Answer Key, Problem Set 4 (With explanations)

Transcription

1 Chemistry 1 Mines, Spring 018 Answer Key, Problem Set 4 (With explanations) ;. NT1; 3. NT; *; ; 6. NT3; ; 8. NT4; 9. NT5; 10. NT6; More Equilibrium Problems (extension of PS3 concepts and sills) For the uation below, K c = at 500 K N O 4 (g) NO (g) If a reaction vessel initially contains an N O 4 concentration of M at 500 K, what are the uilibrium concentrations of N O 4 and NO at 500 K? Answers: [NO4 = M; [NO = M Strategy: Same as in (end of PS3) Execution of Strategy: NO4(g) NO(g) [NO K (at 500 K) [N O 4 Initial concentrations are given as , and 0 M. Since Q is again zero, Q < K and forward reaction must occur to reach uilibrium. Thus I ll let x = [NO4 that reacts and the ICE table becomes: [NO4 (M) [NO (M) Initial Change - x +x Equilibrium x x Substitute in and solve for x: [NO x K x [N O x x x 0.513x x (4)(4)( ) (4) x or A negative x here does not mae sense (can t have a negative [NO, and we now forward, not reverse reaction, must occur to reach uilibrium in this case. So use the 1 st one. [NO M [NO ( ) M M [NO Chec: ( 0.51) [N O K (good!) 4 PS4-1

2 . NT1. Assume that you dump an excess of solid SrF (s) into some water, and the following "reaction" occurs until uilibrium is reached (what process is being described? It isn't technically a "chemical change" but we can treat it lie it is): SrF (s) Sr + (aq) + F - (aq) ; K = 4.3 x 10-9 (at some T) (a) What will be the concentrations of Sr + (aq) and F - (aq) at uilibrium? Answer: [Sr x 10-3 M; [F -.05 x 10-3 M Reasoning / Wor: Recognize that this is just lie any other process in which uilibrium is attained via "reaction". You now the initial concentrations of dissolved species (here, zero), and you now the K for the process. You just need to figure out "how much reaction" occurs before the system reaches uilibrium. I will let x = the [Sr + that forms as the system reaches uilibrium. The appropriate relationships (stoichiometry and algebraic) are summed up in the ICE table below. Don't forget to deal appropriately with the stoichiometry of the uation! Note that since SrF(s) does not appear in the Q or K expression, we need not worry about its actual amounts. In fact, you could actually just leave that column out of the ICE table. I included it only to show that there is sufficient solid present initially such that uilibrium is reached BEFORE all of the solid "runs out" (i.e., dissolves). From the balanced uation, we have: "SrF(s)" [Sr + (M) [F - (M) I (initial) some 0 0 C (change in) - less than all of it! +x +x E (at uilibrium) still some left! x x K [Sr Substituting in the algebraic expressions figured out via the ICE table (or uivalent reasoning), and using the given value for K you get: 4.3 x 10-9 = (x)(x) = 4x 3 [F - x 3 = x 10-9 Thus: Note: x x 10 x 1.04 x [Sr + = 1.0 x 10-3 M, and [F - =.05 x 10-3 M Significant figures are a bit hard to properly handle in this this ind of problem (where algebra is used to solve for x); if you eep only two, then when you substitue bac into the K expression, you end up with K = 4.0 x 10-9 rather than 4.3 x 10-9, so I ept three [which gives K = 4.9 x But two would be oay as well. (b) How many moles of solid SrF (s) "reacted" (i.e., "dissolved") per liter of solution formed? Answer: 1.0 x 10-3 M Reasoning: Since one mole of SrF(s) dissolves ("reacts") per each mole of Sr + that ends up in solution (1:1 stoichiometry between SrF(s) and Sr + ), the number of moles of SrF(s) that dissolved per liter of (eventual) solution formed must just ual the number of moles of Sr + that formed, per liter, which was just "x in this problem! Note: We call "1.0 x 10-3 M" the "molar solubility" of SrF (s) at this temperature, since it represents the maximum number of moles of the solid that would dissolve in 1 liter of solution. You could not get greater than this number of moles of SrF (s) to dissolve at this temperature because uilibrium will be reached at this point. PS4-

