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Chem 40S Unit 4 Notes
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Chem 40S Unit 4 Notes

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  • Initially the reaction is at equilibrium - both the forward and the reverse rates are equal. At the instant when more reactant NO 2 is added, the forward rate increases. As the reactant is consumed in the reaction, the forward rate decreases to a constant value. Initially the reverse rate is unchanged. However, as more product is formed, the rate of the back (reverse) reaction increases to the new constant value. Initially the reaction is at equilibrium - both the forward and reverse rates are equal. At the instant when reactant NO 2 is removed, the forward rate decreases. As more NO 2 is produced through the back reaction, the forward rate increases to a new constant value. Initially the reverse rate is unchanged. However, since product is no longer being formed at the same rate, the back rate decreases as the amount of product decreases (and more reactant is formed).
  • Initially the reaction is at equilibrium - both the forward and the reverse rates are equal. At the instant when more reactant NO 2 is added, the forward rate increases. As the reactant is consumed in the reaction, the forward rate decreases to a constant value. Initially the reverse rate is unchanged. However, as more product is formed, the rate of the back (reverse) reaction increases to the new constant value. Initially the reaction is at equilibrium - both the forward and reverse rates are equal. At the instant when reactant NO 2 is removed, the forward rate decreases. As more NO 2 is produced through the back reaction, the forward rate increases to a new constant value. Initially the reverse rate is unchanged. However, since product is no longer being formed at the same rate, the back rate decreases as the amount of product decreases (and more reactant is formed).
  • Initially the reaction is at equilibrium - both the forward and the reverse rates are equal. At the instant when more reactant NO 2 is added, the forward rate increases. As the reactant is consumed in the reaction, the forward rate decreases to a constant value. Initially the reverse rate is unchanged. However, as more product is formed, the rate of the back (reverse) reaction increases to the new constant value. Initially the reaction is at equilibrium - both the forward and reverse rates are equal. At the instant when reactant NO 2 is removed, the forward rate decreases. As more NO 2 is produced through the back reaction, the forward rate increases to a new constant value. Initially the reverse rate is unchanged. However, since product is no longer being formed at the same rate, the back rate decreases as the amount of product decreases (and more reactant is formed).
  • Initially, the system is at equilibrium - the rates of the forward and reverse reactions are equal. When the temperature is increased, both the forward and the reverse rates increase. Because the reaction is exothermic, the reverse rate goes up more than does the forward reaction. Initially, the system is at equilibrium - the rates of the forward and reverse reactions are equal. When the temperature is lowered, both the forward and reverse rates decrease. Because the reaction is exothermic, the reverse rate goes down less than does the forward reaction.
  • Initially the reaction is at equilibrium - the concentrations of reactant NO 2 and product N 2 O 4 are constant. At the instant when more reactant NO 2 is added, the [NO 2 ] increases abruptly. As the reactant is consumed in the reaction, its concentration decreases to a constant value. Initially [N 2 O 4 ] is unchanged. However, as reaction proceeds, more product is formed, and [N 2 O 4 ] increases to a new constant value, a new equilibrium position. Initially the reaction is at equilibrium - the concentrations of reactant NO 2 and product N 2 are constant. At the instant when reactant NO 2 is removed, the [NO 2 ] decreases abruptly. As more NO 2 is produced through the back reaction, its concentration increases to a new constant value. Initially the [N 2 O 4 ] is unchanged. However, since product is no longer being formed at the same rate, its concentration decreases to a new constant value, a new equilibrium position is established.
  • At the instant when product N 2 O 4 is added, [N 2 O 4 ] goes up abruptly. Then, as the product concentration is decreased by reaction, the reverse rate decreases till it reaches a constant value. At the instant when product N 2 O 4 is added, the forward rate is unchanged. As the 'reactant' concentration increases through the reaction, the forward rate increases to its new constant value. At the instant when product N 2 O 4 is removed, the reverse rate goes down. Then, as more product is formed by reaction, the reverse rate increases to a new constant value. At the instant when product N 2 O 4 is removed, the forward rate is unchanged. Then, as more reactant is used up in producing more of the product, the forward rate decreases to a new constant value.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C decreases when the temperature increases. When the temperature is increased, the system is not at equilibrium under the new conditions. [NO 2 ] increases to establish a new equilibrium. Initially, the system is at equilibrium - [NO 2 ] and [N 2 O 4 ] are constant. The reaction is exothermic, K C increases when the temperature decreases. When the temperature is decreased, the system is not at equilibrium under the new conditions. [N 2 O 4 ] increases to establish a new equilibrium.
