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Chemical Systems & Equilibrium

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  1. Chemical Systems & Equilibrium Unit 4

  2. Dynamic Equilibrium • dynamic equilibrium = a balance between forward and reverse processes occurring at the same rate

  3. Party Analogy • 30 people at a house party • 8pm: • 16 people in the kitchen • 14 people in the living room • 10pm: • 16 people in the kitchen • 14 people in the living room • Different people but same number in each room

  4. At Equilibrium: • Closed system – no matter/energy/pressure changes • No macroscopic changes • Reactants and products both present (and usually in different amounts) • [reactant] = constant, [product] = constant • Can be approached from both sides Rate of forward reaction = rate of reverse reactions

  5. Dynamic Equilibrium • Example: Closed bottle of pop • CO2 gas leaving dissolved state and entering gas state • CO2 gas ALSO, leaving gas state and entering liquid state • No visible change • CO2(g)CO2(aq)

  6. N2O4(g) 2NO2(g) Equilibrium Double Arrow • equilibrium is symbolized with an equation containing a forward (→) and a reverse (←) arrow combined into:

  7. Equilibrium Double Arrow • forward reaction = in an equilibrium equation, the left-to-right reaction • reverse reaction = in an equilibrium equation, the right-to-left reaction CO2(g)CO2(aq) Forward Reverse

  8. Drinking Bird Equilibrium • • At rest the vapor and the liquid inside the tube are in an equilibrium • Wet head of bird with water – as the water evaporates from around the head, it takes energy with it, head cools down = vapor inside the head cools and contracts = vacuum = pulls liquid up

  9. 3 Types of Equilibrium 1. Solubility Equilibrium (dissolving process) 2. Phase Equilibrium (change of state) 3. Chemical Reaction Equilibrium (reactants ⇆ products)

  10. Types of Equilibrium #1 • solubility equilibrium = a dynamic equilibrium between a solute and a solvent in a saturated solution in a closed system I2(s) I2(aq)

  11. Solubility Equilibrium • Saturated solution = a solution containing the maximum quantity of a solute • Beyond the solubility limit, any added solute will remain solid and not dissolve

  12. Solubility Equilibrium • kinetic molecular theory states that particles are always moving and colliding • even if no changes are observed • Dissolution = the process of dissolving

  13. (a) When the solute is first added, many more ions dissociate from the crystal than crystallize onto it. (b) As more ions come into solution, more ions also crystallize. (c) At solubility equilibrium, solute ions dissolve and crystallize at the same rate.

  14. Digesting a Precipitate • Allow precipitates to sit for long periods of time before filtering • The longer you wait the more pure the crystal, also the larger the crystal • If precipitate forms quickly, impurities maybe trapped in the precipitate

  15. Types of Equilibrium #2 • phase equilibrium = a dynamic equilibrium between different physical states of a pure substance in a closed system • closed system = a system that may exchange energy but NOT matter with it’s surroundings H2O(l) H2O(g) H2O(s) H2O(l)

  16. Phase Equilibrium

  17. Types of Equilibrium #3 • chemical reaction equilibrium a dynamic equilibrium between reactants and products of a chemical reaction in a closed system • reversible reaction = a reaction that can achieve equilibrium in the forward or reverse direction

  18. Chemical Reaction Equilibrium In a Closed System • N2O4(g) 2 NO2(g)

  19. Reversible Reactions • The same dynamic equilibrium composition is reached whether we start from pure N2O4(g), pure NO2(g), or a mixture of the two, provided that environment, system and total mass remain the same.

  20. Calculating the Equilibrium Constant • The equilbrium constant, Keq, is the ratio of equilibrium concentrations at a particular temp • Kc for solution-phase systems or Kp for gas-phase systems • Keq= [C]c[D]d for the eqn [A]a[B]baA+bBcC+dD Note: The equilibrium constant depends ONLY on the concentration of gases (not liquids/solids)

  21. [H2(g)][I2(g) ] [N2(g)][H2(g) ]3 [0.11][1.91]3 [N2O4(g)] K = K = K = K = [HI(g) ]2 [NH3(g) ]2 [0.25]2 [NO2(g) ]2 Questions: Equilibrium Law Expression 1. Write the equilibrium law expression for the following: a) 2NO2(g) ↔ N2O4(g) b) 2HI(g) ↔ H2(g) + I2(g) 2. A reaction vessel contains NH3, N2 and H2 gas at equilibrium at a certain temperature. The equilibrium concentrations are [NH3] = 0.25mol/L, [N2] = 0.11mol/L and [H2] = 1.91 mol/L. Calculate the equilibrium constant for the decomposition of ammonia. 2NH3(g) ↔ N2(g) + 3H2(g) K = 12.3

  22. [NO2(g)]2 K = [NO (g) ]2 [O2 (g) ] Questions: Equilibrium Law Expression 3. Nitryl chloride gas, NO2Cl, is in equilibrium at a certain temperature in a closed container with NO2 and Cl2 gases. At equilibrium, [NO2Cl] = 0.00106mol/L and [NO2] = 0.0108mol/L. If K = 0.558, what is the equilibrium concentration of Cl2? 4. Write a balanced equation for the reaction with the following equilibrium law expression:

  23. Heterogeneous Equilibria • homogeneous equilibria = equilibria in which all entities are in the same phase • Reactants and products are all gas or all aqueous • heterogeneous equilibria= equilibria in which reactants and products are in more than one phase • Reactants and products are in different phases

  24. N2O4(g)2NO2(g) Kc = Kp = P2 [NO2]2 NO2 [N2O4] P N2O4 Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.

  25. CaCO3(s) CaO(s) + CO2(g) Kc = [CaO(s) ][CO2(g)] [CaO(s)] Kc [CaCO3(s)] [CaCO3(s)] Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. [CaCO3(s)] = constant [CaO(s) ] = constant Kc= [CO2(g)] = [CO2(g)] The concentration of solids and pure liquids are considered to be constant and are not included in the expression for the equilibrium constant.

