Answer Key, Problem Set 3 (full explanations and work)

Transcription

1 Chemistry 1 Mines, Spring, 018 Answer Key, Problem Set (full explanations and work) 1. NT1;. NT;. 1.4; 4. 1.;.1.; ; ; ; ; *; 11. NT; 1. NT4; The Equilibrium Condition and the Law of Mass Action (concepts) 1. NT1. Consider a reaction represented by: A(g) + B(g) C(g) Assume a particular trial s initial concentrations were [A] 0 = 9 M; [B] 0 = 9 M, and [C] 0 = 0 M, and that the final value for [B], when the system reaches uilibrium ([B] ), is 8 M (see left trace below). 9 B Concen -tration (M) A Rate Rf C Time Time (a) Copy the left plot down on your paper and add the traces for [A] and [C], being careful to consider the stoichiometry of the reaction. Answer: See blue traces on plot. A must be lost at x the rate of loss of B. Since B went down by 1 M (from 9 to 8), A must go down by M (i.e., down from 9 to 6). C must form at x the loss of B. Thus it must go up by M (from 0 to ). (b) Copy the right plot down on your paper and add a trace for the reverse rate (R r ). Answer: See blue traces on plot. Rr must start at zero since there is no C to begin with. It must increase at time goes by (since [C] increases with time) end at the value of Rf since once they become ual, the system reaches uilibrium. (c) What two things must always be ual when a chemical system is at uilibrium? Answer: Rf and Rr must be ual at uilibrium. That is the definition of (dynamic) uilibrium: Rate forward Rate reverse. (Note: It is also true that the value of Q K at uilibrium, but this question was designed with the rate idea in mind, since that is the definition of dynamic uilibrium.) (d) Do you think that the value of the uilibrium constant for the reaction uation in question is somewhat large or somewhat small? Give your reasoning. Answer: Somewhat small. One reactant only goes down in value by ~11% (B), and the other by only ~%. That means more of the reactants remain at uilibrium than reacted. Fewer products than reactants at uilibrium means reactant favored and a small value of K.. NT. (a) State the law of mass action. The Law of Mass Action states that at a given temperature, when a given reaction system reaches dynamic uilibrium, regardless of what the initial concentrations of reactants and/or products were, the value of the reaction quotient, Q, always has the same value. NOTE: Q is defined in terms of a balanced chemical uation that represents the forward reaction that occurs in the system: c d [C] [D] For: a A + b B c C + d D (A, B, C, and D are gases or solution species): Q a b Since the value of Q at uilibrium is always a constant, it is given a special name and symbol: the uilibrium constant, K, and so in mathematical form, the Law of Mass Action is simply: [A] [B] [C] [A] c a [D] [B] d b K for a system at uilibrium (i.e., if [ ] s are uilibrium ones) PS-1

