answer key, problem set 3 (full explanations and work)

Chemistry 122 Mines, Spring, 2020 (5/e)

PS3-1

Answer Key, Problem Set 3 (full explanations and work) 1. NT1; 2. NT2; 3. NT3; 4. 16.4; 5. 16.23; 6.16.35; 7. 16.36; 8. 16.39; 9. 16.46; 10. 16.47; 11. 16.50*; 12. NT4; 13. NT5; 14. 16.30

-----------

The Equilibrium Condition and the Law of Mass Action (concepts)

1. NT1. Consider a reaction represented by: 3 A(g) + B(g) 2 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 equilibrium ([B]eq), is 8 M (see left trace below).

(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 3x the rate of loss of B. Since B went down by 1 M (from 9 to 8), A must go down by 3 M (i.e., down from 9 to 6). C must form at 2x the loss of B. Thus it must go up by 2 M (from 0 to 2).

(b) Copy the right plot down on your paper and add a trace for the reverse rate (Rr).

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 equal, the system reaches equilibrium.

(c) What two things must always be equal when a chemical system is at equilibrium?

Answer: Rf and Rr must be equal at equilibrium. That is the definition of (dynamic) equilibrium: Rate forward Rate reverse.

(Note: It is also true that the value of Q K at equilibrium, but this question was designed with the rate idea in mind, since that is the definition of dynamic equilibrium.)

(d) Do you think that the value of the equilibrium constant for the reaction equation 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 ~33%. That means more of the reactants remain at equilibrium than reacted. Fewer products than reactants at equilibrium means “reactant favored” and a small value of K.

2. NT2. (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 equilibrium, 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 equation that represents the forward reaction that occurs in the system:

For: a A + b B c C + d D (A, B, C, and D are gases or solution species): ba

dc

Q[B][A]

[D][C]

Since the value of Q at equilibrium is always a constant, it is given a special name and symbol: the equilibrium constant, K, and so in mathematical form, the Law of Mass Action is simply:

Kbeqeq

eqeq

[B][A]

[D][C]a

dc

for a system at equilibrium (i.e., if [ ]’s are equilibrium ones)

Time

B

Time

Rf

Rate Concen-tration

(M)

9

3

A

C

Answer Key, Problem Set 3

PS3-2

(b) True or false: K changes as a reaction occurs, but Q does not. Give reasoning.

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 equilibrium, 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).

(c) 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 equilibrium is established, Q will equal K. (See part (a)).

(d) True or false: The equilibrium concentrations of reactants and products depend on their initial concentrations. Give reasoning.

True. If you start with different initial concentrations, then you (nearly always) end up with different final (equilibrium) concentrations because of stoichiometry (mass conservation). As a trivial example (just to demonstrate the idea), if you start with 3 M of A in a reaction where A

B and K 2, the final equilibrium state will be [A] 1 M and [B] 2 M. However, if you

started with 1 M of A, the equilibrium state will end up with [A] 1/3 M and [B] 2/3 M. Both

equilibrium states are characterized by the same value of Q ( K [2 in this case]), but the systems contain different final amounts because of the different starting amount(s).

(e) 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 equilibrium positions / states. Explain.

True. As noted above, it is the value of the reaction quotient that is constant for any given reaction system at equilibrium. But there are an infinite number of combinations of values that will satisfy the equation given in part (a) for any given chemical equation. For example, if K =

36 for a system in which [A][B]

[D][C]2

K , the following equations are all true (given a bit of

uncertainty in each concentration value) even though the values for [A], [B], [C], and [D] are different in each case:

etc. 0.253

41.6836

45

62036

1.7771

8136

11

1636

2222

Hopefully you can see that mathematically there are an infinite number of combinations of values that will satisfy a given law of mass action equation. Thus there are an infinite number of equilibrium “positions” or “states” for a given reaction system even though they are all characterized by the same value of K.

(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.

Q, being the reaction quotient, has the form: ba

dc

Q[B][A]

[D][C] . As such, if all concentrations are 1

M, you get 1c∙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!


