Slide 1

Chemical Equilibrium (some early goals/topics) Concept / Condition of equilibrium Law of Mass Action Reaction Quotient, Q vs. Equilibrium Constant, K Meaning of K? Relationships between Ks of related reaction equations Equilibrium Problems! Slide 2 Concept / Condition (Board work) Copyright Houghton Mifflin Company. All rights reserved.132 Slide 3 Copyright Houghton Mifflin Company. All rights reserved.133 Figure 13.2 H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) [CO 2 ] or [H 2 ] [CO] or [H 2 O] Time Concentration EquilibriumAt some T (where equilibrium favors products; rxn goes until nearly completion) Slide 4 Copyright Houghton Mifflin Company. All rights reserved.134 Figure 13.4 As the Concentrations of the Reactants Decrease, the Forward Reaction Slows Down (and as [Ps] increases, R rev increases!) Slide 5 Copyright Houghton Mifflin Company. All rights reserved.135 Figure 13.3 a-d Gaseous Equilibrium Mixture of Reactants and Products 7422035574220355 CO H2H2 H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Slide 6 Copyright Houghton Mifflin Company. All rights reserved.136 Figure 13.2 (Mines) H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) [CO 2 ] or [H 2 ] [CO] or [H 2 O] Time Concentration Equilibrium At a different T (where equilibrium favors reactants; rxn goes very little before equilibrium is established) Slide 7 Copyright Houghton Mifflin Company. All rights reserved.137 Figure 13.1 a-d A Molecular Representation of the Reaction 2 NO 2 (g) N 2 O 4 (g) as a particular system reaches equilibrium Although the same number of each kind of molecule are present in (c) and (d), they are not the same exact moleculeschemical equilibrium is dynamic. The rate at which N 2 O 4 is made equals the rate at which it goes away (R f = R r ) Slide 8 The Reaction Quotient, Q Q is defined to be the following quotient. It is defined for a balanced chemical equation: aA + bB cC + dD Copyright Houghton Mifflin Company. All rights reserved.138 (A, B, C, D are gases or solution species) Ps (products) Rs (reactants) Q is defined for any system, whether it is at equilibrium or not Slide 9 Law of Mass Action "When a reactive system* comes to equilibrium at a given temperature, the value of Q is always the same (even when all concentrations are different!). This constant value (of Q at equilibrium) is called the equilibrium constant, K." Copyright Houghton Mifflin Company. All rights reserved.139 Slide 10 Law of Mass ActionFollow up By "reactive system", I mean any system with any amounts of a combination of species A, B, C, and/or D where a reaction can occur according to a balanced equation, e.g., aA + bB cC + dD. In other words, each "reactive system" must be defined by a balanced chemical equation. Copyright Houghton Mifflin Company. All rights reserved.1310 Slide 11 Law of Mass ActionFollow up It does not matter what (initial) amounts you start with in your system, some net reaction will occur until equilibrium is established, at which point the value of Q will always be the (same) value of K. Copyright Houghton Mifflin Company. All rights reserved.1311 Slide 12 Law of Mass ActionSummary Copyright Houghton Mifflin Company. All rights reserved.1312 For a balanced chemical equation: aA + bB cC + dD Value depends on the state system is in. Varies for any reaction system from 0 to Value is constant for a given reaction system & equals Q when system reaches (or is at) equilibrium. Value can be between 0 and It can not be zero. Value varies with T (discuss later) Slide 13 Copyright Houghton Mifflin Company. All rights reserved.1313 Example N 2 + 3 H 2 2 NH 3 Q = K Q < K (Q increasing w/time) NOTE: As forward reaction occurs, Q increases until it equals K Slide 14 Copyright Houghton Mifflin Company. All rights reserved.1314 Q at equilibrium equals a constant value, K! -------------------------------- Q = K at eq Slide 15 Q vs. K Tells you Direction of Reaction Because Q is Ps over Rs: When forward reaction occurs, Q increases When reverse reaction occurs, Q decreases If Q is bigger than K: The numerator is too big too many Ps So Q needs to decrease to become equal to K So reverse reaction occurs If Q is smaller than K: not enough Ps; so forward reaction occurs! Copyright Houghton Mifflin Company. All rights reserved.1315 Slide 16 Q vs. K Tells you Direction of reaction If Q equals K: System is at equilibrium! Revisit Earlier Example (next slide) Copyright Houghton Mifflin Company. All rights reserved.1316 Slide 17 Copyright Houghton Mifflin Company. All rights reserved.1317 K = 0.0602 NOTE: Q I (= 0) < K; not