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Equilibrium

Equilibrium. L. Scheffler Lincoln High School 2009. 1. Equilibrium Systems. Many chemical reactions are reversible. Such reactions do not go to completion. There is a state of balance between the products and the reactants

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Equilibrium

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  1. Equilibrium L. Scheffler Lincoln High School 2009 1

  2. Equilibrium Systems • Many chemical reactions are reversible. • Such reactions do not go to completion. • There is a state of balance between the products and the reactants • When the concentration of neither the reactants nor the products is changing, the system is in equilibrium. 2

  3. Chemical Equilibrium Chemical equilibrium occurs in chemical reactions that are reversible. In a reaction such as: CH4(g) + H2O(g)  CO(g) + 3H2 (g) The reaction can proceed in both directions CO(g) + 3H2 (g)  CH4(g) + H2O(g) 3

  4. Equilibrium Conditions • At equilibrium the rate of reaction in the forward direction and the rate in the reverse direction are equal 4

  5. An Equilibrium System CH4(g) + H2O(g) CO(g) + 3H2 (g) • After some of the products are created, the products begin to react to form the reactants. • At equilibrium there is no net change in the concentrations of the reactants and products. • The concentrations do not change but they are not necessarily equal. 5

  6. H2O NaCl (s)  NaCl (aq) Dynamic Equilibrium • At equilibrium two opposing processes are taking place at equal rates. • In other words the rate in the forward direction = the rate in the reverse direction • Examples H2O (l)  H2O (g) CO (g) + 2 H2 (g)  CH3OH (g) 6

  7. Equilibrium Conditions H2O + CO  H2 + CO2 7

  8. Law of Mass Action • Given the reaction aA + bB  cC + dD The rate in the forward direction is rate forward = kf [A]a [B]b The rate in the reverse direction is rate reverse = kr [C]c [D]d 8

  9. Law of Mass Action • At equilibrium these rates are equal rate forward = rate reverse kf [A]a [B]b = kr [C]c [D]d • The ratio of the rate constants is 9

  10. Writing Equilibrium Expressions • N2(g) + 3 H2 (g) 2NH3 (g) • 2 SO2(g) + O2 (g)  2SO3 (g) • H2(g) + Br2 (g) 2 HBr(g) • 2N2O (g) 2 N2 (g) + O2 (g) 10

  11. Answers Check your work against the following: 11

  12. Reaction Quotient • The equilibrium constant is a constant ratio only when the system is in equilibrium. • If the system it not at equilibrium the ratio is known as a Reaction Quotient • If the reaction quotient is equal to the equilibrium constant then the system is at equilibrium 12

  13. The Meaning of the Equilibrium Constant • K>>1: The reaction is product-favored; equilibrium concentrations of products are greater than equilibrium concentrations of reactants. • K<<1: Thereaction is reactant-favored; equilibrium concentrations of reactants are greater than equilibrium concentrations of products. 13

  14. Calculating Equilibrium Constants Nitrogen dioxide decomposes at high temperatures according to this equation: 2 NO2 (g)  2 NO (g) + O2(g) If the equilibrium concentrations are as follows: [NO2]= 1.20 M, [NO] = 0.160, and [O2] = 0.080 M; calculate the equilibrium constant. 14

  15. Calculating Equilibrium Constants The equilibrium equation for the oxidation of sulfur dioxide is as follows: 2SO2(g) + O2 (g) 2 SO3 (g) If the equilibrium concentrations are as follows: [SO2 ]= 0.44 M, [O2] = 0.22, and [SO3] = 0.78 M , Calculate the equilibrium constant 15

  16. Practice Problem 1 The equilibrium equation for the carbon monoxide with steam to produce hydrogen gas is as follows: CO2(g) + H2(g)  CO (g) + H2O (g) If the equilibrium concentrations are as follows: [CO] = 1.00 M, [H2O] = 0.025, [CO2] = 0.075 M and [H2] = 0.060 M, calculate the equilibrium constant. 16