PS3-3

NOTE: Q = K means “the system is at equilibrium” with whatever concentrations there are right now in the system. Those concentrations need not be (and generally are not) equal to one another nor equal to 1 M.

(g) True or false: If Q > K, there must be 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 equilibrium. Q > K basically just means that there are more products than there could be at equilibrium 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 equation whose K value equals 5.6 x 10-8, and a reactive system having a value of Q equal to 1.4 x 10-5. Clearly, in this case, Q > K, but in the system when Q = 1.4 x 10-5, there are not more products than reactants—there are more reactants than products!

* This isn’t strictly true for all reaction equations for math reasons I don’t want to get into now. It is

strictly true only for an equation such as A B, but only “mostly true” for most chemical reaction

equations 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

equilibrium. K will depend on the reaction type (nature of the reaction) and reaction equation (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 equilibrium”.* This is the case when the equilibrium 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 equilibrium”. Reverse reaction (net) will occur as this system reaches equilibrium. This is the case when Q is larger than the equilibrium constant (Q > K). It is not just an issue of the value of Q; the value of K also matters here—it’s Q relative to K that determines whether a given system is at equilibrium, “before” equilibrium, or “past” equilibrium.

Note: No matter what the K is for a given reaction/balanced equation, 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 equilibrium. It only represents a relative

comparison between the two states.

* See previous problem for the qualification associated with this slight overgeneralization.

(i) True or false: Reactions with large K values are fast. Explain.

False. It may be fast or it may be slow. A large K indicates only that once equilibrium 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 equilibrium). It is the rate constant (kf) that is proportional to rate, not K. It is easily possible to have a reaction


PS3-4

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,

r

f

k

kK . You can see from this equation that if kf is,

say, 10-10 and kr is 10-18, K will be 108. 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).

3. NT3. Consider a reaction represented by the following equation: A(g) B(g)

In each scenario (a-c) below, state whether: (i) the reaction is product favored or reactant favored; (ii) the reaction mixture is product heavy (too many products to be at equilibrium) or reactant heavy (too few products to be at equilibrium); (iii) the reaction mixture has more products than reactants or more reactants than products; and (iv) forward or reverse reaction will occur (net) as the system reaches equilibrium. Provide brief reasoning. Hopefully in doing this problem you will clarify the differences of each of these “conditions”!

(a) Q = 5 x 109; K = 9 x 107 (b) Q = 3 x 10-5; K = 2 x 10-8 (c) Q = 4 x 10-7; K = 6 x 10-4

Answers:

(a) (i) Rxn is product favored, since K > 1.

(ii) Mixture is “product heavy”, since Q > K.

(iii) Mixture has more products than reactants, since Q > 1.

(iv) Reverse reaction will occur, since Q > K.

NOTE: Just because a reaction is “product favored” doesn’t mean that forward reaction has to

occur!

(b) (i) Rxn is reactant favored, since K < 1.

(ii) Mixture is “product heavy”, since Q > K.

(iii) Mixture has more reactants than products, since Q < 1.

(iv) Reverse reaction will occur, since Q > K.

NOTE: Just because there are more reactants than products doesn’t mean that forward reaction

has to occur!

(c) (i) Rxn is reactant favored, since K < 1.

(ii) Mixture is “reactant heavy”, since Q < K.

(iii) Mixture has more reactants than products, since Q < 1.

(iv) Forward reaction will occur, since Q < K.

NOTE: Just because a reaction is reactant favored doesn’t mean that reverse reaction has to

occur! Also, the reason that forward reaction occurs here is NOT because there are “more

reactants than products” (Q < 1)—it’s because Q < K. (See (b) above)

Strategy/Reasoning

Key Concepts:

1) Recall that Q and K both have “products in the numerator” and “reactants in the denominator”.

2) Q reflects the composition of the reaction mixture “now” (whatever the concentrations are in the

mixture, regardless of whether the system is at equilibrium), while K reflects the composition of the

reaction mixture (only) at equilibrium.

Based on the above, as well as the concept of “equilibrium”, the following three questions/comparisons

are distinct:

A) Is Q greater than or less than 1?