  26. CaCO3(s)CaO(s) + CO2(g) PCO 2 does not depend on the amount of CaCO3 or CaO PCO2 = Kp

  27. equilibrium equilibrium equilibrium N2O4(g) 2NO2(g) Start with NO2 Start with N2O4 Start with NO2 & N2O4 Equilibrium favors the reactant side

  28. CHECKPOINT The reaction at 200C between ethanol and ethanoic acid produces ___________________ and __________________. Write the equation for this reaction Determine the equilibrium constant expression for the reaction

  29. Calculating Equilibrium Concentrations (when given one concentration) Sample Problem: When ammonia is heated it decomposes: 2NH3(g)↔N2(g)+ 3H2(g) When 4.0 mol of ammonia is introduced in a 2.0L container and heated. The equilibrium amount of ammonia is 2 0 mol. Determine the equilibrium concentrations of the other two entities. STEP 1: Determine the concentration (initial and equilibrium) for known values STEP 2: Setup an ICE Table STEP 3: Determine the value of X STEP 4: Use x value to determine the other quantities

  30. Determine the concentrations [NH3]initial = 4.0mol/2.0L = 2.0mol/L [NH3]equilibrium = 2.0mol/2.0L = 1.0mol/L Setup ICE Table

  31. Determine the value of X [NH3](g)equil= 2 0mol / L - 2x [NH3](g)equil= 1.0mol/L (from calculations in Step 1) 2.0mol/L – 2x = 1.0mol/L -2x = - 1.0mol/L x = 0.5mol/L Use X to determine other quantities

  32. constant

  33. Reversible Reactions • For a given overall system composition, the same equilibrium concentrations are reached whether equilibrium is approached in the forward or the reverse direction • What about Keq will it be the same in fwd/rev?

  34. Heat + N2O4(g) 2NO2(g) Equilibrium Tubes Brown Colourless The effects of temperature on equilibrium ENDOTHERMIC Rxn Hot Very Cold Cold

  35. N2O4(g) 2NO2(g) NO2 is one of the chemicals in smog! • In the summer on hot, windless days an orange haze is seen over the horizon, this is NO2 • In the winter, the smog doesn't go away, it is just less noticeable. The cooler temperatures lead to more N2O4 and less NO­2 which we can't see as well! Brown Colourless

  36. Qualitative Changes in Equilibrium Systems You should be familiar with your own body’s attempt at maintaining equilibrium or “homeostasis”: • If body T too high  sweat, surface blood vessels dilate • If body T too low  shiver, surface blood vessels constrict • If blood CO2 levels ↑  breathe deeper & faster • If blood sugar levels ↑  insulin released to remove excess glucose

  37. Le Châtelier’s Principle • When a chemical system at equilibrium is disturbed by a change in a property, the system adjusts in a way that opposes the change. • In other words: If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

  38. Le Châtelier’s Principle Le Chatelier’s Principle: if you disturb an equilibrium, it will shift to undo the disturbance. equilibrium shift = movement of a system at equilibrium, resulting in a change in the concentrations of reactants and products •

  39. Le Châtelier’s Principle • System starts at equilibrium. • A change/stress is then made to system at equilibrium. • Change in concentration • Change in temperature • Change in volume/pressure • System responds by shifting to reactant or product side to restore equilibrium.

  40. Le Châtelier’s Principle Change in Reactant or Product Concentrations • Adding a reactant or product shifts the equilibrium away from the increase. • Removing a reactant or product shifts the equilibrium towards the decrease. • To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product.

  41. Le Châtelier’s Principle Change in Reactant or Product Concentrations • If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 • That is, the system must consume the H2 and produce products until a new equilibrium is established. • Equilibrium shifts to the right. • Therefore, [H2] and [N2] will decrease and [NH3] increases. N2(g)+ 3H2(g)↔2NH3(g)

  42. N2(g) + 3H2(g) 2NH3(g) Equilibrium shifts left to offset stress Add NH3 Change in Reactant or Product Concentrations

  43. aA + bBcC + dD Change in Reactant or Product Concentrations Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left

  44. Le Châtelier’s Principle Effect of Temperature Changes • The equilibrium constant is temperature dependent. • For an endothermic reaction, H > 0 and heat can be considered as a reactant. • For an exothermic reaction, H < 0 and heat can be considered as a product.

  45. Effect of Temperature Changes

  46. Effect of Temperature Changes Adding heat (i.e. heating the vessel) favors away from the increase: • if H = + (Endothermic), adding heat favors the forward reaction, • if H = - (Exothermic), adding heat favors the reverse reaction. Removing heat (i.e. cooling the vessel), favors towards the decrease: • if H = + (Endothermic), cooling favors the reverse reaction, • if H = - , (Exoothermic), cooling favors the forward reaction.

  47. Gas Law – Boyle’s LawRelationship: Pressure & Volume • As pressure on a gas increases, the volume of the gas decreases

  48. Le Châtelier’s Principle Effects of Volume and Pressure • As volume is decreased pressure increases. • The system shifts to decrease pressure. • An increase in pressure favors the direction that has fewer moles of gas. • Decreasing the number of molecules in a container reduces the pressure. • In a reaction with the same number of product and reactant moles of gas, pressure has no effect. • Only a factor with gases.

  49. A (g) + B (g) C (g) Effects of Volume and Pressure Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas