 HI 1.0×10-3 M Initial 2.0×10-3 M 0M Change 1.87×10-3 M Equilibrium Use Equilibrium Values Find a way to determine them and use them in the K expression Given information Fill in any information given 33/66 Determining the “Change” The balanced equation gives the stoichiometric change for the reaction. 2 H2 (g) + O2 (g) 2 H2O (g) For every 2 H2’s that react, 1 O2 will react and 2 H2O will form. H2 O2 H2O Initial -2x Change -x +2x Equilibrium Mg(OH)2 (s) Mg2+ (aq) + 2 OH- (aq) For every 1 Mg(OH)2 that reacts, 1 Mg2+ and 2 OH- will form. Mg(OH)2 Mg2+ OH- Initial Change -x +x +2x Equilibrium 34/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 17
  18. 18. AP Chemistry Rapid Learning Series - 20 Finding Equilibrium Constant - 1 Example: H2 (g) + I2 (g) 2 HI (g) A mixture of 1.0 × 10-3 M H2 and 2.0 × 10-3 M I2 is placed in a container. At equilibrium, the mixture shows the concentration of HI is 1.87 × 10-3 M. Find the equilibrium constant. H2 Initial Change Equilibrium 1.0×10-3 I2 M -x 6.0×10-5 HI 2.0×10-3 M 0M -x M +2x 1.1×10-3 M 1.87×10-3 M If you know an initial, change and equilibrium for one species, you can solve for the “x”. l f h “ ” x= 35/66 1.87 × 10 −3 M − 0 M 2 x = 9.4×10-4 M Once the “x” is known, the equilibrium concentrations can be determined. Finding Equilibrium Constant - 2 Example: H2 (g) + I2 (g) 2 HI (g) A mixture of 1.0 × 10-3 M H2 and 2.0 × 10-3 M I2 is placed in a container. At equilibrium, the mixture shows the concentration of HI is 1.87 × 10-3 M. Find the equilibrium constant. H2 Initial Change Equilibrium I2 HI 1.0×10-3 M 2.0×10-3 M 0M -x 6.0×10-5 M -x +2x 1.1×10-3 M 1.87×10-3 M Once all equilibrium concentrations are known, you can write the equilibrium constant expression and solve for K. ilib i i d l f K K= [ HI ]2 [ H 2 ][ I 2 ] K = 53 (1.87 ×10 −3 M ) 2 K= (6.0 ×10 −5 M )(1.1× 10 −3 M ) 36/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 18
  19. 19. AP Chemistry Rapid Learning Series - 20 Definition: Quadratic Formula Quadratic formula – formula used to find “x” “ ” in an equation that eq ation contains “x” and “x2”. − b ± b 2 − 4ac x= 2a where ax 2 + bx + c = 0 37/66 Quadratic Formula Example When working equilibrium problems, you may end up with equations that contain an “x” and an “x2 Example: Solve for x (“x” represents “concentration”) 1x 2 + 5 x − 8 = 0 x= − b ± b 2 − 4ac 2a From above: a = 1, b = 5, c = -8 x= − 5 ± 52 − (4 ×1× −8) 2× 2 ×1 Using the + give x = 1.27 Using the – gives x = -6.27 x= − 5 ± 57 2 A negative answer isn’t possible for concentration 38/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 19
  20. 20. AP Chemistry Rapid Learning Series - 20 Finding Equilibrium Concentrations #1 Example: H2 (g) + I2 (g) 2 HI (g) A flask is filled with 1.0 M H2 and 2.0 M I2. The value of K at this temperature is 50.5. What are the concentrations of H2, I2 and HI in the flask at equilibrium? H2 I2 HI Initial 1.0 M 2.0 M 0M Change -x -x Equilibrium 1.0 M - x +2x 2.0 M - x 0 + 2x No equilibrium concentrations are known, so the “x” cannot be determined yet. Therefore, only expressions may be written for the d i d Th f l i b i f h equilibrium concentrations. 4x2 [ HI ]2 ( 2 x) 2 50.5 = 50.5 = K= [ H 2 ][ I 2 ] (1.0 − x)(2.0 − x) 2.0 − 3.0 + x 2 50.5 × (2.0 − 3.0 + x 2 ) = 4 x 2 101 − 151.5 + 50.5 x 2 = 4 x 2 0 = −46.5 x + 151.5 − 101 2 39/66 Finding Equilibrium Concentrations #2 Example: H2 (g) + I2 (g) 2 HI (g) A flask is filled with 1.0 M H2 and 2.0 M I2. The value of K at this temperature is 50.5. What are the concentrations of H2, I2 and HI in the flask at equilibrium? H2 I2 HI Initial 1.0 M 2.0 M 0M Change -x -x +2x 1.0 M - x 2.0 M - x 0 + 2x Equilibrium Use quadratic formula to solve: q x= 40/66 − 151.5 ± 151.12 − (4 × −46.5 × −101) 2 × −46.5 x = 0.94 M or 2.31 M 0 = −46.5 x 2 + 151.5 − 101 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 20
