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### Question 1

The following process occurs at constant pressure. Which of the following statements are correct regarding the work done on the system (consider only PV work from gases) for

Fe2S3(s) + 6HNO3(aq) → 2Fe(NO3)3(aq) + 3H2S(g).

Select one or more:

### Question 2

The following process occurs at constant pressure. Which of the following statements are correct regarding the work done on the system (consider only PV work from gases) for

CH4(g) + 2O2(g) → CO2(g) + 3H2O(l).

Select one or more:

### Question 3

Calculate the work (in kJ) when 2.00 moles of hydrogen gas are produced from the reaction of sodium in excess water at 298 K at constant pressure:

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

### Explanation

Hint, is the work (-) or (+)?

PV= nRT

w=-pV

w=-nRT

w=-ΔnRT

w=-(2.0)(8.314)(298)=4955.144 J=-4.96kJ

### Question 4

Calculate w (in kJ) when 181 g iron (III) oxide (MM = 159.7 g/mol) reacts with excess carbon to produce carbon dioxide gas at 359 K:

2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g)

### Explanation

Hint, is the work (-) or (+)?

181g * 1 mol/159.7 g/mol = 1.133*3 mol CO2/2 mol Fe2O3= 1.70 mol CO2

w= -1.70 (8.314)(359)=-5074.0342 J = -5.07 kJ

### Question 5

Calculate the work (in kJ) when 1.70 moles of methane react with excess oxygen at 474 K:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

### Explanation

Hint, what would the answer be for 1 mole of carbon dioxide? Rescale your answer. Pay attention to the sign for the work (-) or (+).

w= -[1.7 - (1.7 + 3.4)] (8.314)(474) = 13398.8424 J = 13.398 kJ

### Question 6

A reaction at 3.16 atm consumes 6.70 L of a gas adiabatically (q = 0). Calculate the change in internal energy of the system (in kJ).

Hint given in feedback.

### Explanation

Hint, calculate the work in kJ. Is ΔV (-) or (+)?

ΔE  = q + w= 0 - pV

ΔE= -pV

= -3.16(6.70)

=(21.172) L *atm * 101.3 J / L*atm =2144.7236 J =2.145 kJ

### Question 7

A reaction at 1.61 atm produces 15.7 L of a gas and evolves 82.5 kJ of heat (exothermic). Calculate the change in internal energy of the system (in kJ).

### Explanation

ΔH = E + PV

PV = (1.61)(15.7) L * atm * 101.3 J/ L*atm = 2560.5601 J = 2.560 kJ

E= -82.5 kJ - 2.560= -85.060

### Question 8

1.456 g of an unknown hydrocarbon (72.6 g/mol) burns in bomb calorimeter in excess oxygen. The heat capacity of the calorimeter,Cv, = 5.107 kJ/ºC and ΔT =7.230 ºC. Find ΔE for this reaction in kJ/mol.

Hint given in feedback.

### Question 9

Using the technique of the previous problem ΔE was found to be -2,000.00 kJ/mol of an unknown liquid hydrocarbon at 298 K. In another experiment it was determined that for each mole of hydrocarbon, 8 moles of oxygen gas are consumed and 8 moles of CO2 gas and 5 moles of H2O liquid are produced. Find ΔH per mole of this hydrocarbon (in kJ) at 298 K. Please use six significant figures, that is, to hundredths.

Hint given in feedback.

Not used to answer question: This question is a bit awkward and unrealistic for molar amounts, but allows for random numbers. An example of a possible reaction is:
C2H4(OH)2(l) +2.5O2(g) →2CO2(g)+ 3H2O(l)

### Explanation

Recall ΔH = ΔE+PΔV = ΔE + ΔnRT for gases.
A comment on the last 2 problems. Normally, ΔH is calculated, then ΔE because it is easier to do experiments at constant pressure.

ΔH= ΔE + pΔV  = ΔE + nRT = ΔE + 0(RT) = -2000

### Question 10

Arrange CBr4, C2Br6, C3Br8 in order from least to greatest entropy.

Select one:

### Question 11

Arrange CH3F, CH3Cl, and CH3I in order from least to greatest entropy.

Select one:

### Question 12

Arrange H2, O2, and N2 in order from least to greatest entropy.

Select one:

### Question 13

Check those changes where the entropy increases, that is, ΔS is positive.

(All responses must be correct for credit.)

Select one or more:

### Question 14

Check those changes where the entropy increases, that is, ΔS is positive.

(All responses must be correct for credit.)

Select one or more:

### Question 15

Check those changes where the entropy increases, that is, ΔS is positive.

(All responses must be correct for credit.)

Select one or more:

### Question 16

Check those changes where the entropy increases, that is, ΔS is positive.

(All responses must be correct for credit.)

Select one or more:

### Question 17

Calculate the increase of entropy (in J/K) when 54 g of ice melts at 0 ºC and 1 atm. (The heat of fusion for ice is 6,000 J/mol.)

### Explanation

ΔS=-ΔHsys/T  ( @ constant pressure and temperature)

ΔH=ΔE+pΔV

V=0

54 g * 1 mol/18 g/mol = 3 mol

ΔH=q

q=mLf= 3(6,000)=18000

ΔS= 18000/273 K= 65.93 J/K

### Question 18

Calculate the change in entropy (in J/K) when a 68.8 g of water is heated from 16.2 ºC to 78.7 ºC at 1 atm. (The specific heat is 4.184 J/(g-K).)

Notice that entropy and heat capacity have the same units.

### Explanation

ΔS=mc ln(T2/T1) = (3.822)(4.184)ln(78.7+273/12.2+273)=    56.32

68.8 g * 1 mol / 18 g/mol = 3.822

### Question 19

Calculate the change in entropy (in J/K) when a 66.6 g of nitrogen gas is heated at a constant pressure of 1.50 atm from 26.4 ºC to 65.2 ºC. (The molar specific heats are Cv is 20.8 J/(mol-K) and Cp is 29.1 J/(mol-K) .)

### Explanation

ΔS=mc ln(T2/T1) = (2.37)(29.1)ln (65.2+273/26.4+273)=8.358

66.6g * 1 mol/ 28 g/mol = 2.37 mol

### Question 20

In the previous problem the gas is heated at constant pressure. This means that as the gas warms, it also expands. Would the change in entropy be the same, if the gas were heated at constant volume and then allowed to expand such that the final temperature, pressure, and volume were the same as the constant pressure case?

Select one:

### Question 21

Calculate the change in entropy (in J/K) for the following reaction at 298 K:

2NO(g) +O2(g) → 2NO2(g)

Use the data in Appendix II (Tro); Appendix 4 (Zumdahl)

(When no amounts are given, it is for the reaction as written in moles. In this case 2 moles NO and 1 mole oxygen produce 2 moles NO2.)

### Explanation

2NO  + O2 ------> 2NO2

### Explanation

ΔS= [197.7 + 130.7] - [5.7 + 188.8]

=133.9

### Question 22

Calculate the change in entropy (in J/K) for the following reaction at 298 K:

C(s) + H2O(g) → CO(g) + H2(g)

Use entropy for graphite (diamond is too expensive).

Notice that the entropy of the solid is much smaller than the entropy of the gases!

### Explanation

C (s)  + H2O (g)  --------->     CO (g)       + H2(g)

ΔS= [197.7 + 130.7]  -[5.7 +188.8]

=133.9