2 Answer Key, Problem Set (b) True or false: K changes as a reaction occurs, but Q does not. Give reasoning. (c) False. Q changes as a reaction occurs, but K does not. Since Q is made up of concentration terms, with those of the products in the numerator and those of the reactants in the denominator: If forward reaction occurs ([products] increase and [reactants] decrease and so), Q must increase. If reverse reaction occurs ([products] decrease and [reactants] increase and so), Q must decrease. Since K is the value of Q only when the system is at uilibrium, and that value is found to always be the same (at a given temperature) [by experiment summarized by the Law of Mass Action (see part (a) above)], the value of K does not change as a reaction occurs (assuming temperature is constant, of course). True or false: The value of K depends on the initial concentrations of reactants and products. Give reasoning. False. The value of K is constant for a given reaction system, as long as T is kept constant. It matters not what position the system starts off in. When uilibrium is established, Q will ual K. (See part (a)). (d) True or false: The uilibrium concentrations of reactants and products depend on their initial concentrations. Give reasoning. (e) True. If you start with different initial concentrations, then you (nearly always) end up with different final (uilibrium) concentrations because of stoichiometry (mass conservation). As a trivial example (just to demonstrate the idea), if you start with M of A in a reaction where A B and K, the final uilibrium state will be [A] 1 M and [B] M. However, if you started with 1 M of A, the uilibrium state will end up with [A] 1/ M and [B] / M. Both uilibrium states are characterized by the same value of Q ( K [ in this case]), but the systems contain different final amounts because of the different starting amount(s). True or false: There is only one value of K for a particular system at a particular T, but there are an infinite number of uilibrium positions / states. Explain. True. As noted above, it is the value of the reaction quotient that is constant for any given reaction system at uilibrium. But there are an infinite number of combinations of values that will satisfy the uation given in part (a) for any given chemical uation. For example, if K = [C] [D] 6 for a system in which K, the following uations are all true (given a bit of [A][B] uncertainty in each concentration value) even though the values for [A], [B], [C], and [D] are different in each case: Hopefully you can see that mathematically there are an infinite number of combinations of values that will satisfy a given law of mass action uation. Thus there are an infinite number of uilibrium positions or states for a given reaction system even though they are all characterized by the same value of K. etc. (f) True or false: When all concentrations of reactants and products are 1 M, Q = K. Explain. False. When all concentrations of reactants and products are 1 M, Q = 1. c d [C] [D] Q, being the reaction quotient, has the form: Q. As such, if all concentrations are 1 a b M, you get 1 c 1d in the numerator, which will always be 1, no matter what c and d are (or if there are other products, E, F, etc., it won t matter, since all values will be 1 M), and the same is true for the denominator. 1/1 = 1! PS-

3 Answer Key, Problem Set NOTE: Q = K means the system is at uilibrium with whatever concentrations there are right now in the system. Those concentrations need not be (and generally are not) ual to one another nor ual to 1 M. (g) True or false: If Q > K, there are more products than reactants. Explain. False. In order for there to be more products than reactants, the quotient must be larger than 1.* This has nothing to do with K really, since the statement says nothing about being at uilibrium. Q > K basically just means that there are more products than there could be at uilibrium given the number of reactants. That is, this statement corresponds to product heavy rather than product favored. An example should make this clear (I should have put this in the lecture!). Let s assume a reaction uation whose K value uals.6 x 10-8, and a reactive system having a value of Q ual to 1.4 x Clearly, in this case, Q > K, but in the system when Q = 1.4 x 10 -, there are not more products than reactants there are more reactants than products! * This isn t strictly true for all reaction uations for math reasons I don t want to get into now. It is strictly true only for an uation such as A B, but only mostly true for most chemical reaction uations and systems. Ask me if you are curious about this qualification. Please note that: Q reflects the relative amounts of Products to Reactants in any particular system. K reflects the relative amounts of Products to Reactants when the system is at uilibrium. K will depend on the reaction type (nature of the reaction) and reaction uation (and T). It can t be varied (if T is kept constant). Q can be varied since it just depends on the concentrations of reactants and product are present at a given time. (h) What is the difference between the terms Product favored and Product heavy (which I may have referred to as too many products )? Explain, relating these phrases to the quantities K, Q, and/or 1 as needed. [NOTE: Product heavy is a term I came up with; it does not appear in Tro]. Product favored means that there are more products than reactants at uilibrium.* This is the case when the uilibrium constant is large (K > 1). Note: Some reactive systems/chemical reactions will have a great tendency to occur in the forward direction and they will have a large K value. Some will not. It depends on the nature of the reaction. Product heavy means that in the system right now, there are too many products relative to the amount of reactants present for this system to be at uilibrium. Reverse reaction (net) will occur as this system reaches uilibrium. This is the case when Q is larger than the uilibrium constant (Q > K). Note: No matter what the K is for a given reaction/balanced uation, there can always be a system with Q > K. This condition (of Q > K) says nothing in absolute terms about either the state the system is in right now or what the system will look like when it reaches uilibrium. It only represents a relative comparison between the two states. * See previous problem for the qualification associated with this slight overgeneralization. (i) Does the term Reactant favored mean the same thing as Reactant heavy (which I may have referred to in class as too many reactants or not enough products )? Explain, relating these phrases to the quantities K, Q, and/or 1 as needed. [NOTE: Reactant heavy is a term I came up with; it does not appear in Tro]. Reactant favored means that there are more reactants than products at uilibrium.* This is the case when the uilibrium constant is small (K < 1). Note: Some reactive systems/chemical reactions will have little tendency to occur in the forward direction and they will have a small K value. Some will not. It depends on the nature of the reaction. PS-

