hw02 - thermo 2-solutions

kumar (kk24268) – HW02 - Thermo 2 – mccord – (51580) 1

This print-out should have 25 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 points

Which of the following will best help deter-mine the direction of heat flow in a system?

1. temperature correct

2. pressure

3. internal energy

4. work

5. enthalpy

Explanation:

Heat flows spontaneously from a hotterbody to a colder body. In other words, heatflows from a body of high temperature to abody of lower temperature.

002 10.0 points

In the manufacture of nitric acid by the oxi-dation of ammonia, the first product is nitricoxide. The nitric oxide is then oxidized tonitrogen dioxide:

2NO(g) + O2(g) −→ 2NO2(g)

Calculate the standard reaction enthalpyfor the reaction above (as written) using thefollowing data:N2(g) + O2(g) −→ 2NO(g)

∆H◦ = 180.5 kJN2(g) + 2O2(g) −→ 2NO2(g)

∆H◦ = 66.4 kJ

1. −100.3 kJ/mol rxn

2. −690.72 kJ/mol rxn




6. −114.1 kJ/mol rxn correct


Explanation:

Using Hess’ Law and the given standard re-action enthalpies, the first reaction is reversedand added to the second:

2NO(g) −→ N2(g) + O2(g)

∆H◦ = −180.5 kJ

N2(g) + 2O2(g) −→ 2NO2(g)

∆H◦ = 66.4 kJ

2NO(g) + O2(g) −→ 2NO2(g)

∆H◦ = −114.1 kJ

003 10.0 points

What mass of ethanol (C2H5OH(ℓ)) must beburned to supply 500 kJ of heat? The stan-dard enthalpy of combustion of ethanol at 298K is −1368 kJ ·mol−1.

1. 126 g

2. 29.7 g

3. 16.8 g correct

4. 2.74 g

5. 10.9 g

Explanation:

500 kJ

1368 kJ/mol= 0.365497 mol of ethanol

(0.365497 mol)(46 g/mol)= 16.8129 g ethanol

004 10.0 points

Carbon monoxide reacts with oxygen to formcarbon dioxide by the following reaction:

2CO(g) + O2(g) → 2CO2(g)

∆H for this reaction is −135.28 kcal.


How much heat would be released if 12.0moles of carbon monoxide reacted with suffi-cient oxygen to produce carbon dioxide? Useonly the information provided in this ques-tion.

1. 811.68 kcal correct

2. 135.28 kcal

3. 270.56 kcal

4. 1623.36 kcal

5. 541.12 kcal

6. 405.84 kcal

Explanation:

∆H = −135.28 kcal nCO = 12.0 mol(

−135.28 kcal

1 mol rxn

)(1 mol rxn

2 mol CO

)

×(12 mol CO) = −811.68 kcal

so 811.68 kcal were released.

005 10.0 points

Burning 1 mol of methane in oxygen to formCO2(g) and H2O(g) produces 803 kJ of en-ergy. How much energy is produced when 3mol of methane is burned?

1. 803 kJ

2. 2,409 kJ correct

3. 1,606 kJ

4. 268 kJ

Explanation:

n1 = 1 mol q1 = 803 kJn2 = 3 mol

If we know howmuch heat is evolved when 1mole of methane is combusted, then we knowhow much heat would be evolved if 3 mol ofmethane were combusted:(

803 kJ

1 mol methane

)

(3 mol methane)

= 2409 kJ

006 10.0 points

Calculate the standard reaction enthalpy forthe reaction.

