hw05 - equil 2-solutions

kumar (kk24268) – HW05 - Equil 2 – mccord – (51580) 1

This print-out should have 15 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 points

A 2.000 liter vessel is filled with 4.000 molesof SO3 and 6.000 moles of O2. When thereaction

2 SO3(gas) ⇀↽ 2 SO2(gas) + O2(gas)

comes to equilibrium a measurement showsthat only 1.000 mole of SO3 remains. Howmany moles of O2 are in the vessel at equilib-rium?

1. None of these is correct.

2. 7.500 mol correct

3. 12.000 mol

4. 7.000 mol

5. 3.750 mol

Explanation:

Initially,

[SO3] =4 mol

2 L= 2 M [O2] =

6 mol

2 L= 3 M

2SO3 (g) ⇀↽ 2 SO2 (g) + O2 (g)ini, M 2 0 3∆, M −2x 2x xeq, M 2− 2x 2x 3 + x

At equilibrium,

[SO3]eq =1 mol

2 L= 0.5 M, so

2− 2x = 0.5

−2x = −1.5

x = 0.75

Thus

[O2] = 3 + x = 3.75 M

mol O2 = (3.75 mol/L) (2 L) = 7.5 mol .

002 10.0 points

Suppose the reaction

A ⇀↽ B

has an equilibrium constant of 1.0 and the ini-tial concentrations of A and B are 0.5 M and0.0 M, respectively. Which of the following isthe correct value for the final concentration ofA?

1. 0.500 M

2. None of these is correct.

3. 1.50 M

4. 1.00 M

5. 0.250 M correct

Explanation:

K = 1.0 [A]ini = 0.5 M[B]ini = 0 M

A ⇀↽ Bini, M 0.5 0.0∆, M −x xeq, M 0.5− x x

K =[B]

[A]= 1.0

x

0.5− x= 1.0

x = 0.25 M

[A] = 0.5− x = 0.25 M

003 10.0 points

Suppose the reaction

H2(g) + I2(g) ⇀↽ 2HI(g)

has an equilibrium constant Kc = 49 and theinitial concentration of H2 and I2 is 0.5 Mand HI is 0.0 M. Which of the following isthe correct value for the final concentration ofHI(g)?

1. 0.389 M


2. 0.778 M correct

3. 0.599 M

4. 0.250 M

5. 0.219 M

Explanation:

Kc = 49 [H2]ini = 0.5 M[I2]ini = 0.5 M [HI]ini = 0 M

H2(g) + I2(g) ⇀↽ 2HI(g)Ini, M 0.5 0.5 0∆, M −x −x +2 x

Equil, M 0.5− x 0.5− x 2 x

Kc =[HI]2

[H2] [I2]

49 =(2x)2

(0.5− x)2

7 =2x

0.5− x7(0.5− x) = 2x

3.5− 7x = 2x

3.5 = 9x

x =3.5

9= 0.389 M

Looking back at our equilibrium values, wesee that the final concentration of HI is equalto 2x, so 2(0.389) = 0.778 M.

004 10.0 points

At T = 900◦C, Kc = 225 for the gas-phasereaction

A + B ⇀↽ C+D

Starting with 2.54 moles each of A and B ina 5.00 liter container, what will be the equilib-rium concentration of C at this temperature?

Correct answer: 0.47625 M.

Explanation:

[A] =2.54 mol

5 L= 0.508 M T = 900◦C

[B] =2.54 mol

5 L= 0.508 M Kc = 225

A + B ⇀↽ C + Dini, M 0.508 0.508 0 0∆, M −x −x x xeq, M 0.508− x 0.508− x x x

[C][D]

[A][B]= 225

x2

(0.508− x)2= 225

x

0.508− x= 15

x = 7.62− 15x

x = [C] = 0.47625 M

005 10.0 points

The system

H2(g) + I2(g) ⇀↽ 2HI(g)

is at equilibrium at a fixed temperature witha partial pressure of H2 of 0.200 atm, a partialpressure of I2 of 0.200 atm, and a partial pres-sure of HI of 0.100 atm. An additional 0.24atm pressure of HI is admitted to the con-tainer, and it is allowed to come to equilib-rium again. What is the new partial pressureof HI?

Correct answer: 0.148 atm.

Explanation:

PI2 = PH2= 0.2 atm PHI = 0.1 atm

H2(g) + I2(g) ⇀↽ 2HI(g)

Kp =(PHI)

2

PH2· PI2

=(0.1 atm)2

(0.2 atm) (0.2 atm)= 0.25

new PHI = (0.1 + 0.24) atm = 0.34 atm

Adding the products shifts the equilibrium tothe left.