4 Answer Key, Problem Set Reactant heavy means that in the system right now, there are too many reactants relative to the amount of products present for this system to be at uilibrium. Forward reaction (net) will occur as this system reaches uilibrium. This is the case when Q is smaller than the uilibrium constant (Q < K). Note: No matter what the K is for a given reaction/balanced uation, there can always be a system with Q < K. This condition (of Q < K) says nothing in absolute terms about either the state the system is in right now or what the system will look like when it reaches uilibrium. It only represents a relative comparison between the two states. * See previous problem for the qualification associated with this slight overgeneralization. (j) True or false: Reactions with large K values are fast. Explain. False. A large K indicates only that once uilibrium is established, there will be a large amount of products relative to reactants. It does not indicate how fast the forward reaction occurs (i.e., how long it will take to actually reach uilibrium). It is the rate constant (kf) that is proportional to rate, not K. It is easily possible to have a reaction with a very large K that has a very tiny value of kf because of an extremely large activation energy barrier. NOTE: It is easy to derive that for an elementary reaction, K kf. You can see from this uation that if k f is, k r say, and k r is 10-18, K will be In such a case, the reaction is quite product favored even though the forward rate constant is tiny (and thus the forward rate will be small at typical concentrations of reactant). Writing Equilibrium Constant Expressions [All in Mastering] The Meaning of K What is the significance Answers: The (value of the) uilibrium constant reflects the tendency of a reaction to occur in the forward direction (where forward is indicated by the way the balanced chemical uation is written). This is because the uilibrium constant s value reflects the composition of any uilibrium state in terms of products relative to reactants. For example, a large K indicates that once uilibrium is established, there will be a great amount of products relative to reactants, which means the reaction tends to favor products at uilibrium and thus has a great tendency to occur in the forward direction (as written). A small K indicates that once uilibrium is established, there will be a small amount of products relative to reactants, which means the reaction tends to favor reactants at uilibrium and thus has very little tendency to occur in the forward direction (as written) When [a] reaction [represented by the following chemical uation] comes to uilibrium (rest of Q not typed into key) A(g) + B(g) C(g) K c = 1.4 x 10 - Answers: Reactants (will be greater). The uilibrium constant is fairly small (<< 1), meaning the numerator (with [C] in it) will be very small compared to the denominator (which has [A][B] in it). Thus, reactant favored (at uilibrium). The initial concentrations do not matter because Q = K once uilibrium is established. Thus, even though the specific values may differ, in any uilibrium state, the value of [C] will ual 1.4 x 10 - x [A][B] (which means [A][B] >> [C][C], and thus the concentrations of the reactants will be greater). (Technically, it is the product of the two concentrations that must be larger than [C], so in reality, [C] could be greater than one of the two (A or B), but if so, the concentration of the other one would be much much greater than [C].) PS-4