CH4(g) + H2O(g) → CO(g) + 3H2(g)

given2H2(g) + CO(g) → CH3OH(ℓ)

∆H◦ = −128.3 kJ ·mol−1

2CH4(g) + O2(g) → 2CH3OH(ℓ)∆H◦ = −328.1 kJ ·mol−1

2H2(g) + O2(g) → 2H2O(g)∆H◦ = −483.6 kJ ·mol−1

1. +206.1 kJ ·mol−1 correct

2. +216 kJ ·mol−1

3. +42.0 kJ ·mol−1

4. +155.5 kJ ·mol−1

5. +412.1 kJ ·mol−1

Explanation:

We need to reverse the first reaction, halvethe second, halve and reverse the third andadd the results:

CH3OH(ℓ) → 2H2(g) + CO(g)∆H = 128.3 kJ/mol

CH4(g) + 0.5O2(g) → CH3OH(ℓ)∆H = −164.05 kJ/mol

H2O(g) → H2(g) + 0.5O2(g)

∆H = +241.8 kJ/mol

CH4(g) + H2O(g) → CO(g) + 3H2(g)

∆H = 206.05 kJ/mol

007 10.0 points

The value of ∆H for the reaction

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(ℓ)

is −2220 kJ/mol rxn. How much heat isgiven off when 11.0 g of propane gas (C3H8)is burned at constant pressure?

1. 25.96 kJ


2. 1665.0 kJ

3. 2220.0 kJ

4. 22420.0 kJ

5. 555.0 kJ correct

6. 6660.0 kJ

7. 50.5 kJ

Explanation:

∆H = −2220 kJ/mol mC3H8= 11 g

q =(

−2220 kJ

mol rxn

)( 1 mol rxn

1 mol C3H8

)

×

(1 mol C3H8

44 g C3H8

)

(11 g C3H8)

= −555 kJ or 555 kJ released

008 10.0 points

Calculate the standard enthalpy change forthe reaction

2HCl(g) + F2(g) → 2HF(ℓ) + Cl2(g)

given

4HCl(g) + O2(g) → 2H2O(ℓ) + 2Cl2(g)

∆H0 = −202.4 kJ/mol rxn

1

2H2(g) +

1

2F2(g) → HF(ℓ)

∆H0 = −600.0 kJ/mol rxn

H2(g) +1

2O2(g) → H2O(ℓ)

∆H0 = −285.8 kJ/mol rxn

1. ∆H0 = −1015.4 kJ/mol rxn correct

2. ∆H0 = −516.6 kJ/mol rxn

3. ∆H0 = +1587.2 kJ/mol rxn

4. ∆H0 = −1088.2 kJ/mol rxn

5. ∆H0 = +1116.6 kJ/mol rxn

6. ∆H0 = +516.6 kJ/mol rxn

7. ∆H0 = −1116.6 kJ/mol rxn

8. ∆H0 = +1088.2 kJ/mol rxn

9. ∆H0 = +1015.4 kJ/mol rxn

10. ∆H0 = −1587.2 kJ/mol rxn

Explanation:

The first equation needs to be multiplied by1

2in order to get the equation we’re interested

in. Thus its ∆H0 is multiplied by1

2as well.

The second equation needs to be multipliedby two in order to get the equation we’reinterested in. We also multiply its ∆H0 bytwo.The third equation needs to be reversed, so

the sign of its ∆H0 should change.Then we add the equations to get the equa-

tion we’re interested in. We also add the ad-justed ∆H0 values to get the answer, −1015.4kJ/mol rxn.

009 10.0 points

Calculate the standard enthalpy of formationof bicyclo[1.1.0]butane

H

H

C

C

CH2H2C

given the standard enthalpies of formation of717 kJ ·mol−1 for C(g) and 218 kJ ·mol−1 forH(g) and the average bond enthalpies of 412kJ ·mol−1 for C H and 348 kJ ·mol−1 forC C.

1. +175 kJ ·mol−1


2. −472 kJ ·mol−1

3. +312 kJ ·mol−1

4. −124 kJ ·mol−1

5. −36 kJ ·mol−1 correct

Explanation:

We can write an equation in which we com-pletely decompose bicyclo [1,1,0] butane:C4H6(bicyclobutane, g) → 4C(g) + 6H(g)

∆H◦

rxn = 5 (BEC−C) + 6 (BEC−H)(The comment on the right comes from

dissecting the structure given in the questionand noting how many of each kind of bond ispresent). To find ∆H for this reaction we canuse Hess’ Law with formation enthalpies:

∆H◦

rxn =[

4∆H◦

f C(g) + 6∆H◦

f H(g)

]

−

[

1∆H◦

f C4H6(g)

]

= [4 (717 kJ/mol) + 6 (218 kJ/mol)]

−∆H◦

f C4H6(g)

= 4176 kJ/mol−∆H◦

f C4H6(g)

Now we use the comment on which bondswere broken:

∆H◦

rxn = 6BEC−C + 6BEC−H

= 5 (348 kJ/mol) + 6 (412 kJ/mol)

= 4212 kJ/mol

We can set the two sides of the equationequal since they represent the same reaction:

4212 kJ/mol = 4176 kJ/mol−∆H◦

f C4H6(g)

∆H◦

f C4H6(g)= −36 kJ/mol .

010 10.0 points

Calculate the enthalpy change for the reaction

2 SO2(g) + O2(g) → 2 SO3(g)

∆Hf for SO2(g) = −16.9 kJ/mol;∆Hf for SO3(g) = −21.9 kJ/mol.

1. +5.0 kJ/mol rxn

2. +10.0 kJ/mol rxn



5. −10.0 kJ/mol rxn correct

Explanation:

Reactants:∆Hf SO2(g) = −16.9 kJ/mol∆Hf O2(g) = 0 kJ/molProducts:

∆Hf SO3(g) = −21.9 kJ/mol

∆Hrxn =∑

n∆Hf products

−

∑

n∆Hf reactants

= (2 mol)(−21.9 kJ/mol)

− (2 mol)(−16.9 kJ/mol)

− (1 mol)(0 kJ/mol)

= −10.0 kJ/mol rxn

011 10.0 points

Consider the following substances:

HCl(g) F2(g) HCl(aq) Na(s)

Which response includes ALL of the sub-stances listed that have ∆H0

f = 0?

1. HCl(g), Na(s) and F2(g)

2. HCl(g), Na(s), HCl(aq) and F2(g)

3. Na(s)

4. HCl(g) and Na(s)

5. Na(s) and F2(g) correct

Explanation:

∆H0f = 0 for elements in their standard

states. This would be true for both Na(s) andF2(g).

012 10.0 points


If the products of a reaction have a higherenergy than the reactants, then the reaction

1. is not spontaneous.

2. must be spontaneous.

3. is endothermic. correct

4. is exothermic.

Explanation:

Energy must flow into the system for theproducts to have a higher energy than the re-actants. ∆H will be positive and the reactionwill be endothermic.

013 10.0 points

Calculate the average S F bond energy inSF6, using the following ∆Hf values:

SF6(g) = −1209 kJ/molS(g) = 279 kJ/molF(g) = 79 kJ/mol

1. 981 kJ/mol bonds

2. 1209 kJ/ mol bonds

3. 1962 kJ/mol bonds

4. 289.0 kJ/mol bonds

5. 582 kJ/mol bonds

6. 196 kJ/mol bonds

7. 1565 kJ/mol bonds

8. 416 kJ/mol bonds

9. 202 kJ/mol bonds

10. 327 kJ/mol bonds correct

Explanation:

The S F bond energy (in SF6) is definedas the ∆H for the reaction in which ONEMOLE of S F bonds in SF6 are broken togive gaseous atoms. Each molecule of SF6

has SIX S F bonds, so each mole of SF6 has

SIX moles of S F bonds. Thus we need to

break the S F bonds in only1

6mol of SF6.

So we need the ∆H for the reaction

1

6SF6(g) → F(g) +

1

6S(g)

Use Hess’s Law and the given values of ∆Hf

to calculate ∆H for this reaction.

014 10.0 points

Using the bond energy data provided, cal-culate ∆H for the following reaction.

H2(g) + Cl2(g) −→ 2HCl(g)

Bond Energy

Bond

(

kJ

mol

)

H H 436Cl Cl 242H Cl 432

1. 186 kJ ·mol−1

2. −246 kJ ·mol−1

3. 246 kJ ·mol−1

4. −186 kJ ·mol−1 correct

Explanation:

∆H0rxn =

∑

BErct −

∑

BEprod

= (436 kJ/mol + 242 kJ/mol)

− 2 (432 kJ/mol)

= −186 kJ/mol

The negative sign means the reaction isexothermic by 186 kJ/mol.