See Stratgy/Reasoning below for full reasoning/explanation. Please do not memorize what is to the left without understand why it is so. These quantities “mean” something!


PS3-5

This indicates whether there are more products than reactants (numerator > denominator) or

whether there are fewer products than reactants “right now” (whatever state the system is in)

Q > 1 means “more products than reactants” (right now)

Q < 1 means “fewer products than reactants” (right now)

B) Is K greater than or less than 1?

This indicates whether there are more products than reactants (numerator > denominator) or

whether there are fewer products than reactants at equilibrium (regardless of whether the

system is at equilibrium “right now” or not).

K > 1 means “more products than reactants” at equilibrium (and thus “product favored”)

K < 1 means “fewer products than reactants” at equilibrium (and thus “reactant favored”)

C) Is Q greater than or less than K?

This indicates whether or not the system as it is right now has more products relative to

reactants than it would have if the system were at equilibrium, or fewer products relative to

reactants that it would have if the system were at equilibrium. Only this comparison

indicates whether forward or reverse reaction will occur (net) as the system

approaches/reaches equilibrium!

Q > K means “too many products right now (compared to R’s) for the system to be at

equilibrium” (and thus “product heavy”)

Q > K means “too few products right now (compared to R’s) for the system to be at

equilibrium” (and thus “reactant heavy”)

Technical comment: Although the interpretations in A and B above are technically only “absolutely true” for certain

stoichiometries (e.g., A B; A + B C + D), we use the relationships generally at this level (where they are generally true, but not “absolutely” true). The relationship in C, however, is technically true no mater what the stoichiometry of the reaction equation.

Writing Equilibrium Constant Expressions

[All in Mastering]

The Meaning of K

4. 16.4. What is the significance of the equilibrium constant? What does a large equilibrium constant tell us about a

reaction? A small one?

Answers: The (value of the) equilibrium constant reflects the tendency of a reaction to occur in the “forward direction” (where “forward” is indicated by the way the balanced chemical equation is written). This is because the equilibrium constant’s value reflects the composition of any equilibrium state in terms of products relative to reactants. For example, a large K indicates that once equilibrium is established, there will be a great amount of products relative to reactants, which means the reaction tends to “favor products at equilibrium” and thus has a great tendency to occur in the forward direction (as written). A small K indicates that once equilibrium is established, there will be a small amount of products relative to reactants, which means the reaction tends to “favor reactants at equilibrium” and thus has very little tendency to occur in the forward direction (as written).


PS3-6

5. 16.23 When [a] reaction [represented by the following chemical equation] comes to equilibrium, will the

concentrations of the reactants or products be greater? Does the answer to this question depend on the initial concentrations of the reactants and products?

A(g) + B(g) 2 C(g) Kc = 1.4 x 10-5

Answers: Reactants (will be greater). The equilibrium constant is fairly small (<< 1), meaning the numerator (with [C]2 in it) will be very small compared to the denominator (which has [A][B] in it). Thus, “reactant favored (at equilibrium)”. The initial concentrations do not matter because Q = K once equilibrium is established. Thus, even though the specific values may differ, in any equilibrium state, the value of [C]2 will equal 1.4 x 10-5 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]2, 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].)

Equilibrium Problems (basically four “types”)

6. 16.35 Consider the reaction [equation!]:

CO(g) + 2 H2(g) CH3OH(g)

An equilibrium mixture of this reaction at a certain temperature was found to have [CO] = 0.105 M, [H2] = 0.114 M, and [CH3OH] = 0.185 M. What is the value of the equilibrium constant (Kc) at this temperature?

Answer: 136 or 1.36 x 102

Reasoning / Work:

This is a “plug in” type of equilibrium problem. The concentrations given in the problem are clearly denoted as being equilibrium concentrations. So just write the equilibrium constant expression and plug in the concentration values to get K:

210 x1.36 or.57.513

114)(0.105)(0.

0.185

][H[CO]

OH][CH2

2

3 136 .K

eqeq

eq

c 2


NH4HS(s) NH3(g) + H2S(g)

An equilibrium mixture of this reaction at a certain temperature was found to have [NH3] = 0.278 M and [H2S] = 0.355 M. What is the value of the equilibrium constant (Kc) at this temperature?