3 3. NT. Assume that the following uation describes the only possible chemical change that can occur inside of the box shown: N O(g) + 3 O (g) N O 4 (g) (a) If the box mared initial state represents the system at a point in time at which Q > K, which option [(a)-(e) best represents what the system might loo lie after uilibrium is established? Answer: (d) Short Reasoning: 1) Q > K reverse rxn occurs NO4 s are lost must be either (b) or (d). ) 3 O s are gained for every NO4 s lost it must be (d) ((b) has same # of O s as initial state! Also, (d) has fewer N O 4 s, more N O s and 3 more O s than initial. Checs.) Full Explanation: The condition Q > K means Q is too big for the system to be at uilibrium. Thus there are too many products (or too few reactants ) (since products are in the numerator of the K expression (and reactants are in the denominator)). Thus some reverse reaction will occur to reach uilibrium. Thus the number of molecules of NO4 should decrease as time goes by, and the number of molecules of NO and O should increase. If you count up the molecules of NO4 in all the boxes, you will find that the number is the same in (a), larger in (c) and (e), and smaller in (b) and (d). Thus the correct answer must be either (b) or (d). At this point you may have wondered how you could eliminate one of these. You must loo to mae sure that the changes in the number of each ind of molecule are consistent with the stoichiometric recipe given by the balanced uation! In (b), the number of molecules of NO4 left is two, whereas in the initial state there were three, so that means that one molecule of NO4 reacted. Based on the balanced uation, exactly one molecule of NO and three-halves of a molecule of O would have to be formed. Obviously that can t happen! You can t have half a molecule! Furthermore, count up the number of molecules of O in box (b). There are actually the same number (six) as in the original box. This could not happen either, so this box fails the stoichiometry test (even though it seemed qualitatively correct more or less.) In box (d), there is only one molecule of NO4 left, which means that two must have reacted. By stoichiometry, three O s and two NO s should have formed. If you count them up and compare to the original box, that is exactly the case, and so box (d) could represent the system at uilibrium. Perhaps ICE tables can help mae the point above clearer: Box (d) NO (molecules) O (molecules) NO4 (molecules) I (initial) C (change in) E (at uilibrium) Changes are consistent with balanced uation PS4-3

4 Box (b) NO (molecules) O (molecules) NO4 (molecules) I (initial) C (change in) E (at uilibrium) 7 6 Changes are N consistent with balanced uation (b) Once the system in part (a) of this problem reaches uilibrium, will the P total in the box be greater, smaller, or the same as the initial P total? Assume T and V remain constant. Give reasoning. Answer: Ptotal will be greater (because reverse reaction results in more moles of gas in the box) Full Reasoning: If T and V are constant, P ngas. So to assess Ptotal, we need to loo at what happens to ngas(total) as reaction occurs to reach uilibrium. From (a), we now that reverse reaction occurs to reach uilibrium. Looing at the balanced uation, reverse reaction results in more gas molecules being produced than used up because + 3 = 5 molecules of gas get made for each that get used up (ngas = - 5 = -3 in the forward direction; ngas = 5 = +3 in the reverse direction). Thus Ptotal must increase as this system reaches uilibrium, and Ptotal will be greater at uilibrium than initially [The uation for! a particular reaction has an uilibrium constant of K p = A reaction mixture is prepared in which all the reactants and products are in their standard states. In which direction will the reaction proceed? NOTE: In order to do this problem, you must realize (recall?) that the standard state of any gas (Gas A) is the state in which the partial pressure of A is 1 atm. Stating that a reaction mixture is at standard state (conditions) is the same as saying that all gases in that mixture have partial pressures ual to 1 atm. (It also means that all solution species are at a concentration of 1 M.) The Mastering Problem in the preceding question (Q5) indicates this, but I decided to write this here as well. Answer: It will proceed in the reverse direction. Reasoning: At standard state conditions, Q = 1 (here, we d say Qp = 1). Since 1 > 0.50, Qp > Kp, which means too many products (to be at uilibrium) [i.e., product heavy and reverse reaction occurs to reach uilibrium. Recall that I said in class that it is useful to thin of starting a reaction (theoretically) at a point where all reactants and products are present at 1 M so that Q = 1. This way, if the reaction has K > 1, forward reaction will occur from the standard state and there will be more products than reactants at uilibrium ( product favored ). If K < 1, then reverse reaction will occur to get to uilibrium and there will be more reactants than products at uilibrium ( reactant favored ). Well, the same is true for gases if you start at standard state conditions (all PA s at 1 atm)! x is small approximation Consider the reaction [uation: SO Cl (g) SO (g) + Cl (g) ; K c =.99 x 10 at 7 C If a reaction mixture initially contains M SO Cl, what is the uilibrium concentration of Cl at 7 C? Answers: [SOCl = M; [SO Cl.9 x 10-4 M Strategy: Same as in 15.5 (near the end of PS3), except that the small x approximation will be used to simplify the solving for x. Execution of Strategy: SOCl(g) SO(g) + Cl(g) ; [SO [Cl K.99 x 10 (at 7 C) [SO Cl PS4-4