  21. 21. AP Chemistry Rapid Learning Series - 20 Finding Equilibrium Concentrations #3 Example: H2 (g) + I2 (g) 2 HI (g) A flask is filled with 1.0 M H2 and 2.0 M I2. The value of K at this temperature is 50.5. What are the concentrations of H2, I2 and HI in the flask at equilibrium? H2 I2 HI Initial 1.0 M 2.0 M 0M Change -x -x +2x 1.0 M - x 2.0 M - x Equilibrium 0 + 2x Using 2.31 M for “x” would result in negative values for equilibrium concentration, which isn’t possible. isn t Therefore, 0.94 M is the correct value for “x” Use it to find equilibrium concentrations x = 0.94 M or 2.31 M [ H 2 ]eq = 1.0 M − 0.94 M = 0.06 M [ I 2 ]eq = 2.0 M − 0.94 M = 1.06 M 41/66 [ HI ]eq = 2 × 0.94 M = 1.88M Making Approximations for Tiny K’s Approximations can be made when the equilibrium constant is very small. Very tiny equilibrium constant Only a very small amount of reactants react and produce products The concentration of the reactant before is approximately the same as the concentration after For K’s that are ×10-5 and smaller you can approximate any number being added or subtracted by an “x” as the number itself. e.g. 0.25 M + x ≅ 0.25 M because “x” is so tiny. 42/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 21
  22. 22. AP Chemistry Rapid Learning Series - 20 Finding Equilibrium Concentrations #4 Keq = 1.6 × 10-5. If the Example: 2NOCl (g) 2 NO (g) + Cl2 (g) initial concentration of NOCl is 0.50 M, what are the concentrations of all three at equilibrium? NOCl Change Cl2 0.50 M Initial NO 0M 0M -2x Equilibrium +2x +x Approximations: 0.50 M 0 M + 2x 0M+x 0 M + 2x 0.50 M - 2x 0M+x K i very small—you may approximate that 0.50 M – 2 ≅ 0 50 M is ll i t th t 0 50 2x 0.50 K= [ NO ]2 [Cl2 ] [ NOCl ]2 1.6 × 10 −5 = 3 [ 2 x ]2 [ x ] [0.50 M ]2 1.6 ×10 −5 = 1.6 ×10 −5 × 0.25 =x 4 4 x3 0.25 x = 0.01 M 43/66 Definition - Solubility Solubility – Amount of a solid y that will dissolve in a solution. The maximum initial amount of solid that would go to 0 moles per liter. liter 44/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 22
  23. 23. AP Chemistry Rapid Learning Series - 20 Finding Solubility Example: Find the solubility of CaF2 if the Ksp is 4.0×10-11 CaF2 (s) Ca2+ (aq) + 2 F- (aq) CaF2 Initial Ca2+ F- x 0M 0M Change -x +x Equilibrium 0M 0M+x +2x 0 M + 2x You’re trying to find out how much solid can go to “0”, so set it’s equilibrium value to 0. K sp = [Ca +2 ][ F −1 ]2 4.0 ×10 −11 = [ x][2 x]2 4.0 ×10 −11 = 4 x 3 3 4.0 ×10 −11 =x 4 x = 2.2×10-4 M The “x” is the maximum solid that would dissolve—that is the “solubility” 45/66 Le Chatelier’s Principle 46/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 23
  24. 24. AP Chemistry Rapid Learning Series - 20 Definition—Le Chatelier’s Principle Le Chatelier’s Principle – If a system at equilibrium is disturbed, it will shift to re-establish equilibrium. hift t t bli h ilib i A system will try to undo whatever you’ve done. 47/66 Increasing Concentrations How does adding a reactant or product affect a system at equilibrium? Reaction shifts to right Adding a reactant Q becomes too small (get rid of extra reactants and make more products) Reaction shifts to left Adding a product Q becomes too large (get rid of extra products and make more reactants) 48/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 24