enough Ps, forw rxn occurs Q II (= ) > K; too many Ps, rev rxn occurs Q III (= 4.5) > K; too many Ps, rev rxn occurs Recall Prior Example! (Q vs K determines which direction) Slide 18 Simulation Copyright Houghton Mifflin Company. All rights reserved.1318 Slide 19 More Board Work (some topics on outline handout) Meaning of K (large vs small) Interpretation of Q vs. K comparison Equilibrium constant expressions Types of Equilibrium Problems Relationships between Ks of reaction equations Copyright Houghton Mifflin Company. All rights reserved.1319 Slide 20 Dont mix up Tendency of reaction with Rate of reaction! The Law of Mass Action says nothing about the rate at which equilibrium will be established. It can be extremely slow or extremely fast, and is not related to the value of K! Copyright Houghton Mifflin Company. All rights reserved.1320 Slide 21 Copyright Houghton Mifflin Company. All rights reserved.1321 Figure 13.7 (Zumdahl) A large K does not mean a fast rate! Si boards in our computers (as well as human flesh!) have a tremendous tendency to react with O 2 and turn to sand (or CO 2 and H 2 O)! Thank goodness these are slow!! Slide 22 Write the K expressions (Law of Mass Action) for Each Equation (1) 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) K = (2) 2 H 2 O 2 (aq) 2 H 2 O (l) + O 2 (g) K = Copyright Houghton Mifflin Company. All rights reserved.1322 (Omit pure liquid, H 2 O; see next slide) Slide 23 1323 Figure 14.6 A Heterogeneous Equilibrium. If C (s) is added to the system at the left, the equilibrium will not be disturbed (it will remain at equilibrium). Slide 24 (Continued) (3) 4 KO 2 (s) + 2 H 2 O (l) + 2 CO 2 (g) 4 K + (aq) + 2 HCO 3 - (aq) + 2 OH - + 3 O 2 (g) K = (4) PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) K = Copyright Houghton Mifflin Company. All rights reserved.1324 1 = Slide 25 Equilibrium Problems (on board) Including Definition of K P Back to Yellow Equilibrium Outline Handout to do: E. Relationships between Ks Copyright Houghton Mifflin Company. All rights reserved.1325 Slide 26 (from pink sheet) 1326 Slide 27 (Q3 from salmon PS3 sheet) Copyright Houghton Mifflin Company. All rights reserved.1327 Slide 28 (Q2 from salmon PS3 sheet) Copyright Houghton Mifflin Company. All rights reserved.1328 Slide 29 Le Chteliers Principle If a stress is applied to a system at equilibrium, the system will shift, if possible, in the direction that at least partially alleviates the stress Stress means something that causes the system to not be at equilibrium anymore Shifting means Net reaction occurs in either the forward (shift right) or reverse (shift left) Copyright Houghton Mifflin Company. All rights reserved.1329 Slide 30 Le Chteliers Principle (cont.) Possible* stresses: Increase or decrease the concentration of a reactant or product Add or remove some reactant or product. Changing the pressure of the system (by changing the volume of it) Changing the temperature of the system Copyright Houghton Mifflin Company. All rights reserved.1330 *Be aware that whats written above is somewhat of a simplification, so sometimes what appears to be a stress is not actually a stress!! Ill address this later Slide 31 Le Chteliers Principle IdeaAdding or Removing a Species If the concentration of a reactant is increased, system will react to decrease the amount of that reactant (shifts right; forward rxn occurs) If the concentration of a reactant is decreased, system will react to make more of that reactant (shifts left; reverse rxn occurs) If the concentration of a product is increased, system will react to decrease the amount of that product (shifts left; rxn occurs) Copyright Houghton Mifflin Company. All rights reserved.1331 Slide 32 Le Chteliers Principle IdeaAdding or Removing a Species Caution! Only truly applies to concentration changes Does not apply to addition/removal of pure solids or liquids. Copyright Houghton Mifflin Company. All rights reserved.1332 Recall: If C (s) is added to the system at the left, the equilibrium will not be disturbed (it will remain at equilibrium). *Addition of a solid is not a true stress! System not knocked out of equilibrium! Slide 33 Copyright Houghton Mifflin Company. All rights reserved.1333 Figure 13.8 (Zumdahl). A shift due to addition of a species. (a) An Initial Equilibrium Mixture of N 2, H 2, and NH 3 (b) Addition of N 2 (system not at equilibrium) (c) The New Equilibrium Position (which contains more N 2, less H 2, and more NH 3 than in (a)) N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Slide 34 Fig. 14.9 Le Chateliers Principle: Changing