5 Initial concentrations are given as 0.175, and 0 M (well, one must infer that the other species initial concentrations are zero because the problem does not state that there are any of them). Since Q is zero, Q < K and forward reaction must occur to reach uilibrium. Thus I ll let x = [SOCl that reacts and the ICE table becomes: [SOCl (M) [SO (M) [Cl (M) Initial Change - x +x +x Equilibrium x x x Substitute in and solve for x: [SO [Cl K [SO Cl xx x.99 x 10 Since K is less than 10-4 and is larger than 0.01, it is reasonable to at least try the small x approximation. That is, assume that x << such that x Thus: xx x.99 x 10 Assume x xx x 0.175(.99 x 10 ) 5.3 x x 10 x x x 10-4 A negative x here does not mae sense (can t have a negative [SO, and we now forward, not reverse reaction, must occur to reach uilibrium in this case. So use the positive one x 10 Chec assumption: x % (< 5% so OK) [SOCl M (clearly, the assumption was good!) [SO Cl.9 x 10-4 M -4-4 [Cl.87 x x 10 Chec: K [SO.99 x 10 [SO Cl (good!) 6. NT3. Assume the same chemical uation as in the prior problem (14.56). Assume the same initial reaction mixture (0.175 M SO Cl ), except that in this (new) system, the initial concentration of SO (g) is M instead of 0 M. The initial concentration of Cl is still 0 M. (a) Calculate the uilibrium concentration of Cl at 7 C. Answers: [SOCl = M; [SO M; [Cl 5.3 x 10 M Execution of Strategy: Since Q is still zero (since [Cl = 0), Q < K and forward reaction must occur to reach uilibrium. Thus I ll let x = [SOCl that reacts and the ICE table becomes: [SOCl (M) [SO (M) [Cl (M) Initial Change - x + x + x Equilibrium x x x PS4-5