5 Answer Key, Problem Set Equilibrium Problems (basically four types ). 1. Consider the reaction [uation!]: Answer: 16 or 1.6 x 10 Reasoning / Work: CO(g) + H (g) CH OH(g) This is a plug in type of uilibrium problem. The concentrations given in the problem are clearly denoted as being uilibrium concentrations. So just write the uilibrium constant expression and plug in the concentration values to get K: [CH 6xOH] 0.18 K or c [CO] [H] (0.10)(0.114) Consider the reaction [uation!]: Answer: or 9.87 x 10 - NH 4 HS(s) NH (g) + H S(g) Reasoning / Work: Same exact strategy as in the prior problem. However, this problem involves a pure solid, which must be omitted from the K expression: [NH 77x ] [HS] K (0.78)(0.) or c Consider the reaction [uation!]: NO(g) + Br (g) NOBr(g) ; K p = 8.4 at 98 K Answer: 0.08 atm or 4 torr Reasoning / Work: This is another plug in type of problem, but with a couple of twists. Two of the three uilibrium partial pressures are given, along with Kp, so the third should be calculable by simply plugging into the uilibrium constant expression. Some rearrangement (algebra) is needed since K is not the unknown, but it is nonetheless a plug in situation. However, by convention, Kp s are calculated using partial pressures using 1 atm as the reference (or standard) state. As such, although we input numbers without units, the values have to be associated with the pressure in atmospheres, not torr. Thus, the pressures here need to first be converted from torr to atmospheres before substitution: 1atm 1atm P NO 108 torr x atm ; P Br 16 torr x atm 760 torr 760 torr K p P P P NOBr NO Br PNOBr 8.4 P (8.4)(0.141) (0.167) NOBr (0.141) (0.167) 9810-NOTE: 1 atm = 760 torr (exactly), so 760 has no uncertainty (i.e., an infinite number of SFs). P NOBr atm or (in torr) : 0.08 atm x 760 torr/atm 4 torr PS-

6 Answer Key, Problem Set Consider the reaction [uation!]: CO(g) + H (g) CH OH(g) Answer: 7. Strategy: This is problem in which you are effectively given the initial concentrations and one uilibrium concentration, and you are asked to determine K. Thus, you must recognize that knowing one species initial and uilibrium concentrations allows you to figure out its change (amount reacted or formed). Once that is known, the changes in all of the other species are related by the stoichiometry (the coefficients in the balanced uation). Thus, the final uilibrium concentrations can be determined (we typically use an ICE table to help achieve these calculations). Then one can plug the [ ] s into the uilibrium constant expression. The only twist here is that you are not technically given concentrations you are given masses and a volume, so you must calculate the concentrations using 1 st semester Gen. Chem. ideas (g mol with molar masses, then mol / L to get M). Execution: Initial reaction mixture: 1mol 6.9 g CO x ( ) g CO.4 g H 1mol mol CO.19 L.19 L x mol H M H ( x 1.008) g H M CO (initially) (initially) Since no CHOH is not mentioned at all when the problem states the contents of the reaction mixture, one must assume that there is none there initially. Equilibrium composition: 1mol.19 L 8.6 g CHOH x mol CHOH M CHOH (4 x ) g CHOH (at uilibrium) Fill in what is known into an ICE table: [CO] (M) [H] (M) [CHOH] (M) Initial Change Equilibrium C = E I (final initial) = = Determine the Change row values (from the right-most one and then using the coeff. ratios [stoich]): (Remember that the signs of the changes in the reactants must be the opposite of the signs of the changes in the products, because as products are made, reactants are used, and vice versa) [CO] (M) [H] (M) [CHOH] (M) Initial Change (0.001) Equilibrium PS-6

7 Answer Key, Problem Set Determine the remaining Equilibrium row values (from the I and C values) [CO] (M) [H] (M) [CHOH] (M) Initial Change (0.001) Equilibrium = 0.10 Substitute uilibrium values into K expression: [CH OH] Kc 7... [CO] [H ] (0.10)(0.1196) 0.6 (0.001) = E = I + C (initial + change) Consider the reaction [uation!]: NH 4 HS(s) NH (g) + H S(g) Answer: more solid will form (because reverse reaction will occur; Q > K) Reasoning / Work: This is a Q vs. K kind of problem. You are given initial concentrations and a value of K, and you are asked, in effect, Which direction will reaction occur as uilibrium is established? (because you are asked whether solid will form or get used up). If you calculate Q and compare it to K, you can determine whether the system is product heavy ( too many products to be at uilibrium ) or product deficient ( too few products to be at uilibrium ). Q c [NH ][HS] (0.166)(0.166) which is greater than Kc (= 0.008) [Given] If Q (= products / reactants ) is too large, then there are too many products to be at uilibrium and so reverse reaction will occur to establish uilibrium ( shifts left ). more solid will form Nitrogen dioxide dimerizes according to the reaction [uation!]: NO (g) N O 4 (g) ; K p = 6.7 at 98 K Answer: No, not at uilibrium. Forward reaction occurs (b/c Q < K) Reasoning / Work: This is another Q vs. K type of problem. However, Kp instead of Kc is given even though mol values are given along with volume and temperature. To me, the most straightforward way to proceed is to simply use the ideal gas uation to calculate the partial pressures of the gases (in atm), and then calculate Qp using the Q expression. Then compare to Kp as in the prior problem. (The solution manual authors calculate K c from K p and then use the concentrations to calculate Q c ; that s fine, but ruires more memorization! I am pretty sure that you all know PV=nRT better than the uation that relates K p to K c!) Execution: n PV nrt P RT Thus: V P P NO N O 4 n V NO n V N O 0.0 mol RT. L mol RT. L Latm K atm molk Latm K atm molk PS-7