Answer: 0.0987 or 9.87 x 10-2

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:

2-10 x9.87 0.0987or9.60.098355)(0.278)(0.

S][H][NH 23 .K

eqeq

c1


2 NO(g) + Br2(g) 2 NOBr(g) ; Kp = 28.4 at 298 K

In a reaction mixture at equilibrium, the partial pressure of NO is 108 torr and that of Br2 is 126 torr. What is the partial pressure of NOBr in this mixture?

Answer: 0.308 atm or 234 torr

Reasoning / Work:

This is another “plug in” type of problem, but with a couple of twists. Two of the three equilibrium partial pressures are given, along with Kp, so the third should be calculable by simply plugging


PS3-7

into the equilibrium 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:

1..atm20.14torr 760

atm 1 x torr 108NO P ; 7..atm50.16

torr 760

atm 1 x torr 126

2Br P

2..00.095 7)5(0.161)24(28.4)(0.1 28.47)5(0.161)2(0.14

2

2 2

2

2

2

2

NOBrNOBr

BrNO

NOBrp P

P

PP

PK

torr 234atm 0.308 :torr) (in or 2..80.302..00.095 torr/atm 760 x atm 0.308 NOBrP


CO(g) + 2 H2(g) CH3OH(g)

A reaction mixture in a 5.19-L flask at a certain temperature contains 26.9 g CO and 2.34 g H2. At equilibrium, the flask contains 8.65 g CH3OH. Calculate the equilibrium constant (Kc) for the reaction [equation!] at this temperature.

Answer: 27.3

Strategy:

This is problem in which you are effectively given the initial concentrations and one equilibrium concentration, and you are asked to determine K. Thus, you must recognize that knowing one species’ initial and equilibrium 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 equation). Thus, the final equilibrium concentrations can be determined (we typically use an ICE table to help achieve these calculations). Then one can plug the [ ]eq’s into the equilibrium 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 1st semester Gen. Chem. ideas (g → mol with molar masses, then mol / L to get M).

Execution:

Initial reaction mixture:

CO M 0..50.18 CO 3..mol00.96CO g 16.00) (12.01

mol 1 x CO g 26.9 L 5.19

(initially)

22

2

2 H M 6..30.22 H 0..mol61.1H g 1.008) x (2

mol 1 x H g 2.34 L 5.19

(initially)

Since no CH3OH 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:

OHCH M 1..00.052 OHCH 9..mol90.26mol 1

x OHCH g 8.65 33

3

3

L 5.19

OHCH g 16.00) 12.01 1.008 x (4

Fill in what is known into an ICE table:

[CO] (M) [H2] (M) [CH3OH] (M)

Initial 0.1850 0.2236 0

Change

Equilibrium 0.05201

Determine the “Change” row values (from the right-most one and then using the coeff. ratios [stoich]):

NOTE: 1 atm = 760 torr (exactly), so 760 has no uncertainty (i.e., an infinite number of SFs).

(at equilibrium)

C = E – I (final – initial) =0.05201 – 0 = +0.05201


PS3-8

(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) [H2] (M) [CH3OH] (M)

Initial 0.1850 0.2236 0

Change -0.05201 -2(0.05201) +0.05201

Equilibrium 0.05201

Determine the remaining “Equilibrium” row values (from the I and C values)

[CO] (M) [H2] (M) [CH3OH] (M)

Initial 0.1850 0.2236 0

Change -0.05201 -2(0.05201) +0.05201

Equilibrium 0.1850 – 0.05201 = 0.1330

0.2236 – 2(0.05201) = 0.1196

0.05201

Substitute equilibrium values into K expression:

27.3 .Keqeq

eq

c 3.327.6)90)(0.113(0.13

100.052

][H[CO]

OH][CH2

2

3

2


NH4HS(s) NH3(g) + H2S(g)

At a certain temperature, Kc = 8.5 x 10-3. A reaction mixture at this temperature containing solid NH4HS has [NH3] = 0.166 M and [H2S] = 0.166 M. Will more of the solid form or will some of the existing solid decompose as equilibrium is reached?