6 .9.9Answer Key, Problem Set 4 Substitute in and solve for x: [SO[Cl9x (0.100 x)( x) K OCl ( x) 10[SIn this case, the small x approximation can be made in two places. Assume that x << and x << such that x and x Thus:.9x109x(0.100 x)( x) Assume x (0.100)( x) Assume x ( x) (0.175) 0.175(.99 x 10 ) x 5.3 x Since x is even smaller here than it was in problem 14.56, the assumption clearly was good! [SOCl M (I hope you see why I m not even going to calculate it!) [SO M Cl 5.3 x 10 M Chec: x 10 [SO [Cl K.99 x 10 (good!) [SO Cl (b) Is the [Cl you obtained for this system greater or smaller than the one you obtained for the system in 15.58? By how many times? Try to describe why this is the case (mathematically and conceptually) x x 10 Answer: It is quite a bit smaller! and 437 (rounded), x x 10 which means the value here is a bit more than 400 times smaller. Mathematical analysis: The only difference is the [SO0. When it is at ( nd case), it s lie the numerator in [SO [Cl K get s a head start for any value of x greater than zero, the numerator will [SOCl be larger in the nd case than the 1 st, and so Qnd will always be larger than Q1st (note that the denominators are both essentially constant at so they don t affect the relative values of Q). For example, when M of Cl has formed, the numerators would be: (0.0001)(0.0001) = 1 x 10-8 in the 1 st case, but ( )(0.0001) (0.100)(0.0001) 1 x 10-5 in the nd case (1000 times larger!). This means that Qnd will become the value of K sooner than will Q1st, which means at uilibrium, x will be smaller in the nd case i.e., [Cl will be smaller. Another way to put it is this: as forward reaction occurs (in both cases), Q will increase, but it will always be larger if some SO is present to begin with, so it will reach K after less reaction has occurred. Conceptual analysis: In the initial case, there is none of either product present to start with. The reverse rate is zero but the forward rate is not. The reaction will reach uilibrium after enough products have been formed to mae the reverse rate become ual the forward rate. Now thin about it in the second case, you already start with a significant amount of one of the products before any reaction has occurred. That means once any amount of the nd product is made, the rate of the reaction in the reverse direction will be faster that it would be in the first case. Thus, the reverse rate will become ual to the forward rate after much less reaction has occurred. The generalization is: If one product is present initially the forward reaction is effectively inhibited. Equilibrium will be reached after less reaction has occurred than if no product had been there initially. 10PS4-6

7 Le Châtelier s Principle and related concepts (Q vs K, revisited) Chemical systems represented by each of the following uations are allowed to come to uilibrium and then the volume is changed as indicated. Predict the effect (shift right, shift left, or no effect) of the indicated volume change. Answer (a) CO(g) + H O(g) CO (g) + H (g) (V is decreased) no effect (b) PCl 3 (g) + Cl (g) PCl 5 (g) (V is increased) left (c) CaCO 3 (s) CaCO 3 (s) + CO (g) (V is increased) right Generalized Le Châtelier Reasoning. Brief Reasoning ngas 0; Ptotal can t be changed; (really, Q is not changed by the stress in the first place) ngas 0; V increase leads to Ptotal decrease. System shifts to increase Ptotal by maing more gases (reactants, here); (Q increases because all [ s decrease, but more of them in denominator) ngas 0; V increase leads to Ptotal decrease. System shifts to increase Ptotal by maing more gases (products, here); (Q decreases because only [CO is in Q. If V increases, [CO and Q decrease) If Vvessel is decreased, Ptotal increases. So the system will shift, if possible, in the direction that uses up gas particles (net) so that Ptotal will decrease. That means the shift will occur to the side of the balanced uation (reactants or products) that has fewer moles of gases represented (considering the sum of coefficients of gases on each side). If Vvessel is increased, Ptotal decreases. So the system will shift, if possible, in the direction that produces gas particles (net) so that Ptotal will increase. That means the shift will occur to the side of the balanced uation (reactants or products) that has more moles of gases represented (considering the sum of coefficients of gases on each side). 8. NT Coal can be used to generate hydrogen gas (a potential fuel) by the endothermic reaction [represented by the following uation: C(s) + H O(g) CO(g) + H (g) Assume a reaction mixture of C, H O, CO, and H is at uilibrium. (i) Predict whether each disturbance will result in the formation of additional hydrogen gas, the loss of hydrogen gas, or have no effect on the quantity of hydrogen gas in the system. (ii) In each case, does the disturbance itself (instantaneously, before any reaction/shift can occur) cause an increase, decrease, or no change in the value of Q? of K? H (g) is Answers Disturbance causes in Q (before any reaction occurs) Disturbance causes in K (a) Adding more C to the reaction mixture not changed no change no change (b) Adding more H O to the reaction mixture FORMED a DECREASE no change (c) Raising the T of the reaction mixture FORMED no change an INCREASE (d) Increasing the V of the reaction mixture FORMED a DECREASE no change (e) Adding a catalyst to the reaction mixture not changed no change no change (f) Adding an inert gas to the reaction mixture not changed no change no change Explanations: General Considerations: In order for the stress to be a true stress (i.e., in which the system would shift), it must change either Q or K. The only way to change K is to change the temperature. So if T is not changed, the only way to have a true stress is if the stress changes the value of Q. PS4