8 Answer Key, Problem Set PN O Qp.49.. which is less than Kp (= 6.7 [Given]) [so not at uilibrium] P NO Q is too small need more products (greater numerator) to reach uilibrium forward reaction occurs 11. NT. 1.a&c(+d). Consider the reaction [uation!] and associated uilibrium constant: a A(g) + b B(g) c C(g) K c =.0 Find the uilibrium concentrations of A, B, and C for each value of a, b, and c. [For parts (a) and (c)] assume that the initial concentrations of A and B are each 1.0 M and that no product is present at the beginning of the reaction. For part (d), assume that the initial concentrations of A and B are each 1.0 M and that the initial concentration of C is 4.0 M. (a) a = 1; b = 1; c = (c) a = ; b = 1; c = 1 (set up uation for x; don t solve) (d) a = ; b = ; c = (set up uation for x; don t solve) Answers*: (a) [A] = 0.47 M; [B] = 0.47 M; [C] = 1.06 M (c) (d) x K c.0 * 1 x 1 x 4 x 1 x 1 x K c.0 * * As noted below, there are many correct answers for parts (c) and (d) because the answer here is an algebraic expression whose form will depend on how you define x. However, if you were to actually solve for the final concentrations at uilibrium in all cases, those answers would be the same no matter how x was defined. Strategy: I ignored SFs here for convenience (since we re not solving for x anyway) I will essentially follow the strategy outlined in Tro (starting on p. 6 and shown in Examples 14.9 and 14.10) and demonstrated in class. That is: 1) Write the balanced uation and the uilibrium constant expression ) Calculate, if necessary, initial concentrations, and put into an ICE table (with concentrations, in M). ) Calculate Q and compare to K to determine which reaction (forward or reverse) occurs to get to uilibrium. (This actually isn t technically necessary, but it helps you conceptually in understanding what is going on in the problem and can help you discover, for example, math errors. It also helps you to define your x so that it will be a positive quantity. Alternatively, you can skip this step and just check all possible x roots to find the one that works.) 4) Define an x to represent the concentration of a species that is lost or formed as the system reaches uilibrium. I generally define the x to be a species with a coefficient of 1. However, as you will see in my part (d) below, if there is no coefficient of 1, it is actually simplest to define a x or a x, etc. This just avoids fractions. If you prefer to define an x (and not worry about what I just wrote), that s fine, too. You ll just end up with fractions. ) Utilize the fact that the changes in species concentrations while reaction is taking place are in the ratio of the coefficients in the uation to determine the changes in the other reactants and products in terms of your defined x, and then add those terms to the ICE table (C row). 6) Utilize the idea that the final values (E) must ual the initial ones (I) plus the changes (C) to get expressions with x s in them for the uilibrium concentrations (E) in the table. [Remember, C is a ". = f i, thus C = E I, which means E = I + C.] 7) Looking at the uilibrium constant expression uation (that you should have written earlier, although you didn t really need to do it earlier), which would be of the general form PS-8