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 equilibrium 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 equilibrium”) or “product deficient” (“too few products to be at equilibrium”).

.Qc 5.50.027166)(0.166)(0.S]][H[NH 23 which is greater than Kc (= 0.0085) [Given]

If Q (= “products / reactants”) is too large, then there are “too many products to be at equilibrium” and so reverse reaction will occur to establish equilibrium (“shifts left”).

more solid will form.

11. 16.50 Nitrogen dioxide dimerizes according to the reaction [equation!]:

2 NO2(g) N2O4(g) ; Kp = 6.7 at 298 K

A 2.25-L container contains 0.055 mol of NO2 and 0.082 mol of N2O4 at 298 K. Is the reaction [system] at equilibrium? If not, in what direction will the reaction proceed?

Answer: No, not at equilibrium. 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 equation 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 Kc from Kp and then use the concentrations to calculate Qc; that’s fine, but requires more memorization! I am pretty sure that you all know PV=nRT better than the equation that relates Kp to Kc!)

E = I + C (initial + change)


PS3-9

Execution:

RTV

nPnRTPV Thus:

77..atm90.5K 298 0.08206L 2.25

mol 0.055Kmol

atmL

RTV

nP

NO

NO2

2

12..atm90.8K 298 0.08206L 2.25

mol 0.082Kmol

atmL

RTV

nP

ON

ON42

42

.

P

PQ

NO

ON

p 9.42.0.5977

1290.8

22

2

42 which is less than Kp (= 6.7 [Given]) [so not at equilibrium]

Q is “too small” need more products (greater numerator) to reach equilibrium

forward reaction occurs

12. NT4. 16.52a&c(+d). Consider the reaction [equation!] and associated equilibrium constant:

a A(g) + b B(g) c C(g) Kc = 5.0

Find the equilibrium 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 = 2

(c) a = 2; b = 1; c = 1 (set up equation for x; don’t solve)

(d) a = 3; b = 2; c = 5 (set up equation for x; don’t solve)

Answers*:

(a) [A]eq = 0.47 M; [B]eq = 0.47 M; [C]eq = 1.06 M

(c)

5.0121

xx

xKc 2

*

(d)

5.0

2 13 1

54

23

5

xx

xKc *

* 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 equilibrium in all cases, those answers would be the same no matter how x was defined.

Strategy:

I will essentially follow the strategy outlined in Tro (starting on p. 703 and shown in Examples 16.9 and 16.10) and demonstrated in class. That is:

1) Write the balanced equation and the equilibrium constant expression

2) Calculate, if necessary, initial concentrations, and put into an ICE table (with concentrations, in M).

3) Calculate Q and compare to K to determine which reaction (forward or reverse) occurs to get to equilibrium. (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 equilibrium. 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 “2x” or a “3x”, 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.

I ignored SFs here for convenience (since we’re not solving for x anyway)


PS3-10

5) Utilize the fact that the changes in species’ concentrations while reaction is taking place are in the ratio of the coefficients in the equation 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 equal the initial ones (I) plus the changes (C) to get expressions with x’s in them for the equilibrium concentrations (E) in the table.

[Remember, C is a “". = f – i, thus C = E – I, which means E = I + C.]

7) Looking at the equilibrium constant expression equation (that you should have written earlier, although you didn’t really need to do it earlier), which would be of the general form

beqeq

eqeqK

[B][A]

[D][C]a

dc

[assuming two reactants and two products], substitute the algebraic equilibrium

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 equation (i.e., don’t leave the symbol “K” in there!)

8) If asked to, solve the math equation just generated for x by appropriate means.

9) Substitute the values of x obtained back into the algebraic expressions for the species at equilibrium (E row in the ICE table) and pick the solution that gives “physically meaningful” results (e.g., you can’t have a negative equilibrium concentration!).