8 (a): Since C is a solid, it does not affect the uilibrium at all. Its concentration does not change and it does not appear in the Q or K expression, so Q does not change. So there is no shift, which means (here) that H is neither formed nor lost (net). The system remains at uilibrium. (b): HO is a gaseous reactant here. If a (gaseous) reactant is added, its concentration is increased, and the system will react to get rid of that reactant. So forward reaction will occur (shift right, and more H will form [since H is a product). Since reactants are in the denominator of Q, increasing the concentration of a reactant will decrease Q (instantaneously) leading to a Q < K situation. (c): Le Châtelier approach: An increase in T (lie adding energy ) favors the endothermic reaction (the reaction that uses up or absorbs energy). In this case, the endothermic reaction is the forward reaction, so the system shifts right, and more H is formed (as the system reestablishes uilibrium). In the new uilibrium state, there will be more products and fewer reactants than before the disturbance (because forward reaction occurred!), so K must be larger after the T increase. A diagrammatic depiction of this scenario is as follows: [CO [H Q [H O [CO[H Q [H O T increase caused K to increase (forward rxn endothermic) Q K Q K Q increases (forward rxn occurs) (d): Increasing V decreases Ptotal. System shifts to mae more gas molecules (to increase Ptotal). Here that is to the right because two gas molecules are made for every one lost (ngas 0). Q must have gotten instantly smaller if system shifts right (to increase Q again), and this can be seen to be so since there are two [ s in the numerator and only one in the denominator. (i.e., increasing V maes every species concentration smaller by the same factor. Since there are two factors of concentration in the numerator and only one in the denominator, one of the factors of change in the numerator will cancel with the one in the denominator, but the other one will remain in the numerator thus Q will decrease). (e): A catalyst does not change Q and it does not change K. (It increases both the forward and reverse s by exactly the same factor, so the ratio f/r remains the same). The system never is out of uilibrium. (f): Adding an inert gas does not change Q (the inert gas does not appear in either the numerator or the denominator!) and it does not change K (it does not change either!). It has no effect on any aspect of the uilibrium system. It just increases Ptotal (without consuence). Q [CO Q [H [H O K K = f /r and catalysis effect (or non effect) on uilibrium. Rate vs. Tendency of reaction **NOTE: had a part that involved the effect of a catalyst on uilibrium. Sorry that that slipped in there before this section. Also, NT5 is mostly about Le Chatelier s Principle also, but I put it in this section because of the one part about a catalyst. Clearly there is overlap in these (arbitrary) sections.** 9. NT5. True or false? Correct any false statements by adding, dropping, or changing a few words. (a) When a reactant is added to a system at uilibrium at a given temperature, the system will shift right to reestablish uilibrium. FALSE (strictly speaing), because it will only be true if the reactant is not a pure solid or liquid. CORRECT: When a gaseous (or dissolved) reactant is added to a system at uilibrium at a given temperature, the system will shift right to reestablish uilibrium. PS4-8

9 (b) When a gaseous product is added to a system at uilibrium at a given temperature, the value of K for the reaction will increase as uilibrium is reestablished. FALSE CORRECT: When a gaseous product is added to a system at uilibrium at a given temperature, the value of K for the reaction will remain the same as uilibrium is reestablished. OR CORRECT: When a gaseous product is added to a system at uilibrium at a given temperature, the value of Q for the reaction will decrease as uilibrium is reestablished. Note: The ey word here is as. Initially, Q would be instantly increased. But then as the uilibrium is reestablished, reverse reaction is occurring and while this reaction is occurring Q will be decreasing. (c) When temperature of a reaction mixture at uilibrium is increased, the value of K for the reaction uation will increase. FALSE, because it isn t always true. For exothermic reactions, K will decrease when T increases (see analysis in Problem #8(c) and #9(c), although unfortunately, they only gave endothermic examples. The opposite would be true for exothermic reactions in terms of the shift direction and value of K) CORRECT: When temperature is increased for an endothermic reaction at uilibrium, the value of K for the reaction will increase. OR CORRECT: When temperature is increased for a reaction at uilibrium, the value of K for the reaction will sometimes increase and sometimes decrease. OR CORRECT: When temperature is increased for a reaction at uilibrium, the value of K for the reaction will increase if the reaction is endothermic and decrease if it is exothermic. (d) When the volume of a reaction container is increased for a system at uilibrium at a given temperature, the system will shift left to reestablish uilibrium. FALSE, because it isn t always true. Only if the forward reaction produces fewer moles of gases than are used up (i.e., ngas is negative) CORRECT: When the volume of a reaction container is increased for a system at uilibrium at a given temperature, the system will shift left to reestablish uilibrium if ngas is negative. (e) OR if the reverse reaction produces more moles of gas than are used up OR if the reactants side of the uation has more total moles of gas than the products side. OR something uivalent. Addition of a catalyst will speed up the rate at which uilibrium is established, but it will not change the amounts of reactants and products present at uilibrium. TRUE PS4-9