9 Answer Key, Problem Set [C] [D] K [assuming two reactants and two products], substitute the algebraic uilibrium c a d b concentrations into the K expression, being extremely careful not to make math errors that might arise if you are trying to simplify things in your head! Also substitute in the value of K (which should always be given in such a problem) into your uation (i.e., don t leave the symbol K in there!) 8) If asked to, solve the math uation just generated for x by appropriate means. 9) Substitute the values of x obtained back into the algebraic expressions for the species at uilibrium (E row in the ICE table) and pick the solution that gives physically meaningful results (e.g., you can t have a negative uilibrium concentration!). 10) Check your results by using the numerical values of the uilibrium concentrations to calculate a value of K, and verify that the value obtained is close to (given uncertainties) the given value of K. Execution of Strategy (part (a)): A(g) + B(g) C(g) [C] K [A] [B].0 Initial concentrations are given as 1.0 M, 1.0 M, and 0 M. Thus: Initial Change Equilibrium Since [C]0 = 0, Qc = 0, which means Q < K and forward reaction occurs to reach uilibrium. Let x = [A] that is lost as uilibrium is established (the coefficient of A is 1, so this is convenient but not necessary) Thus, [B] lost also uals x, and the [C] formed is x. Thus (recognizing that the changes in reactants concentrations is negative since they are being lost) we can write: Initial Change - x - x + x Equilibrium Complete the table as follows (E = I + C): Initial Change - x - x + x Equilibrium 1.0 x 1.0 x 0 + x = x Substitute in: [C] K x 1.0 x1.0 x.0 Although one could multiply this all out, combine terms, and use the quadratic uation, if you recognize that the left side is a perfect square, taking the square root of both sides is an easier option: x 1.0 x1.0 x x 1.0 x.0 PS-9 x 1.0 x.0 x 1.0 x.6.. (see note below*)

10 Answer Key, Problem Set * Technically, this should be.6 and I should solve for both x values to make sure I get the correct one. But the -.6 value will lead to a negative x which I know is not valid the way I defined my x, so I will not clutter up the key with that. x.6 -. x x x.6x x.6 x Calculate the uilibrium concentrations using x = 0. M: [A] = [B] = = 0.47 M [C] = (0.) = 1.06 = 1.1 M Check the results (I ll use the unrounded results here [because rounding error causes a discrepancy that I d prefer not to have to worry about)]: [C] K.0.0 Good. (Technically, this should be 0. due to SFs, but I m going to leave SF because they might as well have said 1.00 M or exactly 1 M etc. To end up with 1 SF is somewhat poor problem creation in my opinion and I d prefer to avoid that.) Execution of Strategy (part (c)): A(g) + B(g) C(g) [C] K.0 [A] [B] Initial concentrations are given as 1.0 M, 1.0 M, and 0 M. Thus: Initial Change Equilibrium Since [C]0 = 0, Qc = 0, which means Q < K and forward reaction occurs to reach uilibrium. Let x = [B] that is lost as uilibrium is established (the coefficient of B is 1, so this is convenient but not necessary) Note that [A] lost here (unlike in part (a)) uals x because of the new stoichiometry [coefficient of for A and 1 for B here], and the [C] formed is x. Thus: Initial Change - x - x + x Equilibrium Completing the table yields: Initial Change - x - x + x Equilibrium 1.0 x 1.0 x 0 + x = x Substituting in (be very careful to really substitute in note the changes in the position of the square!): [C] K 1.0 x 1.0 x x.0 Since they asked you not to solve for x, you can leave the uation like this. I can see no benefit to getting the uation into standard form unless you were going to use the quadratic formula. You can better estimate x using this form of the uation, frankly. In a couple of moments, I realized that x = 0. would give K = (zero in the denominator), so I tried x = 0.4, which gave me Q = So clearly, x must be a bit less than 0.4. Not bad for a one-minute estimate! PS-10