10) Check your results by using the numerical values of the equilibrium 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) 2 C(g) eqeq

eqK

[B][A]

[C]2

5.0

Initial concentrations are given as 1.0 M, 1.0 M, and 0 M. Thus:

[A] (M) [B] (M) [C] (M)

Initial 1.0 1.0 0

Change

Equilibrium

Since [C]0 = 0, Qc = 0, which means Q < K and forward reaction occurs to reach equilibrium.

Let x = [A] that is lost as equilibrium is established (the coefficient of A is 1, so this is convenient but not necessary)

Thus, [B] lost also equals x, and the [C] formed is 2x. Thus (recognizing that the changes in reactants’ concentrations is negative since they are being lost) we can write:

[A] (M) [B] (M) [C] (M)

Initial 1.0 1.0 0

Change - x - x + 2x

Equilibrium

Complete the table as follows (E = I + C):

[A] (M) [B] (M) [C] (M)

Initial 1.0 1.0 0

Change - x - x + 2x

Equilibrium 1.0 – x 1.0 – x 0 + 2x = 2x

Substitute in:


PS3-11

5.01.01.0

2

[B][A]

[C]2

xx

xK

eqeq

eq2

Although one could multiply this all out, combine terms, and use the quadratic equation, if you recognize that the left side is a perfect square, taking the square root of both sides is an easier option:

..x

x

x

x

x

x

xx

x3622.

1.0

2 5.0

1.0

2 5.0

1.0

2

1.01.0

2

2

2

2

22

(see note below*)

xxx 3622.-3622.1.03622.2

3622.2.236x 2 x

0.537..20.53624.

3622. 3622.36x24. x

Calculate the equilibrium concentrations using x = 0.53 M:

[A]eq = [B]eq = 1.0 – 0.53 = 0.47 M

[C]eq = 2(0.53) = 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)]:

5.0 205.370.4370.4

601.

[B][A]

[C]2

2

eqeq

eqK Good.

Execution of Strategy (part (c)):

2 A(g) + B(g) C(g) eqeq

eqK

[B][A]

[C]2

5.0

Initial concentrations are given as 1.0 M, 1.0 M, and 0 M. Thus:

[A] (M) [B] (M) [C] (M)

Initial 1.0 1.0 0

Change

Equilibrium

Since [C]0 = 0, Qc = 0, which means Q < K and forward reaction occurs to reach equilibrium.

Let x = [B] that is lost as equilibrium is established (the coefficient of B is 1, so this is convenient but not necessary)

Note that [A] lost here (unlike in part (a)) equals 2x because of the new stoichiometry [coefficient of 2 for A and 1 for B here], and the [C] formed is x. Thus:

[A] (M) [B] (M) [C] (M)

Initial 1.0 1.0 0

Change - 2x - x + x

Equilibrium

Completing the table yields:

[A] (M) [B] (M) [C] (M)

Initial 1.0 1.0 0

Change - 2x - x + x

Equilibrium 1.0 – 2x 1.0 – x 0 + x = x

* Technically, this should be 2.236 and I should solve for both x values to make sure I get the correct one. But the -2.236 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.

(Technically, this should be 0.5 due to SFs, but I’m going to leave 2 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.)


PS3-12

Substituting in (be very careful to really substitute in—note

the changes in the position of the “square”!):

5.0

1.021.0

[B][A]

[C]2

xx

xK

eqeq

eq

2

Execution of Strategy (part (d)):

3 A(g) + 2 B(g) 5 C(g) 2

eqeq

eqK

[B][A]

[C]3

5

5.0

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:

[A] (M) [B] (M) [C] (M)

Initial 1 1 4

Change

Equilibrium

102411

4

[B][A]

[C]3

5

23

5

2Q , which means Q > K and reverse reaction occurs to reach equilibrium.

