hw04 - equil 1-solutions

kumar (kk24268) – HW04 - Equil 1 – mccord – (51580) 1

This print-out should have 14 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 points

An equilibrium in which processes occur con-tinuously, with NO NET change, is called

1. heterogeneous equilibrium.

2. static equilibrium.

3. homogeneous equilibrium.

4. dynamic equilibrium. correct

Explanation:

For a system at dynamic equilibrium, al-though the concentrations of the componentsdo not change, the processes continue to oc-cur in the foward and reverse directions at thesame rate.

002 10.0 points

Explain why equilibium constants are dimen-sionless.

1. They are not really dimensionless butwe must treat them as such in order to beable to take lnK in the expression ∆G0 =−RT lnK.

2. Every concentration or pressure that en-ters intoKc orKp is really divided by the cor-responding concentration or pressure of thesubstance in its standard state. correct

3. They are dimensionless because the pres-sures or concentrations we put in are all forthe substances in their standard states.

4. The statement is not true. Equilibriumconstants have units that involve some multi-ple of atmospheres or moles per liter.

5. They are dimensionless because concen-trations and pressures have no units.

Explanation:

The amount of each component is interms of activity, which is the measuredamount (concentration, pressure) divided bythe amount of that component in its standardstate in that unit. The units in the numera-tor and denominator are identical and cancelout.

003 10.0 points

The expression forKc for the reaction at equi-librium is4 NH3(g) + 5 O2(g) ⇀↽ 4 NO(g) + 6 H2O(g)

1.[NH3]

4 [O2]5

[NO]4 [H2O]6

2. [NH3]4 [O2]

5

3.[NO]4 [H2O]6

[NH3]4 [O2]5correct

4. [NO]4 [H2O ]6

Explanation:

The equation must be written with the ap-propriate formula and correctly balanced. Kc

is the equilibrium constant for species in so-lution and equals the mathematical productof the concentrations of the chemical prod-ucts, divided by the mathematical product ofthe concentrations of the chemical reactants.In this mathematical expression, each concen-tration is raised to the power of its coefficientin the balanced equation. For Kc the molarconcentrations are used for the activities ofthe components.

004 10.0 points

Consider the following reactions at 25◦C:

reaction Kc

2 NO(g) ⇀↽ N2(g) + O2(g) 1× 1030

2 H2O(g) ⇀↽ 2 H2(g) + O2(g) 5× 10−82

2 CO(g) + O2(g) ⇀↽ 2 CO2(g) 3× 1091

Which compound is most likely to dissociateand give O2(g) at 25

◦C?

1. CO2

2. H2O


3. NO correct

4. CO

Explanation:

Only two are dissociation reactions: disso-ciation of NO and dissociation of H2O. Kc isgreater for dissociation of NO.

005 10.0 points

At 600◦C, the equilibrium constant for thereaction

2HgO(s) → 2Hg(ℓ) + O2(g)

is 2.8. Calculate the equilibrium constant forthe reaction

1

2O2(g) + Hg(ℓ) → HgO(s) .

1. −1.7

2. 0.36

3. 1.7

4. 0.60 correct

5. 1.1

Explanation:

006 10.0 points

If Kc = 2.78× 105 for the reaction

A(g) ⇀↽ 2B(g) ,

what is Kc for the reaction written as

2B(g) ⇀↽ A(g) ?

Correct answer: 3.59712× 10−6.

Explanation:

Kc, ini = 2.78× 105

A(g) ⇀↽ 2B(g) , Kc =[B]2

[A]= 2.78× 105

2B(g) ⇀↽ A(g)

K−1c =

[A]

[B]2=

1

Kc=

1

2.78× 105

= 3.59712× 10−6

007 10.0 points

Calculate the equilibrium constant at 25◦C fora reaction for which ∆G0 = −2.92 kcal/mol.

1. −138.228

2. 138.228 correct

3. 1382.28

4. 69.1138

5. 276.455

Explanation:

T = 25◦C+ 273 = 298 K∆G0 = −2920 cal/molAt equilibrium

∆G0 = −RT lnK

−2920 = (−1.987 cal/mol ·K)

× (298.15 K)(lnK)

K = 138.228

008 10.0 points

The reaction

A + B ⇀↽ C+ 2D

has an equilibrium constant of 3.7 × 10−3.Consider a reaction mixture with

[A] = 2.0× 10−2 M

[B] = 1.7× 10−4 M

[C] = 2.4× 10−6 M

[D] = 3.5× 10−3 M

Which of the following statements is defi-nitely true?

1. Heat will be evolved.

2. The system is at equilibrium.

3. The reverse reaction can occur to a


greater extent than the forward reaction untilequilibrium is established.