11 Answer Key, Problem Set Execution of Strategy (part (d)): [C] A(g) + B(g) C(g) K.0 Initial concentrations are given as 1.0 M, 1.0 M, and 4.0 M. I will ignore SFs here for convenience (less bulky ) since we are not going to solve for x anyway. Thus: Initial Change Equilibrium Q [C] , which means Q > K and reverse reaction occurs to reach uilibrium. Since reverse reaction occurs, we know that A and B will be formed (not used up) as uilibrium is reached. Thus Let x = [B] that is formed as uilibrium is established (NOTE: There is no species with a coefficient of 1 here, so instead of defining an x, I decided to define a x. This is convenient but not necessary. For those that may have defined x differently, I ll show other results below) With the loss of B as x, the loss of A must just be x and the formation of C must be x. Thus: Initial Change + x + x - x Equilibrium Completing the table yields: Initial Change + x + x - x Equilibrium x x 4 x Substituting in (be very careful to really substitute in use parentheses and correctly place each exponent): K [C] 4 x 1 x 1 x [A] [B].0 NOTE: There are many correct answers to this problem as written, because there are many ways to define x, and each one will have a different form. For example: 1) If you had not calculated Q to find that reverse reaction actually occurs to reach uilibrium, you may have defined x to be the [B] that reacted. If so, you would have gotten: 4 x 1 x 1 x.0 ) If you had let x = [B] that formed, your answer would have been: ) If you had let x = [A] that formed, your answer would have been: For those interested, using an excel spreadsheet, I was able to quickly estimate x to be 0.06 (to 4 SF). Thus, [C].469 M; [A] M; and [B] 1.61 M. Excel may not be elegant, but it is very practical! (and x would have been a negative quantity) 1 4 x x 1 x 4 x 1 x 1 x.0.0 PS-11

12 Answer Key, Problem Set Finding K's For an Equation from Other K's (For Other Related Equations) 1. NT4. State in words what happens to the value of K when a reaction uation is reversed, and rationalize why this is so conceptually. In other words, why does it make sense that if a reaction (uation) has a really large K, the reverse reaction (uation) will have a really small K? Answer: When a reaction uation is reversed, the uilibrium constant is the reciprocal of the original (forward) reaction uation. Mathematically, this makes sense because when you reverse a reaction, you switch the positions of the (original) reactants and products that is, the original reactants become the products, and the original products become the reactants. Since the uilibrium constant expression has products in the numerator and reactants in the denominator, this switch naturally means that the fraction becomes its reciprocal. Conceptually, this makes sense because if a reaction uation has a really large K, that means that it has a great tendency to occur in the forward direction, meaning that the products would be present much more so than reactants when the system reaches uilibrium. Since the real world obviously does not change just because someone chooses to write the uation representing the reaction in a different way, it makes sense that the K value for the uation written in reverse should be really small, as that would mean that the reactants (which are the same physical species that used to be the products when the uation was written in the original way) are now favored at uilibrium. An example may make this clearer. If C is the product of a reaction whose uation has a large K, then C is present at uilibrium more than the reactant species. If the uation is written in reverse, the real world should not change one bit, so C still will still present more than the other species at uilibrium. But since the uation is now written in reverse, C becomes a reactant, and so the reaction now should be reactant favored at uilibrium, which would mean a small K rather than a large one. This is true if K for the reverse reaction is the reciprocal of the original one Use the reactions [uations!] below and their uilibrium constants to predict [calculate!] the uilibrium constant for the reaction [uation!], A(s) D(g) Answer: 4.7 x 10-4 Strategy: 1) Figure out a way to manipulate the given uations in such a way as to get two uations that will sum up to the target uation (this is analogous to how one would do a Hess s Law type problem in Gen. Chem. I). **A handout has now been posted on the course URL (Week 4, PS0 folder) to help you with this kind of manipulation of given uations.** ) Determine the uilibrium constant values for the new uations from Step 1 using the relationships discussed in class, on the yellow handout, and in the text (and in prior problems on this set). ) Take the product of the two new uations K s to get the K value for the sum. Execution of Strategy: Noting that one needs two A s as reactants in the target uation, multiply uation (1) [which is the only uation that has A in it] by to get: A(s) B(g) + C(g) K1 K1 x K1 = 0.04 ( x 10 - ) Noting that one needs three D s as products in the target uation, and that uation () has three D s as reactants, reverse uation () to get: B(g) + C(g) D(g) K K- K ( 4...x 10-1 ) Verifying that the sum of the above two uations does ual the target uation (the B and C parts will cancel out leaving A(s) D(g) as desired), the K for the sum should ual the product of the two new K s: K1 + K1 K 0.04 x x x 10-4 PS-1