Since reverse reaction occurs, we know that A and B will be formed (not used up) as equilibrium is reached. Thus…

Let 2x = [B] that is formed as equilibrium is established (NOTE: There is no species with a coefficient of 1 here,

so instead of defining an “x”, I decided to define a “2x”. 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 2x, the loss of A must just be 3x and the formation of C must be 5x. Thus:

[A] (M) [B] (M) [C] (M)

Initial 1 1 4

Change + 3x + 2x - 5x

Equilibrium

Completing the table yields:

[A] (M) [B] (M) [C] (M)

Initial 1.0 1.0 4

Change + 3x + 2x - 5x

Equilibrium 1.0 + 3x 1.0 + 2x 4 – 5x

Substituting in (be very careful to really substitute in—use parentheses and correctly place each exponent):

5.02 13 1

54

[B][A]

[C]3

5

23

5

2xx

xK

eqeq

eq

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 equilibrium, you may have defined x

to be the [B] that reacted. If so, you would have gotten:

5.02131

54

23

5

xx

x (and x would have been a negative quantity)

Since they asked you not to solve for x, you can leave the equation like this. I can see no benefit to getting the equation into “standard form” unless you were going to use the quadratic formula. You can better estimate x using this form of the equation, frankly. In a couple

of moments, I realized that x = 0.5 would give K = (zero in the denominator), so I tried x = 0.4, which gave me Q = 16.6. So clearly, x must be a bit less than 0.4. Not bad for a one-minute estimate!

For those interested, using an excel spreadsheet, I was able to quickly

estimate x to be 0.3062 (to 4 SF). Thus, [C]eq 2.469 M; [A]eq 1.919 M;

and [B]eq 1.612 M. Excel may not be “elegant”, but it is very practical!


PS3-13

Answer: 4.75 x 10-4

2) If you had let x = [B] that formed, your answer would have been:

5.0

11

4

3

5

2

3

5

2

2

xx

x

3) If you had let x = [A] that formed, your answer would have been:

5.0

11

4

2

5

23

5

3

3

xx

x

Finding K's For an Equation from Other K's (For Other Related Equations)

13. NT5. State in words what happens to the value of K when a reaction equation is reversed, and rationalize why this is so

conceptually. In other words, why does it make sense that if a reaction (equation) has a really large K, the reverse reaction (equation) will have a really small K?

Answer: When a reaction equation is reversed, the equilibrium constant is the reciprocal of the original (forward) reaction equation.

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 equilibrium 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 equation 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 equilibrium. Since the real world obviously does not change just because someone chooses to write the equation representing the reaction in a different way, it makes sense that the K value for the equation 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 equation was written in the original way) are now favored at equilibrium. An example may make this clearer. If C is the product of a reaction whose equation has a large K, then C is present at equilibrium more than the reactant species. If the equation is written in reverse, the real world should not change one bit, so C still will still present more than the other species at equilibrium. But since the equation is now written in reverse, C becomes a reactant, and so the reaction now should be “reactant favored” at equilibrium, 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.

14. 16.30 Use the reactions [equations!] below and their equilibrium constants to predict [calculate!] the equilibrium

constant for the reaction [equation!], 2 A(s) 3 D(g)

A(s) ½ B(g) + C(g) K1 = 0.0334

3 D(g) B(g) + 2 C(g) K2 = 2.35

Strategy:

1) Figure out a way to manipulate the given equations in such a way as to get two equations that will sum up to the “target” equation (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, PS03 folder) to help you with this kind of manipulation of “given equations”.**

2) Determine the equilibrium constant values for the “new” equations from Step 1 using the relation-ships discussed in class, on the yellow handout, and in the text (and in prior problems on this set).

3) Take the product of the two “new” equations’ 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 equation, multiply equation (1) [which is the only equation that has A in it] by 2 to get:

2 A(s) B(g) + 2 C(g) K1’ K1 x 2 K1

2 = 0.03342 ( 1.115..x 10-3)


PS3-14

Noting that one needs three D’s as products in the target equation, and that equation (2) has three D’s as reactants, reverse equation (2) to get:

B(g) + 2 C(g) 3 D(g) K2’ K-2 K2-1 2.35-1 ( 4.255..x 10-1)

Verifying that the sum of the above two equations does equal the target equation (the B and 2 C parts will cancel out leaving 2 A(s) 3 D(g) as desired), the K for the sum should equal the product of the two “new” K’s:

K1’+2’ K1’K2’ 0.03342 x 2.35-1 4.747..x 10-4 4.75 x 10-4

answer key, problem set 3 (full explanations and work)

Documents