  • Transcript

    • 1. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium. • Typically when we think of what happensEquilibrium during a chemical reaction we think of theIntroduction reactants getting totally used up so that none are left and ending up with only products. Also, we generally consider chemical reactions as one-way events. You may well have learned during earlier science classes that this is one way to distinguish chemical change from physical changes - physical changes (such as the melting and freezing of ice) are easily reversed, but chemical changes cannot be reversed (pretty tough to un-fry an egg). • In this unit we will see that this isnt always the case. We will see that many chemical reactions are, in fact, reversible under the right conditions. And because many reactions can be reversed, our idea of a reaction ending with no reactants left, only products, will need to be modified.
    • 2. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium. • The amount of water on the earthAbout (including the atmosphere) staysChemical relatively constant. Water evaporatesEquilibrium or sublimes into the gas phase. Water in the gas phase returns to the earth in the form of precipitation and dew. There is a balance struck between the various phases of water, solid, liquid and gas. This is known as the water cycle. • A basketball team plays a game. During the game, players enter and leave the game. The number of players on the floor never changes. • What do these have in common?
    • 3. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium.ChemicalEquilibrium Equilibrium is the state at which the rate of the forward reaction equals the rate of the reverse reaction. At the point of equilibrium, no more measurable or observable changes in the system can be noted.
    • 4. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium. • Equilibrium occurs when the rateReview of the forward process is equal to the rate of the reverse process in a reversible reaction. • Equilibrium can only occur in a closed system. • Equilibrium is dynamic. • Liquid-vapour equilibrium occurs when liquid molecules enter and leave the liquid state at the same rate. • Solution equilibria occur in saturated solutions in which solute remains undissolved. • Physical equilibria are those occurring in physical changes.
    • 5. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant.Equilibrium
    • 6. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium.Reversible • In our study of physicalReactions processes, equilibrium could only be established in a closed system. This also holds for chemical reactions in equilibrium. • Lets take a look at the conversion of nitrogen dioxide gas, NO2, into dinitrogen tetraoxide, N2O4 by the reaction: 2 NO2(g) ↔ N2O4(g)
    • 7. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium. • This reaction is then written asCharacteristics follows:of ChemicalEquilibrium 2 NO2(g) ↔ N2O4(g) • The double arrow indicates the reaction is reversible. • In reversible reactions: –conversion to products is the forward reaction –conversion to reactants is the reverse reaction • For a system at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.
    • 8. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium. • For the reactionDefining aA + bB ↔ cC + dDChemicalEquilibrium • at equilibrium, the graph of concentration vs time may appear as below:
    • 9. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium. • OR at equilibrium, the graph ofDefining concentration vs time may appear asChemical below:Equilibrium • the concentrations of reactants and products remain constant over time because the rate of the forward reaction is equal to the rate of the reverse reaction.
    • 10. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium.Equilibrium • In 1864, Cato Guldberg andLaw Peter Waage proposed the law of mass action or the equilibrium law. The studied many systems at equilibrium and found there was a relationship between the concentration of reactants and products at equilibrium. They suggested the equilibrium law be a ratio of product concentrations to reactant concentrations. The value of this ratio is called the equilibrium constant.
    • 11. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium. • They proposed, for the reactionEquilibrium aA + bB ↔ cC + dDLaw • the forward and reverse reactions were elementary reactions. This means rateforward = kf[C]c[D]d ratereverse = kr[A]a[B]b since rateforward = ratereverse kr[A]a[B]b = kf[C]c[D]d
    • 12. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium. • By rearranging the expression toEquilibrium solve for rate constants,Law • The ratio of rate constants was condensed to one constant, Kc, called the equilibrium constant. The law of mass action or equilibrium law then becomes
    • 13. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium.Using the Example 1 For the reactionEquilibriumLaw H2(g) + F2(g) ↔ 2HF(g) 1.00 moles of hydrogen and 1.00 moles of fluorine are sealed in a 1.00 L flask at 150°C and allowed to react. At equilibrium, 1.32 moles of HF are present. Calculate the equilibrium constant.