  25. 25. AP Chemistry Rapid Learning Series - 20 Decreasing Concentrations How does removing a reactant or product affect a system at equilibrium? Removing a reactant Q becomes too large Removing a product Q becomes too small Reaction shifts to left (make more reactants) Reaction shifts to right (make more products) 49/66 Changes in Pressure How does changing the pressure affect a system at equilibrium? Decrease volume Increase volume Pressure increases Reaction shifts to the side with least moles of gas to decrease pressure Pressure decreases Reactions shifts to the side with the most moles of gas to increase pressure 50/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 25
  26. 26. AP Chemistry Rapid Learning Series - 20 Definition - Endo & Exothermic Reactions Endothermic Reaction – The reaction gy products have takes in energy…the p more energy than the reactants. Energy is a reactant in the reaction. Exothermic Reaction – The reaction gives off energy…the products have less energy than the reactants. Energy is a product in the reaction. 51/66 Temperature and Endothermic For endothermic, think of temperature (or energy) as a reactant. Reaction shifts to right Increase temperature of endothermic reaction Increasing a reactant Decrease temperature of endothermic reaction Remove a reactant (get rid of extra reactants and make more products) Reaction shifts to left (make more reactants) 52/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 26
  27. 27. AP Chemistry Rapid Learning Series - 20 Temperature and Exothermic For exothermic, think of temperature (or energy) as a product. Increase temperature of exothermic reaction Decrease temperature of exothermic reaction Reaction shifts to left Increasing a product Remove a product (get rid of extra products and make more reactants) Reaction shifts to right (make more products) 53/66 Changes that Have No Effect Some changes have no effect because they do not affect the value of “Q”. Adding a pure solid or liquid reactant or g p q product. Increasing pressure by adding an inert gas. Changing the volume of a reaction with an equal number of moles of gas on each side of the reaction. Adding a catalyst A catalyst will speed up how fast equilibrium is established—but not the number of reactants and products once it’s at equilibrium. 54/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 27
  28. 28. AP Chemistry Rapid Learning Series - 20 Le Chatelier’s Examples Example: Which way will the reaction shift for each of the following changes: NH4Cl (s) Removing NH4Cl NH3 (g) + HCl (g) No change (it s a solid) (it’s Adding HCl (Adding a product) Adding Ne (g) No change (it’s an inert gas) Decreasing volume (Goes to side with least gas moles) Example: Which way will the reaction shift for each of the following changes: 2 SO2 (g) + O2 (g) 2 SO3 (g) an exothermic reaction Increasing volume (Goes to side with most gas moles) Raising temperature (Energy is a product) Adding O2 (Adding a reactant) Removing SO2 (Removing a reactant) 55/66 Le Chatelier in Industry When companies need to make large amounts of product, a reaction with a very small K is a problem. Small K Small ratio of products to reactants Lots of reactants left over (wasting money) and few products made (not making y) money) They can push the reaction towards the products. e.g. Remove the products as they’re made, adjust pressure or temperature as needed to push it to the right. 56/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 28
  29. 29. AP Chemistry Rapid Learning Series - 20 Haber Process The reaction to produce ammonia, NH3, is very important to manufacturing. N2 + 3 H2 2 NH3 (an exothermic reaction) In order for the reaction to occur at a reasonable rate, the temperature must be very high. But when the temperature is high, the equilibrium constant is very low. A compromise is made and a catalyst is added to increase the rate at the lower temperature. The reaction yields 20%...the leftover reactants are recycled and put back into the reaction again. 57/66 Equilibrium & The AP Exam 58/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 29