Concentration Slide 35 Quick Quiz Before We Move Forward: From Pink Sheet (Q3) Copyright Houghton Mifflin Company. All rights reserved.1335 Slide 36 Le Chteliers Principle IdeaChanging Pressure (with a volume change) If the pressure is increased, system will react to decrease the pressure. How can it do this? Only by decreasing the total moles of gases in the container! Sometimes not possible. (No true stress) If the pressure is decreased, system will react to increase the pressure. Caution: The principle only works if the pressure change is due to a volume change! Tricky question: What if you add some Ar(g) to increase the pressure? Copyright Houghton Mifflin Company. All rights reserved.1336 Slide 37 Fig. 14.9 Le Chateliers Principle: Changing Pressure (via a volume change!) V is decreased: P total increases Reaction will occur to decrease P total (if possible) Reaction that decreases total moles of gas will occur [e.g., you need to look at n gas ] For this reaction equation, making products will decrease n gas Shifts Right Slide 38 V is increased: Everything exact opposite of prior slide! Slide 39 Copyright Houghton Mifflin Company. All rights reserved.1339 Figure 13.9 (Zumdahl). A shift resulting from a V decrease (a) A Mixture of NH 3 (g), N 2 (g), and H 2 (g) at Equilibrium (b) The Volume is Suddenly Decreased (c) The New Equilibrium Position for the System Containing More NH 3 and Less N 2 and H 2 (system reacts to make less moles of gas) N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Slide 40 Le Chteliers Principle IdeaChanging Temperature If the temperature is increased, system will react to decrease the temperature? Okay, not quite. That doesnt work. If the temperature is increased, thats like adding energy to the system system will react to absorb (use up?) the added energy (convert it into potential energy) The endothermic reaction (whether it be forward or reverse) will occur! NEED TO CONSIDER SIGN OF H. Copyright Houghton Mifflin Company. All rights reserved.1340 Slide 41 Le Chteliers Principle IdeaChanging Temperature If the temperature is decreased, thats like removing energy from the system system will react to produce (give off?) energy (lower PE and give off heat) The exothermic reaction (whether it be forward or reverse) will occur! NEED TO CONSIDER SIGN OF H. Copyright Houghton Mifflin Company. All rights reserved.1341 Slide 42 T Effect on an Equilibrium System (older slidesbit of review) Depends on whether reaction is exo- or endothermic! Exo: H 0 Copyright Houghton Mifflin Company. All rights reserved.1342 Many ways to approach this The standard Le Chatelier argument o If T increases, its like adding KE, and system shifts to absorb the added energy o Treat energy like a reactant or product (but its not!?) Helps you see which way system shifts, but o Must also follow up to realize that with T changes, K changes (this is not true for other Le Chat. stresses) Slide 43 T Effects, continued Slightly different way to look at this: Copyright Houghton Mifflin Company. All rights reserved.1343 Learning that when T increases, the endothermic reaction** is what occurs to reach a new equilibrium, and it is the endothermic reactions** K that increases o **Sometimes this is the forward direction, and sometimes this is the reverse direction. So be careful when you analyze these problems! Slide 44 Copyright Houghton Mifflin Company. All rights reserved.1344 Photo 13.5 a-b (Zumdahl) LeChatelier's Principle II (in ice)(over flame) 1) Write the balanced equation for the reaction that occurs (net) when the T is decreased. (Blue = N atom; red = O atom) 2) Is the reaction that occurred endo or exothermic? 3) Is H for the balanced equation you wrote positive or negative? Slide 45 1345 This reaction is endothermic (in forward direction) so an increase in T: 1)Makes system shift right, 2)Makes K increase. (This is the real reason that the system shifts right!!) H > 0 ; system absorbs energy when rxn occurs Slide 46 Copyright Houghton Mifflin Company. All rights reserved.1346 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) ; H < 0 This reaction is exothermic (in forward direction) so an increase in T: 1)Makes system shift left (less products are made), 2)Makes K decrease. (This is the real reason that the system shifts left!!) This reaction uses up gases (net) [ n gas < 0], so it makes more products (with same K) as V is decreased (P total is increased) Slide 47 Le Chteliers Principle: a Q vs. K view (the real story) Copyright Houghton Mifflin Company. All rights reserved.1347 At (New) Equilibrium (with same K, but