    • 14. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium.Using the Example 1 For the reactionEquilibriumLaw H2(g) + F2(g) ↔ 2HF(g)   H2 + F2 2 HF   I 1.00 1.00 0 Insert initial concentrations Since the equilibrium [HF] is 1.32 mol/L, it increases by that amount. However, C + 1.32 according to the stoichiometry [H2] and [F2] will decrease by one half that amount (see below) E  
    • 15. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium.Using the Example 2 For the reactionEquilibriumLaw N2(g) + O2(g) ↔ 2 NO(g) the equilibrium constant is 6.76. If 6.0 moles of nitrogen and oxygen gases are placed in a 1.0 L container, what are the concentrations of all reactants and products at equilibrium?
    • 16. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium.Using the Example 2 For the reactionEquilibriumLaw N2(g) + O2(g) ↔ 2 NO(g)   N2 + O2 2 NO I 6.0 6.0 0 Insert initial concentrations Since we do not know the equilibrium concentrations of any of the species we C insert x for the amount of N2 and O2 consumed and 2x for the amount of NO produced. E  
    • 17. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium. • When reactants and products areReaction added into a container it is good toQuotient, Q know whether equilibrium has been reached. If equilibrium has not been reached it is helpful to know which reaction, forward or reverse, is favoured in order for equilibrium to be achieved. • The reaction quotient , Q, or trial Kc enables us to determine this information. The reaction quotient is determined by using the equilibrium law and using either initial concentrations or those determined during experimental trials.
    • 18. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium. • To determine which reaction isReaction favoured and in which direction theQuotient, Q system is moving, Q is compared to Kc. • If Q = K, the system is at equilibrium. The forward and reverse rates are equal and the reactant and product concentrations remain constant. • If Q > K, the system is NOT at equilibrium. There is too much product, so the reverse reaction is favoured to bring the reactant-product ratio to equal K by increasing reactant concentration.
    • 19. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium. • If Q < K, the system is NOT atReaction equilibrium. The concentration ofQuotient, Q reactants is too large, so the forward reaction is favoured. This results in decreasing reactant concentrations and increasing product concentrations, bringing their ratio to a value equal to K.
    • 20. Outcome 4-01 Relate the concept of equilibrium to physical and chemical systems.Include: conditions necessary to achieve equilibrium.Using the Example 3 For the reactionEquilibriumLaw N2(g) + O2(g) ↔ 2 NO(g) It was found that 8.50 moles of nitrogen, 11.0 moles of oxygen and 2.20 moles of nitrogen monoxide were in a 5.00 L container. If the equilibrium constant is 0.035, are the following concentrations at equilibrium? If not, which reaction is favored and which concentrations are increasing and which are decreasing?
    • 21. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant.LeChatelier’sPrinciple
    • 22. LeChatelier’sPrinciple
    • 23. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. • Henri Louis Le Chatelier (1850 -Le 1936) was a French chemist andChatelier’s a mining engineer. He spent muchPrinciple of his time studying flames in order to prevent mine explosions. He also invented two new ways of measuring very high temperatures. • In 1884 Le Chatelier proposed the Law of Mobile Equilibrium, more commonly called Le Chateliers Principle. The principle states: When a system at equilibrium is subjected to a stress, the system will adjust so as to relieve the stress.
    • 24. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant.LeChatelier’sPrinciple
    • 25. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. • Le Chateliers Principle. TheLe principle states:Chatelier’sPrinciple When a system at equilibrium is subjected to a stress, the system will adjust so as to relieve the stress. • Stresses include: –Changing concentration –Changing pressure –Changing temperature –Adding a catalyst
    • 26. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. • Le Chateliers Principle. TheLe principle states:Chatelier’sPrinciple When a system at equilibrium is subjected to a stress, the system will adjust so as to relieve the stress. • Stresses include: –Changing concentration –Changing pressure –Changing temperature –Adding a catalyst
    • 27. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. • In a system at equilibrium, a change in the concentration ofChanging products or reactants present atConcentration equilibrium, constitutes a stress. • At equilibrium, the ratio of product to reactant concentrations is constant. • Adding more reactant, or removing product, upsets the established equilibrium. • The stress is relieved by forming more product, or using up reactant.