  17. Calculating Equilibrium Concentrations 17

  18. Equilibrium Calculations –Using I. C. E. Models Equilibrium constants and concentrations can often be deduced by carefully examining data about the initial and equilibrium concentrations Initial Change Equilbrium 18

  19. Equilibrium CalculationsICE Model problem 1 Hydrogen and iodine are in equilibrium with Hydrogen iodide to this reaction: H2 + I2 2HI Suppose that 1.5 mole of H2 and 1.2 mole of I2 are placed in a 1.0 dm3 container. At equilibrium it was found that there were 0.4 mole of HI. Calculate the equilibrium concentrations of [H2] and [I2] and the equilibrium constant. 19

  20. Equilibrium CalculationsICE Model Problem 1 -- Solution Hydrogen and iodine are in equilibrium with Hydrogen iodide to this reaction: H2 + I2 2HI Suppose that 1.5 mole of H2 and 1.2 mole of I2 are placed in a 1.0 dm3 container. At equilibrium it was found that there were 0.4 mole of HI. Calculate the equilibrium concentrations of [H2] and [I2] and the equilibrium constant. ICESince 2x = 0.4, x = 0.2 [H2 ] 1.5 - x 1.5- x [H2 ] = 1.5 – 0.2 = 1.3 [ I2 ] 1.2 - x 1.2 –x [I2 ] = 1.2 – 0.2 = 1.0 [HI ] 0 +2x 0.4 Keq = [HI]2 = (0.4)2 = 0.123 [H2 ] [ I2 ] (1.3) (1.0) 20

  21. Equilibrium CalculationsICE Model Problem 2 Sulfur dioxide reacts with oxygen to produce sulfur trioxide according to this reaction: 2 SO2 + O2 2SO3 Suppose that 1.4 mole of SO2 and 0.8 mole of O2 are placed in a 1.0 dm3 container. At equilibrium it was found that there were 0.6 dm3 of SO3. Calculate the equilibrium concentrations of [SO2] and [O2] and the equilibrium constant. 21

  22. Equilibrium CalculationsICE Model Problem 2 - Solution Sulfur dioxide reacts with oxygen to produce sulfur trioxide according to this reaction: 2 SO2 + O2 2SO3 Suppose that 1.4 mole of SO2 and 0.8 mole of O2 are placed in a 1.0 dm3 container. At equilibrium it was found that there were 0.6 dm3 of SO3. Calculate the equilibrium concentrations of [SO2] and [O2] and the equilibrium constant. ICESince 2x = 0.6, x = 0.3 [SO2 ] 1.4 -2x 1.4-2x [SO2 ] =1.4 – 2( 0.3) = 0.8 [O2 ] 0.8 -x 0.8 –x [O2 ] = 0.8 – 0.3 = 0.5 [SO3] 0 +2x 0.6 Keq = [SO3]2 = (0.6)2 = 0.281 [SO2 ]2 [O2 ] (0.8)2(0.5) 22

  23. Calculating Equilibrium Concentrations from Keq What is the concentration for each substance at equilibrium for the following gaseous reaction C2H6 C2H4 + H2KC = 1.01 if the initial concentration of ethene, C2H4 and that of hydrogen are both 0.300 M? 1.01x = 0.09 - 0.6x +x2 x2 - 1.61x +0.09 = 0 x = 1.61+ (1.61)2-4(1)(0.09) 2(1) x = 0.0580 and 1.552. The second root is extraneous. so [C2H6] = 0.0580 and [H2 ] = [C2H4] = 0.242 23

  24. Le Chatelier’s Principle • Le Chatelier's Principle states: When a system in chemical equilibrium is disturbed by a change of temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change of variable. • A change imposed on an equilibrium system is called a stress • The equilibrium always responds in such a way so as to counteract the stress 24