4. The forward reaction can occur to agreater extent than the reverse reaction untilequilibrium is established. correct

5. No conclusions about the system can bemade without additional information.

Explanation:

Q =[C] [D]2

[A] [B]=

(2.4× 10−6 M) (0.0035 M)2

(0.02 M) (0.00017 M)

= 8.64706× 10−6

SinceQ < K the foward reaction is favored.

009 10.0 points

The reaction

N2(g) + 3H2(g) ⇀↽ 2NH3(g) ,

has an equilibrium constant of 4.0 × 108 at25◦C. What will eventually happen if 44.0moles of NH3, 0.452 moles of N2, and 0.108moles of H2 are put in a 10.0 liter container at25◦C?

1. Nothing; the system is at equilibrium.

2. More N2 and H2 will be formed.

3. More NH3 will be formed. correct

Explanation:

K = 4.0× 108 [NH3] =44.0 mol

10 L

[N2] =0.452 mol

10 L[H2] =

0.108 mol

10 L

Q =[NH3]

2

[N2] [H2]3

=(4.40 M)2

(0.0452 M) (0.0108 M)3

= 3.4× 108

Since Q < K equilibrium will shift to theright, forming more NH3.

010 10.0 points

A 10.0 L vessel contains 0.0015 mole CO2 and0.10 mole CO. If a small amount of carbon isadded to this vessel and the temperature israised to 1000◦C

CO2(g) + C(s) ⇀↽ 2CO(g) ,

will more CO form? The value of Kc for thisreaction is 1.17 at 1000◦C. Assume that thevolume of the gas in the vessel is 10.0 L.

1. No, the rate of the reverse reaction willincrease to produce more CO2.

2. Yes, the rate of the forward reaction willincrease to produce more CO. correct

3. Unable to determine this from the dataprovided.

Explanation:

[CO] =0.1 mol

10 L[CO2] =

0.0015 mol

10 LCarbon, being a solid, has no effect on equi-

librium.

[Q] =[CO]2

[CO2]=

(

0.1

10.0M

)2

(

0.0015

10.0M

)

= 0.666667 < Kc = 1.17

Therefore equilibrium will shift to the right.

011 10.0 points

Kc = 2.6× 108 at 825 K for the reaction

2H2(g) + S2(g) ⇀↽ 2H2S(g)

The equilibrium concentration of H2 is 0.0020M and that of S2 is 0.0010 M. What is theequilibrium concentration of H2S?

1. 10 M

2. 1.02 M correct

3. 0.10 M

4. 0.0010 M


Explanation:

Kc = 2.6× 108 [H2]eq = 0.0020 M[S2]eq = 0.0010 M

2H2(g) + S2 ⇀↽ 2H2S

Kc =[H2S]

2

[H2]2 [S2]

[H2S] =√

Kc [H2]2 [S2]

=√

(2.6× 108) (0.0020 M)2 (0.0010 M)

= 1.0 M

012 10.0 points

A mixture of PCl5(g) and Cl2(g) is placedinto a closed container. At equilibrium it isfound that [PCl5] = 0.7 M, [Cl2] = 0.41 Mand [PCl3] = 0.1 M.

PCl5 ⇀↽ PCl3 +Cl2

What is the value of Kc for the reaction?

1. 0.0292857

2. 0.0585714 correct

3. 0.117143

4. 0.175714

5. 100

Explanation:

[PCl5] = 0.7 M [Cl2] = 0.41 M[PCl3] = 0.1 M

Kc =[Cl2] [PCl3]

[PCl5]=

(0.41 M)(0.1 M)

0.7 M

= 0.0585714 M

013 10.0 points

At 1000 K the equilibrium pressure of thethree gases in one mixture

2 SO2(g) + O2(g) ⇀↽ 2 SO3(g)

were found to be 0.562 atm SO2, 0.101 atmO2, and 0.332 atm SO3. Calculate the valueof Kp for the reaction as written.

1. 0.171

2. 0.289

3. 3.46 correct

4. 2.64

5. 5.83

Explanation:

PSO3= 0.332 atm PSO2

= 0.562 atmPO2

= 0.101 atm

Kp =P 2SO3

P 2SO2

· PO2

=(0.332)2

(0.562)2(0.101)= 3.46

014 10.0 points

Consider the reaction

2HgO(s) ⇀↽ 2Hg(ℓ) + O2(g) .

What is the form of the equilibrium constantKc for the reaction?

1. Kc =[Hg]2 [O2]

[HgO]2

2. Kc = [Hg]2 [O2]

3. Kc =[O2]

[HgO]2

4. None of the other answers is correct.

5. Kc = [O2] correct

Explanation:

Solids and liquids are not included in theKexpression.

hw04 - equil 1-solutions

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