H2(g) + I2(g) ⇀↽ 2HI(g)ini, atm 0.200 0.200 0.34∆, atm +x +x −2xeq, atm 0.200 + x 0.200 + x 0.34− 2x

(0.34− 2 x)2

(0.2 + x)2= 0.25

0.34− 2 x

0.2 + x=

√0.25

0.34− 2 x = (0.5)(0.2) + 0.5 x

2.5 x = 0.24

x = 0.096

PHI = 0.34 atm− (2) (0.096 atm) = 0.148 atm

006 10.0 points

At 990◦C, Kc = 1.6 for the reaction

H2(g) + CO2(g) ⇀↽ H2O(g) + CO(g)

How many moles of H2O(g) are presentin an equilibrium mixture resulting from theaddition of 1.00 mole of H2, 2.00 moles ofCO2, 0.75 moles of H2O, and 1.00 mole of COto a 5.00 liter container at 990◦C?

1. 1.7 mol

2. 1.1 mol correct

3. 1.4 mol

4. 1.0 mol

5. 0.60 mol

6. 0.80 mol

Explanation:

Kc = 1.6

[H2] =1.00 mol

5 L[CO2] =

2.00 mol

5 L

[H2O] =0.75 mol

5 L[CO] =

1.00 mol

5 L

Q =[H2O][CO]

[H2][CO2]=

(

0.75mol

5 L

)(

1mol

5 L

)

(

1mol

5 L

)(

2mol

5 L

)

= 0.375 < Kc = 1.6

Therefore equilibrium moves to the right.H2(g) + CO2(g) ⇀↽ H2O(g)+ CO(g)0.2 0.4 0.15 0.2−x −x x x

0.2− x 0.4− x 0.15 + x 0.2 + x

(0.15 + x)(0.2 + x)

(0.2− x)(0.4− x)= 1.6

x2 + 0.35x+ 0.03

x2 − 0.6x+ 0.08= 1.6

x2 + 0.35x+ 0.03 = 1.6x2 − 0.96x+ 0.128

0.6x2 − 1.31x+ 0.098 = 0

x =1.31±

√

(1.31)2 − 4(0.6)(0.098)

1.2= 7.756× 10−2 M

molH2O = 5.00 L

×(

0.15 + 7.75× 10−2) mol

L= 1.1molH2O

007 10.0 points

What happens to the concentration ofNO(g) when the total pressure on the equilib-rium reaction3NO2(g) + H2O(ℓ) ⇀↽

2HNO3(aq) + NO(g)is increased (by compression)?

1. decreases

2. increases correct

3. Unable to determine

4. remains the same

Explanation:


Increasing the total pressure on the sys-tem by decreasing its volume will shift theequilibrium toward the side of the reactionwith fewer numbers of moles of gaseous com-ponents. If the total number of moles ofgas is the same on the product and reactantsides of the balanced chemical equation, thenchanging the pressure will have little or no ef-fect on the equilibrium distribution of speciespresent. The amount and concentration ofNO will increase.

008 10.0 points

Consider the reaction

2 SO2(g) + O2(g) ⇀↽ 2 SO3(g)

where ∆Hrxn = −198 kJ. The amount ofSO2(g) at equilibrium increases when

1. the volume is increased. correct

2. the temperature is decreased.

3. more oxygen is added.

4. SO3 is removed.

5. the pressure is increased.

Explanation:

2 SO2(g) + O2(g) ⇀↽ 2 SO2(g) + heat

According to Le Chatelier’s principle, theamount of reactant SO2(g) is increased whenthe equilibrium shifts to the left. This willhappen when another reactant (O2) is re-moved. For an exothermic reaction decreasingtemperature removes heat and sends equilib-rium to the right. Increasing pressure sendsthe equilibrium in the direction that has fewernumbers of moles of gas. Increasing the vol-ume is equivalent to decreasing gas pressure.

009 10.0 points

For an exothermic reaction, what would hap-pen to the numerical value of Kc, if we in-crease the temperature at constant pressure?

1. Kc would either increase or decrease, de-pending on the number of moles of gas in-volved.

2. Kc would either increase or decrease, de-pending on the concentrations.

3. Kc would increase.

4. Kc would decrease. correct

5. Kc would not change.

Explanation:

An exothermic reaction gives off heat asproducts are formed, so at a higher tem-perature the reverse endothermic reaction isfavoured, decreasing Kc, which reflects theratio of the products to reactants.

010 10.0 points

Suppose the reaction mixture

N2(g) + 3H2(g) ⇀↽ 2NH3(g)

is at equilibrium at a given temperature andpressure. The pressure is then increased atconstant temperature by compressing the re-action mixture, and the mixture is then al-lowed to reestablish equilibrium. At the newequilibrium,

1. there is more ammonia present than therewas originally. correct

2. there is the same amount of ammoniapresent as there was originally.

3. the nitrogen is used up completely.

4. there is less ammonia present than therewas originally.