different equilibrium states) How Q will change will depend on a variety of things, and in some cases, Q wont change at all. Slide 48 Fig. 14.10 Q > K ! Reverse rxn occurs (this is not new!!) Shifts left (verbiage is new) 1348 Slide 49 Fig. 14.10 Q < K ! Forward rxn occurs (this is not new!!) Shifts right 1349 Slide 50 Le Chteliers Principle: a Q vs. K view (the real story) At (New) Equilibrium (with greater K) At (New) Equilibrium (with smaller K) If T increases, K will increase if endothermic If T increases, K will decrease if exothermic 1350 Slide 51 Copyright Houghton Mifflin Company. All rights reserved.1351 Table 13.3 (Zumdahl) Value of K for Ammonia Synthesis as a Function of Temperature Slide 52 Copyright Houghton Mifflin Company. All rights reserved.1352 Slide 53 Effect of a Catalyst on Equilibrium? Will return to this question after discussing relation of K to k f, k r (next slide) Copyright Houghton Mifflin Company. All rights reserved.1353 Slide 54 Relating K to k f and k f for an elementary reaction (equation) See board. Easy to show that Try to see how this makes sense from the standpoint of what must be true for R f = R r : If k f >> k r, the concentrations of products must be greater than reactants to make rates equal (be at equilibrium) P favored ( @ Equilibrium) Revisit Simulation! http://www.chem.uci.edu/undergraduate/applets/sim/simulation.htm Copyright Houghton Mifflin Company. All rights reserved.1354 Slide 55 Effect of a Catalyst on Equilibrium? See board Recall that a catalyst creates a new pathway for the rxn with lower activation energy. Might think, then, that system would shift right (forward rate increases), but What about reverse reactions rate Also increasesby same exact amount (factor of change)! So a catalyst has no effect on equilibrium concentrations / state: The system just reaches equilibrium faster Forward and reverse rates (at equilibrium) are both faster with a catalyst present. Copyright Houghton Mifflin Company. All rights reserved.1355 Slide 56 The small x approximation Helpful way to solve for x in the problem type where one needs to define an x and use Law of Mass Action equation to solve for x. Not always valid. Must check the assumption afterward See board Copyright Houghton Mifflin Company. All rights reserved.1356 Slide 57 EXAMPLE 14.13 Finding Equilibrium Concentrations from Initial Concentrations in Cases with a Small Equilibrium Constant PROCEDURE FOR... Finding Equilibrium Concentrations from Initial Concentrations in Cases with a Small Equilibrium Constant 1.Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the reactants and products. (In these examples, you must first calculate the concentration of H 2 S from the given number of moles and volume.) 2.Use the initial concentrations to calculate the reaction quotient (Q). Compare Q to K to predict the direction that the reaction will proceed. continued By inspection, Q = 0; the reaction will proceed to the right. To solve these types of problems, follow the procedure outlined below. 2011 Pearson Education, Inc. Consider the reaction for the decomposition of hydrogen disulfide: A 0.500-L reaction vessel initially contains 1.25 10 4 mol of H 2 S at 800 C. Find the equilibrium concentrations of H 2 and S 2. Slide 58 3.Represent the change in the concentration of one of the reactants or products with the variable x. Define the changes in the concentrations of the other reactants or products with respect to x. 4.Sum each column for each reactant and product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. 2011 Pearson Education, Inc. continued Slide 59 5. Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the equilibrium constant. Use the given value of the equilibrium constant to solve the expression for the variable x. In this case, the resulting equation is cubic in x. Although cubic equations can be solved, the solutions are not usually simple. However, since the equilibrium constant is small, we know that the reaction does not proceed very far to the right. Therefore, x will be a small number and can be dropped from any quantities in which it is added to or subtracted from another number (as long as the number itself is not too small). Check whether your approximation was valid by comparing the calculated value of x to the number it was added to or subtracted from. The ratio of the two numbers should be less than 0.05 (or 5%) for the approximation to be valid. If approximation is not valid, proceed to step 5a. 2011 Pearson Education, Inc. Checking the x is small approximation: The approximation does not satisfy the