10 10. NT6. Assume that the reaction uation NO (g) N O 4 (g); H -55. J/mol represents an (a) elementary reaction, in which f and r are the rate constants for the forward and reverse reactions. Starting with the relationship R f R r for a system at uilibrium, and using the rate laws for the forward and reverse reactions (See Ch. 14), derive the following relationship: For elementary reactions, the orders in the rate law ual the coefficients in the elementary reaction uation. [N O f 4 r [NO Thus, Rf = f[no and Rr = r[no4 for the reaction in this problem. uilibrium Rf = Rr (and the concentrations are uilibrium concentrations). Thus: f [NO r [NO4 r [NO4 [NO4 f f (Done!) [NO [NO Since f and r are both constant, their ratio must also be a constant as well. That means that r K f r [NO4 [NO a constant, and if we call that constant K, then K f r (b) Assuming the relationship between K, f, and r is true generally, describe how a reaction having an uation with a very large K can be slow. (Hint: Does f need to be large for K to be large?) Answer: A large K just means that f is large relative to f. f need not, itself, be large as long as it is 7 larger than r. So, for example, if f = 10 s -1 and r = 10-1 s -1 f 10 5, K 10. This 1 r 10 reaction would be significantly product favored, but yet its half life in the forward direction (assuming it were 1 st order) would be approximately t x / s which is roughly days! Clearly this would be considered slow and clearly its K value is large. (c) If T is increased, what should happen to K for this reaction uation? How do you now? Answer: K should decrease. You now by noticing that the sign of Hsys is negative, meaning the reaction is exothermic (and then using Le Châtelier type arguments). (More) Reasoning: The negative sign here (H -55. J/mol) means energy flows out from the system and thus the process is called exothermic. By Le Châtelier s principle (see #8(c)), if T is increased, the endothermic reaction occurs. In this case, the endothermic reaction is the reverse reaction. Thus, after uilibrium is reestablished, there are fewer products and more reactants than before, and thus K is smaller. Thus, K should decrease when T increases for this reaction uation. (d) What should happen to f when T is increased? r? (This is Chapter 14 material here!!) Answer: Both rate constants should increase with T. This relationship is indicated by the E a Arrhenius uation RT Ae, and is explained by the collision model (along with inetic molecular theory). NOTE: there is no f or r in the Arrhenius uation! PS4-10