  25. Le Chatelier’s Principle • A stress is any sudden change in conditions that drives the system out of equilibrium. • When a stress is placed on an equilibrium system, the system will shift in such a way so as to lessen or mitigate the stress and restore equilibrium. • A stress usually involves a change in the temperature, pressure, or in the concentration of one or more of the substances that are in equilibrium. • Le Chatelier's principle predicts the direction of the change in the equilibrium. 25

  26. Applications ofLe Chatelier’s Principle N2 (g) +3 H2 (g)  2NH3(g) DH = -92 kJ • Haber’s process for the production of ammonia is an example of an industrial equilibrium system. We will use this equilibrium as a model to explain the how Le Chatelier’s principle operates with the following stresses: • Change in the concentration of one of the components • Changes in pressure • Changes in temperature • Use of a catalyst 26

  27. Applications of Le Chatelier’s Principle N2 (g) +3 H2 (g)  2NH3(g) DH = - 92 kJ The equilibrium constant for this reaction is Any change in the concentration (or partial pressure) of any component will cause the equilibrium to shift in such a way so as to return the equilibrium constant to its original value. 27

  28. Le Chatelier’s Principle –The Concentration Effect N2 (g) +3 H2 (g)  2NH3(g) DH = - 92 kJ An increase in the concentration of N2 results in a decrease H2 and an increase in NH3 in such a way to keep the equilibrium constant the same 28

  29. Le Chatelier’s Principle – The Concentration Effect N2 (g) +3 H2 (g)  2NH3(g) DH = - 92 kJ Likewise an increase in the concentration ofH2 results in a decrease in N2 and an increase in NH3 in such a way to keep the equilibrium constant the same 29

  30. Le Chatelier’s Principle – The Concentration Effect N2 (g) +3 H2 (g)  2NH3(g) DH = - 92 kJ An increase in the concentration ofNH3 results in a increase in N2 and an increase in H2 in such a way to keep the equilibrium constant the same. 30

  31. Le Chatelier’s Principle – The Temperature Effect N2 (g) +3 H2 (g)  2NH3(g) DH = - 92 kJ • The reaction is exothermic in the forward direction. • An increase in the temperature would trigger a response in the heat consuming (endothermic) direction. • An increase in temperature therefore causes the reaction to shift in the reverse direction. Some NH3 decomposes to N2 and H2. 31

  32. Le Chatelier’s Principle –The Pressure Effect N2 (g) +3 H2 (g)  2NH3(g) DH = - 92 kJ • All molecules in the equilibrium are gases • When the reaction proceeds in the forward direction the number of moles are reduced from 4 to 2. • Pressure is proportional to the number of moles of gas • An increase in pressure therefore causes the reaction to moved in the forward direction. Some N2 and H2 combine to form more NH3 32

  33. Le Chatelier’s Principle – The Effect of Catalysts N2 (g) +3 H2 (g)  2NH3(g) DH = - 92 kJ • Catalysts lower the activation energy • A catalyst affects the forward and the reverse direction equally • There is no change in the equilibrium position from a catalyst • A catalyst decreases the time required for the system to achieve equilibrium 33

  34. Heterogeneous Equilibria • In a heterogeneous equilibrium, the components are in two different phases. • A common form of a heterogeneous equilibrium is the equilibrium which exists between a solid and an aqueous solution • When a substance dissolves in water there is an equilibrium established between the solid an its dissolved ions • Example • AgCl (s)  Ag+ (aq) + Cl- (aq) 34

  35. Heterogeneous Equilibria • AgCl (s)  Ag+ (aq) + Cl- (aq) • In this example the concentration of the solid phase is essentially 1. The equilibrium constant then takes the form • K = [Ag+ ][Cl-]. • This form is known as a solubility product and is usually designated as Ksp 35

  36. Different designations of equilibrium constants • Keq General designation for an equilibrium constant • Kc Equilibrium constant based on concentration • Kp Equilibrium constant based on pressure (gases) • Ksp Solubility product • Ka Acid equilibrium constant • Kb Base equilibrium constant • Kw Ion product of water 36

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