    • 28. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. • If reactant is added or product isChanging removed, we say that theConcentration equilibrium "shifts to the right". • The forward reaction rate increases until equilibrium is reestablished. That is, the forward reaction is favoured until Kc is attained again. • Similarly, adding more product, or removing reactant, causes the system to shift the equilibrium left. The reverse rate is favoured until the product to reactant ratio is equal to Kc once again.
    • 29. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant.LeChatelier’sPrinciple
    • 30. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. • The pressure of a system can beChanging changed by increasing orPressure reducing the volume of the reaction container. Increasing the size of the container reduces the pressure, while decreasing the size of the container increases the pressure of the system. Changing the pressure of a system only affects those equilibria with gaseous reactants and/or products.
    • 31. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. • According to Le ChateliersChanging Principle, increasing the pressurePressure on a system at equilibrium causes the system to shift to reduce its pressure, by reducing the number of particles in the system. That is, shift to the side with fewer molecules. • Conversely, decreasing the pressure on a system causes the system to shift to increase the pressure by increasing the number of particles in the container. That is, shift to the side with more molecules.
    • 32. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant.LeChatelier’sPrinciple
    • 33. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. • Recall from Kinetics thatChanging increasing temperature alwaysTemperature increases the rate of a reaction. However, increasing temperature always increases the rate of an endothermic reaction more than the rate of an exothermic reaction. • According to Le Chateliers Principle, a change in temperature causes a stress on a system at equilibrium. The system attempts to relieve the stress by either replacing lost heat or consuming added heat.
    • 34. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. • To solve equilibrium problemsChanging involving heat changes, considerTemperature heat to be a product (exothermic reactions) or a reactant (endothermic reactions).
    • 35. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. • Recall that adding a catalyst to aAdding a system decreases the activation energy of a reaction. This will causeCatalyst the rate of a reaction to increase. However, a catalyst lowers the activation energy of BOTH forward and reverse reactions equally. • Therefore, adding a catalyst to a system at equilibrium will NOT affect the equilibrium position. However, if a catalyst is added to a system which is not at equilibrium, the system will reach equilibrium much quicker since forward and reverse reaction rates are increased.
    • 36. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. • The graphs we will be Analyzing studying illustrate the rate Experimental changes for the NO2 - N2O4 Data equilibrium system. • We will assume that the system was at equilibrium initially before the stress was applied, and after an instantaneous change, the system was allowed to reestablish equilibrium. The reaction is 2 NO2(g) ↔ N2O4(g) ΔH = -58.0 kJ/mol N2O4
    • 37. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. Rate vs Time Graphs For Changing Reactant Concentration If we add more NO2 to the If we remove some NO2 system the following from the system the graph results. following graph results.
    • 38. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. Rate vs Time Graphs For Changing Product Concentration This time see what the This time see what the graphs look like when graphs look like when product is added. product is removed.
    • 39. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. Rate vs Time Graphs For Changing Product Concentration This time see what the This time see what the graphs look like when graphs look like when product is added. product is removed.
    • 40. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. Rate vs Time Graphs For Temperature Changes This time see what the This time see what the graphs look like there is an graphs look like heat is increase in temperature. removed.
    • 41. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant.Concentration vs Time Graphs For Changing Reactant Concentrations The first graph shows the The next graph shows the effect of adding more removal of the reactant. reactant.
    • 42. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. Concentration vs Time Graphs For Changing Product Concentrations This time see what the This time see what the graphs look like when graphs look like when product is added. product is removed.
    • 43. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant. Concentration vs Time Graphs for Temperature Changes This time see what the This time see what the graphs look like when graphs look like when heat is added. heat is removed.
    • 44. Outcome 4-06 Use Le Chatelier’s principle to predict and explain shifts in equilibrium.Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition ofa catalyst, the addition of an inert gas, the effects of various stresses on the equilibrium constant.
    • 45. Outcome 4-10 Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility. • When a solution is saturated,Solubility there exists an equilibriumEquilibria between the dissolved solute particles and the solid solute particles. • For an ionic compound, such as sodium chloride, we express the equilibrium in terms of a chemical equation: NaCl(s) ↔ Na+(aq) + Cl¯(aq) • This equilibrium is dynamic, since the rate of dissolving of each ion is equal to the crystallization of each ion.