11 (e) How can you reconcile your answers to parts (c) and (d)? At first they may seem contradictory, but thin carefully about what must be true for this to occur. Answer: It s simple! Although f and r both increase in value, that does not mean that they increase by the same factor! To reconcile the fact that K goes down (in this case) when f and r both go up, you just need to assume that f goes up by a smaller factor than r. For example, if f were to double while r quadrupled, K would become half as f f 1 f 1 large: K K1 4 r r r A chemist trying to synthesize a particular compound attempts two different synthesis reactions. The uilibrium constants for the two reactions [reaction uations are 3.3 and. x 10 4 at room temperature. However, upon carrying out both reactions for 15 minutes, the chemist finds that the reaction with the smaller uilibrium constant produces more of the desired product. Explain how this might be possible. Answer: Shortest answer: The reaction with the smaller K is faster, and after 15 minutes, it therefore has produced a greater amount of products. The reaction with the larger K is slower, and must not have reached uilibrium yet. A larger K does not guarantee a greater f, only a larger f/r. In this case, the reaction with the larger K also happens to have a larger Ea, maing it slower. Full(er) analysis. The rate of a reaction is proportional to the rate constant, not the uilibrium constant. Thus, in this scenario, the reaction with the larger K might have a smaller f (as discussed in the prior problem). A large K value simply does not guarantee a large f and a fast rate of reaction! It only indicates that once uilibrium is established, the reaction mixture will be mostly products. The reaction might tae days, years, or millenia to occur! Note that this scenario might arise because of a much larger Ea for the reaction with the larger K. The rate constant for a reaction depends on the activation energy, but the uilibrium constant does not. As noted in class, it is a very good thing (for us) that many reactions with large K s are very slow. For example, humans should burn up (undergo combustion) in the oxygen-rich atmosphere in which we live. Than goodness that the combustion reaction has a sufficiently high Ea that at room temperature, the reaction is extremely slow! The same can be said for our computer chips. The silicon wafers should turn into sand (SiO) in our oxygen-rich atmosphere, but since the Ea for that reaction is so high, the process will tae millennia (or longer?) to occur to any significant extent. Hurray for Ea barriers! *****************FORMAL END OF SET. PROBLEM BELOW IS FOR EXTRA PRACTICE ONLY*************** Consider a system at uilibrium represented by the following uation: C(s) + H O(g) CO(g) + H (g) (i) Predict whether the reaction [system will shift left, shift right, or remain unchanged after each disturbance. (ii) In each case, does the disturbance itself (immediately) cause an increase, decrease, or no change in the value of Q? of K? Shift (L, R, or unchanged) Disturbance causes in Q (before any reaction occurs) Disturbance causes in K (a) C is added to the reaction mixture unchanged no change no change (b) H O is condensed and removed from the reaction mixture LEFT an INCREASE no change (c) CO is added to the reaction mixture LEFT an INCREASE no change (d) H is removed from the reaction mixture. RIGHT a DECREASE no change Explanations: PS4-11

12 First off, the only way to change K is to change the temperature. That is why none of these changes affects K. Since T was not changed in any of these, if a shift occurs, it must be as a result of the disturbance changing Q. (a): Since C is a solid, it does not affect the uilibrium at all (its concentration does not change and it does not appear in the Q or K expression). (b) (d). Since all these species are gases, when they are added or removed, the concentration of the species does change, and so Q will change and the system will shift in the direction that will mae Q become ual to K again (see below). Le Châtelier s principle can be used to predict the direction of the shift without considering the effect on Q. However, in my class, you are also responsible for understanding how Q changes as a result of a stress (and during any net reaction the occurs to reestablish uilibrium). That is, you should understand that the application of Le Châtelier s principle is just a simpler way to analyze what fundamentally is just a Q vs. K ind of situation (i.e., What happens to Q or K, and how do systems react when Q K?) If a (gaseous) reactant is removed (as in (b)), the system will react to replenish that reactant, and so reverse reaction will occur (shift left). Since reactants are in the denominator of Q, decreasing the concentration of a reactant will increase Q (instantaneously) leading to a Q > K situation. [CO [H Q [H O [CO [H [H O Q[CO [H Q [H O Removal of HO causes Q to increase Q decreases (reverse rxn occurs) Q K Q > K Q K If a (gaseous) product is removed (as in (d)), the system will react to replenish that product, and so forward reaction will occur (shift right). Since products are in the numerator of Q, decreasing the concentration of a product will decrease Q (instantaneously) leading to a Q < K situation. [CO [H Q [H O Q [CO [H O [H [CO Q [H O [H Removal of H causes Q to decrease Q increases (forward rxn occurs) Q K Q < K Q K If a (gaseous) product is added (as in (c)), the system will react to get rid of some that product, and so reverse reaction will occur (shift left). Since products are in the numerator of Q, increasing the concentration of a product will increase Q (instantaneously) leading to a Q > K situation. PS4-1