015 10.0 points

The standard molar enthalpy of formationof NH3(g) is −46.11 kJ/mol. What is thestandard molar internal energy of formationof NH3(g)?

1. −46.11 kJ/mol


2. −51.07 kJ/mol

3. −41.15 kJ/mol

4. −43.63 kJ/mol correct

5. −48.59 kJ/mol

6. −38.67 kJ/mol

7. −39.91 kJ/mol

Explanation:

∆H = −46.11 kJ/molT = 298.15 K (standard temperature)

1

2N2(g) +

3

2H2(g) → NH3(g)

ni =

(

1

2+

3

2

)

mol = 2 mol

nf = 1 mol∆n = (1− 2) mol = −1 mol

∆E = q + w, q = ∆H and

w = −P ∆V = −∆nRT

= −(−1 mol)(8.31451 J/mol ·K)

×(298.15 K)

(

kJ

1000 J

)

= 2.47897 kJ

Work is per 1 mole of NH3 formed, so

∆E = ∆H + w

= −46.11 kJ/mol + 2.47897 kJ/mol

= −43.631 kJ/mol

016 10.0 points

The internal energy change is 7 kJ when anideal gas expands from 3.50 to 29.2 liters at aconstant external pressure of 3.47 atm. Whatis the heat absorbed by the gas from its sur-roundings?

Correct answer: 16.0361 kJ.

Explanation:

∆E = 7 kJ P = 3.47 atmVi = 3.5 L Vf = 29.2 L

∆V = 29.2 L− 3.5 L = 25.7 L

∆E = q + w = ∆H − P ∆V

∆H = ∆E + P ∆V

= (7 kJ) +

[

3.47 atm · (25.7 L)

×

101.33 J

1 atm · L·

kJ

1000 J

]

= 16.0361 kJ

017 10.0 points

What is the total motional contribution to themolar internal energy of gaseous BF3?

1. 1.5RT

2. 2.5RT

3. 3RT correct

4. 3.5RT

5. RT

Explanation:

The contribution of each mode of motion to

the total molar internal energy is1

2RT . BF3

is a nonlinear molecule so it has three modesof translational motion and three modes ofrotational motion (assuming no contributionfrom vibration). Therefore,

Um = 6

(

1

2RT

)

= 3 RT

018 10.0 points

For a given transfer of energy, a greater changein disorder occurs when the temperature ishigh.

1. False correct

2. True

Explanation:


From ∆S =q

Tsince T is in the denomina-

tor, ∆S will be larger (more positive) when-ever T is smaller.

019 10.0 points

Entropy is a state function.

1. False

2. True correct

Explanation:

State functions are denoted by capital-ized letters. They are P (ressure), V (olume),T (emperature), S (entropy), G(ibb’s Free en-ergy), H (enthalpy). The change in the valueof a state function is independent of the pathtaken.

020 10.0 points

Place the following in order of increasing en-tropy.

1. solid, liquid, and gas correct

2. gas, liqiud, and solid

3. liqiud, solid, and gas

4. gas, solid, and liqiud

5. solid, gas, and liqiud

Explanation:

Entropy (S) is high for systems with highdegrees of freedom, disorder or randomnessand low for systems with low degrees of free-dom, disorder or randomness.

S(g) > S(ℓ) > S(s) .

021 10.0 points

Which substance has the higher molar en-tropy?

1. KCl(aq) at 298 K and 1.00 atm correct

2. Unable to determine

3. KCl(s) at 298 K and 1.00 atm

4. They are the same

Explanation:

KCl(aq) has ions distributed more ran-domly in solution than ions localized in acrystal lattice, hence a higher molar entropy.

022 10.0 points

Calculate the standard entropy of vaporiza-tion of ethanol at its boiling point 352 K. Thestandard molar enthalpy of vaporization ofethanol at its boiling point is 40.5 kJ ·mol−1.