  30. 30. AP Chemistry Rapid Learning Series - 20 Equilibrium in the Exam Common equilibrium problems: The first free response question is always equilibrium: ilib i Either K, Ksp, Ka, or Kb (Ka and Kb are in the next tutorial) Determine the solubility or solubility product Determine equilibrium concentrations given K Determine K given equilibrium concentrations Use Q to determine which way a reaction will proceed Use Le Chatelier’s Principle to determine which way a reaction will proceed 59/66 Multiple Choice Questions Occasionally, equilibrium is found in the multiple choice section. Example: Which is the correct equilibrium constant expression for HOBr (aq) H+ (aq) + OBr- (aq)? A. B. C. D. E. [HOBr] / [H+][OBr-] [H+][OBr-] / [HOBr] [H+] / [HOBr] [OBr-] / [HOBr] None of the above Answer: B 60/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 30
  31. 31. AP Chemistry Rapid Learning Series - 20 Free Response Questions The equilibrium free response question is often about solubility. At 25°C the value of Ksp for PbCl2 is 1 6 × 10-5 and the value 25°C, 1.6 of Ksp for AgCl is 1.8 × 10-10. A. If 30.0 mL of 0.060 M NaCl is added to 30 mL of 0.030 M Pb(NO3)2, will a precipitate form. Assume that the volumes are additive. B. Calculate the equilibrium value of [Pb+2] in a saturated solution of PbCl2. C. If NaCl is added slowly to a beaker containing equal y g q concentrations of Pb+2 and Ag+, which will precipitate first? 61/66 Answering Free Response Questions At 25°C, the value of Ksp for PbCl2 is 1.6 × 10-5 and the value of Ksp for AgCl is 1.8 × 10-10. A. If 30.0 mL of 0.060 M NaCl is added to 30 mL of 0.030 M Pb(NO3)2, will a precipitate form. Assume that the ( p p volumes are additive. New volume = 60 mL New [NaCl] = 30 mL × 0.060 M / 60 mL = 0.030 M New [Pb(NO3)2] = 30 mL × 0.030 M / 60 mL = 0.015 M PbCl2 Pb+2 + 2 Cl-1 Q = [Pb+2][Cl-1]2 Q = (0.015 M)(0.030 M)2 = 1.35 × 10-5 Q < K, it will not precipitate out 62/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 31
  32. 32. AP Chemistry Rapid Learning Series - 20 Answering Free Response Questions At 25°C, the value of Ksp for PbCl2 is 1.6 × 10-5 and the value of Ksp for AgCl is 1.8 × 10-10. B. Calculate the equilibrium value of [Pb+2] in a saturated solution of PbCl2. PbCl2 Pb+2 + 2 Cl-1 Ksp = [Pb+2][Cl-1]2 1.6 × 10-5 = (x)(2x)2 x = 0.016 M [Pb+2] = 0.016 M C. If NaCl is added slowly to a beaker containing equal concentrations of Pb+2 and Ag+, which will precipitate first? AgCl is less soluble (Ksp is smaller), therefore it will precipitate first. 63/66 Learning Summary The ICE chart is a good technique to use to solve equilibrium problems. Dynamic equilibrium is established when the rates of the forward and reverse reactions are equal. The Th equilibrium constant ilib i t t give the ratio of product: reactants with the stoichiometric ratios as the powers. Le Chatelier’s Principle governs how a reaction at h ti t equilibrium will change when disturbed. The solubility product is the equilibrium constant for a dissolution reaction. 64/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 32
  33. 33. AP Chemistry Rapid Learning Series - 20 Congratulations You have successfully completed the rapid tutorial Equilibrium Rapid Learning Center Rapid Learning Center Chemistry :: Biology :: Physics :: Math What’s N t Wh t’ Next … Step 1: Concepts – Core Tutorial (Just Completed) Step 2: Practice – Interactive Problem Drill Step 3: Recap – Super Review Cheat Sheet Go for it! 66/66 http://www.RapidLearningCenter.com © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 33

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