    • 46. Outcome 4-10 Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility. • Substances chemists believed to beThe Solubility insoluble in water, by experiment, were shown to be slightly soluble.Product That is, with sensitive instruments, chemists we able to show that substances that would not dissolve in water actually do dissolve to a very small extent. • When a sparingly soluble ionic solid is dissolved in water to form a saturated solution the general equilibrium equation is AaBb(s) ↔ aA+(aq) + bB¯(aq) • Where A is a positively charged ion and B is a negatively charged ion.
    • 47. Outcome 4-10 Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility. • At a given temperature, the equilibrium law for this reaction is given asThe SolubilityProduct • However, the term KC[AaBb] can be replaced by a new constant, Ksp, called the solubility product. Ksp = [A+]a[B¯]b • The solubility product constant is the product of ion concentrations in a saturated solution. The solubility product constant takes into account the presence of the solid.
    • 48. Outcome 4-10 Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility. • Recall that the general form of the solubility product expression isThe SolubilityProduct Ksp = [A+]a[B¯]bExpression • for the dissociation equation AaBb(s) ↔ aA+(aq) + bB¯(aq) Example 1 • Write the dissociation equation and the expression for the solubility product constant for calcium hydroxide.
    • 49. Outcome 4-10 Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility. Example 2Calculating • Write the solubility productSolubility expression for Pb3(PO4)2.Product Example 3 • If at equilibrium, the concentration of silver ions is 1.3 x 10-5 mol/L and the concentration of chloride ions is 1.3 x 10-5 mol/L, what is the Ksp of silver chloride?
    • 50. Outcome 4-10 Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility. • Solubility and solubility product are twoSolubility different terms. • Solubility is the maximum amount of solute that can dissolve in a certain amount of solvent at a certain temperature. Solubility has an infinite number of possible values, depending on temperature and other solutes present. • Solubility product is an equilibrium constant and has only one value for a given solid at a particular temperature. Example 4 • The solubility of PbF2 is 0.466 g/L. What is the value of the solubility product?
    • 51. Outcome 4-10 Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility. Example 5Determining • The Ksp of magnesium hydroxideIonConcentration is 8.9 x 10-12. What will be thefrom Ksp equilibrium concentrations of the dissolved ions in a saturated solution of Mg(OH)2?
    • 52. Outcome 4-11 Solve problems involving Ksp. Include: common ion problems.Common Ions • When an ionic compound dissolves in pure water, the initial concentration of each ion is zero. However, if an ionic compound dissolves in a solution that has an ion in common with the compound, this is not the case. • Even though the starting concentrations may not be zero, the product of the ions must still equal the solubility product constant.
    • 53. Outcome 4-11 Solve problems involving Ksp. Include: common ion problems. • For example, how would the solubility ofCommon Ions silver chloride in pure water change if we try dissolving it in tap water? • Lets write out the dissociation equation: AgCl(s) Ag+(aq) + Cl¯(aq) • Tap water often has chlorine added to kill bacteria. The chlorine exists as chloride ions, so when we dissolve silver chloride in tap water, chloride ions are present. • According to Le Chateliers Principle, Adding more chloride ions to a saturated solution would cause the equilibrium to shift to the left to use up the excess product. This would result in more solid formed, and a decreased solubility.
    • 54. Outcome 4-11 Solve problems involving Ksp. Include: common ion problems.Solubility in • Le Chateliers Principle predictsthe Presence that the solubility of an ionic solidof a Common in a solution containing a commonIon ion decreases its solubility. Lets see if this is supported by the calculations. Example 1 • Determine the solubility of silver chloride in pure water and in a solution of 0.10 mol/L sodium chloride. The Ksp of AgCl is 1.7 x 10-10.
    • 55. Outcome 4-11 Solve problems involving Ksp. Include: common ion problems.Solubility in Example 2the Presence • The Ksp of lead (II) chloride, PbCl2,of a Common is 1.6 x 10-5. What is the solubilityIon of lead (II) chloride in a 0.10 mol/L solution of magnesium chloride, MgCl2? Example 3 • The Ksp of lead (II) chloride is 1.6 x 10-5. What is the solubility of lead (II) chloride in a 0.10 mol/L solution of lead (II) nitrate, Pb(NO3)2?

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