Explanation:

LeChatelier’s Principle states that if achange occurs in a system at equilibruim,the system responds to relieve the stress andreach a new equilibrium. Here, the number ofmoles of gaseous reactants is greater than thenumber of moles of products. Increasing thepressure of the above system will result in the


reaction proceeding to reduce that pressureincrease. The system will shift to the right(the side that has fewer moles of gas), so thepressure will be reduced; thus more ammoniawill be produced.

011 10.0 points

Consider the system

2N2O5(g) ⇀↽ 2N2O4(g) + O2(g) + heat

at equilibrium at 25◦C. If the temperaturewere raised would the equilibrium be shiftedto produce more N2O5 or more N2O4?

1. There would be no effect.

2. more N2O4

3. more N2O5 correct

Explanation:

This is an exothermic reaction and so in-creasing temperature provides more heat tothe system, shifting equilibrium to the leftand producing more N2O5.

012 10.0 points

Given the reaction

2NH3(g) ⇀↽ N2(g) + 3H2(g)

at equilibrium, if the pressure is doubled(think of the volume of the container halving),in which direction will the reaction shift?

1. left correct

2. no change

3. right

Explanation:

Increasing pressure shifts the equilibriumin the direction that produces fewer moles ofgas.

013 10.0 points

Given Kp = 4.6 × 10−14 and ∆H0 = 115kJ/mol for the reaction

2Cl2(g) + 2H2O(g) ⇀↽ 4HCl(g) + O2(g)

at 25◦C, what is Kp at 400◦C?

1. 1.4× 10−5

2. 3.9× 10−4

3. 7.9× 10−2

4. 7.7× 10−3 correct

5. 1.3× 102

Explanation:

T1= 298.15 K T2= 673.15 KUse the van’t Hoff equation:

ln

(

K2

K1

)

=∆H0

rxn

R

(

1

T1−

1

T2

)

lnK2 − lnK1 =∆H0

rxn

R

(

1

T1−

1

T2

)

lnK2 =∆H0

rxn

R

(

1

T1−

1

T2

)

+ lnK1

=

(

115000 J/mol

8.314 J/mol ·K

)

×(

1

298.15 K−

1

673.15 K

)

+ ln(

4.6× 10−14)

= −4.86538

K2 = e−4.86538 = 0.0077089

014 10.0 points

For a certain reaction, K = 31 at 300 K andthe reaction is endothermic by 15.6 kJ/mol.What is K at 500 K?

Correct answer: 378.339.

Explanation:

T1 = 300 K K300 = 31T2 = 500 K ∆H = 15.6 kJ/molK500 = ?

lnK2

K1

= ln K2 − ln K1

=∆H

R

(

1

T1−

1

T2

)

lnK2 = lnK1 +∆H

R

(

1

T1−

1

T2

)


lnK500 = ln 31 +15600 J/mol

8.314 J/mol K

×(

1

300 K−

1

500 K

)

= 5.93579

K = e5.93579 = 378.339

015 10.0 points

The system

CO2(g) + H2(g) ⇀↽ H2O(g) + CO(g)

is at equilibrium at some temperature. Atequilibrium a 4.00 L vessel contains 1.00 moleCO2, 1.00 mole H2, 2.40 moles H2O, and 2.40moles CO. How many moles of CO2 must beadded to the system to bring the equilibriumCO concentration to 0.713 mol/L?

Correct answer: 2.029 moles.

Explanation:

V = 4.0 L nCO2= 1.0 mol

nH2= 1.0 mol nH2O = 2.40 mol

nCO = 2.40 molAt equilibrium, the concentrations are

[CO2] = [H2] =1 mol

4 L= 0.25 M

[H2O] = [CO] =2.4 mol

4 L= 0.6 M

First calculate the value of K:

K =[CO] [H2O]

[CO2] [H2]=

(0.6 M) (0.6 M)

(0.25 M) (0.25 M)= 5.76

Set up your equilibrium table knowing thatthe initial CO2 concentration is now higherthan 0.25 M. The reaction will go to the rightand you can get the following...

CO2 + H2 ⇀↽ H2O + CO? 0.25 0.6 0.6−x −x +x +x?− x 0.25− x 0.6 + x 0.6 + xThe final concentration of CO is 0.713, so

0.6 + x = 0.713

x = 0.713− 0.6 = 0.113

and [H2] = 0.25− x = 0.137 M.

We also now know all of the final concen-trations except CO2, which is calculated fromthe equilibrium expression

5.76 =(0.713 M) (0.713 M)

[CO2] (0.137 M)

[CO2] =(0.713 M) (0.713 M)

(5.76 M) (0.137 M)

so that [CO2]fin = 0.6442 mol/L, which corre-sponds to 0.6442 + 0.113 = 0.7572 for theinitial concentration of CO2 BEFORE theequilibrium occurred.That value corresponds to 3.029 moles of

CO2. Now substract the 1.00 mole of CO2

that was already there and you get 2.029moles of CO2 that had to be added.

hw05 - equil 2-solutions

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