1. +115 J·K−1·mol−1 correct

2. − 40.5 kJ·K−1·mol−1

3. +40.5 kJ·K−1·mol−1

4. +513 J·K−1·mol−1

5. − 115 J·K−1·mol−1

Explanation:

∆Hvap = 40500 J ·mol−1 TBP = 352 K

∆Scond =q

T=

∆Hcon

TBP=

∆Hvap

TBP

=40500 J ·mol−1

352 K= +115.057 J ·mol−1

·K−1

023 10.0 points

What is the entropy change for freezing4.64 g of C2H5OH at 158.7 K? ∆H =−4600 J/mol.

Correct answer: −2.91941 J/K.

Explanation:

∆H = −4600 J/mol T = 158.7 Km = 4.64 g

MW = 2 (12.0107 g/mol)

+ 6 (1.00794 g/mol)

+ 8 (15.9994 g/mol)

= 46.0684 g/mol .


The molar change in entropy is

∆S =q

T=

∆H

T

=−4600 J/mol

158.7 K

= −28.9855J

mol ·K,

and the change in entropy is(

−28.9855J

mol ·K

)(

4.64 g

46.0684 g/mol

)

= −2.91941 J/K .

024 10.0 points

Suppose that 140 g of ethanol at 25◦C is mixedwith 235 g of ethanol at 85 ◦C at constantatmospheric pressure in a thermally insulatedvessel. What is the ∆Ssys for the process?The specific heat capacity for ethanol is 2.42J/g K.

Correct answer: 3.50979 J/K.

Explanation:

mc = 140 g mh = 235 gTc,i = 25◦C = 298.15 KTh,i = 85◦C = 358.15 K

CP = 2.42J

g ·K

We use the specific heat here (J/g·K) althoughthe problem can also be worked with the mo-lar heat capacity also. The change in entropyfor each portion of ethanol is calculated viathe equation:

∆S = mCP,s ln

(

T2T1

)

Note that the “hot” side (85◦C) will cooldown to the final temperature and the “cold”side (25◦C) will warm up to the final tempera-ture. The amount of heat that flows from oneto the other is

qc = −qh

mc Cp(ethanol) (Tf − Tc,i)

= −mh Cp(ethanol) (Tf − Th,i)

Tf =mc Tc,i +mh Th,i

mc +mh

=(140 g) (298.15 K) + (235 g) (358.15 K)

140 g + 235 g= 335.75 K

The change in entropy for the “cold”ethanol initially at 25◦C is

∆Sc = (140 g)

(

2.42J

g ·K

)

× ln

(

335.75 K

298.15 K

)

= 40.2393 J/K

and for the “hot” ethanol initially at 85◦C,

∆Sh = (235 g)

(

2.42J

g ·K

)

× ln

(

335.75 K

358.15 K

)

= −36.7295 J/K

The total change in entropy is that of theentire system (there is no change in entropyof the surroundings as the vessel isolates thesystem from the surroundings):

∆Stot = 40.2393 J/K+ (−36.7295 J/K)

= 3.50979 J/K

025 10.0 points

The temperature of 2.00 mol Ne(g) is in-creased from 25◦C to 200◦C at constant pres-sure. Calculate the change in the entropy ofneon. Assume ideal behavior.

1. − 19.2 J·K−1

2. +7.68 J·K−1

3. +9.60 J·K−1

4. − 7.68 J·K−1

5. +19.2 J·K−1 correct


Explanation:

T1 = 25◦C+ 273.15 = 298.15 KT2 = 200◦C+ 273.15 = 473.15 K

For a monoatomic ideal gas, Cp,m = 2.5 R

∆S = nCp,m ln

(

T2T1

)

= (2.00 mol) 2.5 (8.314 J ·mol−1·K−1)

× ln

(

473.15 K

298.15 K

)

= +19.1977 J ·K−1

We expect a positive answer since tempera-ture increased.

hw02 - thermo 2-solutions

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