CHAPTER 6 REGULAR HOMEWORK:

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Question 1

4200 J of heat are added to a system. It does 5900 J of work on the surroundings. What is ΔE in J? Explanation

If the correct answer is positive, it means the system gained energy. If it is negative, then the system lost energy.

+4200J -5900J = -1700 J

Question 2

3500 J of heat are lost from a gas to cooler surroundings. At the same time the gas is compressed and the surroundings do 3100 J of work on the gas. What is ΔE for the gas in J? Explanation

If the correct answer is positive, it means the system gained energy. If it is negative, then the system lost energy.

-3500J +3100J = -400J

Question 3

A gas is compressed from 85.8 L to 1.6 L at 0.50 atmospheres? In addition a total of 5700 J of heat are gained from warmer surroundings. What is the change in internal energy in kJ? Explanation

If the correct answer is positive, it means the system gained energy. If it is negative, then the system lost energy.

w=  -P delta V

-(0.5)(1.6-85.8) * 101.33= 4265.993

+5700 + 4265.993 = 9965.993/1000 = 9.965 kJ

Question 4

The combustion of 1.00 mol of liquid methly alcohol (CH3OH) at constant pressure in excess oxygen is exothermic, giving 727 kJ of heat per mole of CH3OH. Write the balanced combustion reaction using smallest whole numbers. What is the coefficient for oxygen? Explanation

(CH3OH + 3/2 O2  --> CO2  + 2H2O)2         dH= -727 kJ/mol

2CH3OH + 3 O2  --> 2CO2  + 4H2O

Question 5

What is the enthalpy change (in kJ) for burning 16.0 g of methyl alcohol?

Hint given in feedback Explanation

What is the sign of the change in enthalpy for an exothermic reaction? Hint, when a reaction is exothermic,what is the sign (positive or negative) for qsurr? Remember, energy is conserved. So, ΔH +qsurr = 0. In other words when one is positive the other is negative.)

Remember that the 727 kJ is per mole of CH3OH.

16 g CH3OH * 1mol/ 3Lg = 0.5 mol CH3OH * -727kJ/mol = -363.5 kJ/mol

Question 6

A reaction that is used to propel rockets is N2O4(l) + 2N2H4(l) --> 3N2(g) + 4H2O(g). This reaction has the advantage that neither product is toxic, so no dangerous pollution is released. When the reaction consumes 10.0 g of liquid N2O4, it releases 124 kJ of heat. What is the value of ΔH (in kJ per mole N2O4) for the chemical equation as written? Explanation

N2O4(l) + 2N2H4(l) --> 3N2(g) + 4H2O(g)

-124 kJ/ 10g = -12.4 kJ/g * 92g/mol = -1140.8 kJ/mol

Question 7

Ammonia is produced commercially by the direct reaction of the elements. The formation of 1.00 moles of gaseous NH3 by this reaction releases 46.17 kJ of heat. How much energy (in kJ) is released, when 43.0 kg of hydrogen gas, H2, reacts in excess nitrogen gas, N2?

Hint given in feedback Explanation

3H2+N2--> 2NH3

43 kg *1000g/1 kg =43000g H2* 1 mol/Lg= 21500 mol H2 * 2molH3/3molH2

=14333.3molNH3*46.17kJ/1mol=661770kJ/mol

Question 8

The thermochemical equation for the burning of ethyl alcohol is

C2H5OH(l) + 3O2(g) --> 2CO2(g) + 3H2O(l) ΔH = -1,367 kJ

What is the enthalpy change (in kJ) for burning 6.65 g of ethyl alcohol? Explanation

C2H5OH + 3O2-->2CO2 + 3H2O   dH=-1367 kJ

6.65g C2H5OH * 1 mol/ 46 g/mol = 0.145 mol C2H5OH*-1367kJ/mol

=-197.62 kJ/mol

Question 9

A 85.1 g sample of metal at 80.91 oC is added to 48.84 g of water that is initially at 22.51 oC. The final temperature of both the water and the metal is 30.09 oC. The specific heat of water is 4.184 J/(goC). Calculate the specific heat of the metal. Explanation

85.1g(x)(30.09-80.91)+ 48.84g(4.18)(30.09-22.51)=0

x=0.358 J/g C

Question 10

A 52.5 g sample of Al at 93.7 oC is added to 189.6 g of water that is initially at 24.2 oC. The specific heat of Al and water are 0.90 J/(goC) and 4.184 J/(goC), respectively. Calculate the equilibrium temperature. Explanation

(52.5g)(0.9)(x-93.7)+(189.6g)(4.18)(x-24.2)=0

x=28.11 C

Question 11

A 65 mL solution of a dilute AgNO3 solution is added to 60 mL of a base solution in a coffee-cup calorimeter. As AgOH (s) precipitates, the temperature of the solution increases from 23.55 oC to 25.60 oC. Assuming the mixture has the same specific heat (4.184J/goC) and density (1.00 g/cm3 or 1.00 g/mL) as water, calculate the heat (in J) transferred to the surroundings, qsurr.

Hint given in feedback Explanation

1 mL = 1 cm3. Now use the density of water to go from cm3 to g.

q(surr) = -q(sys)

(125g)(4.18)(25.6-23.55)=q

1071.125=q(surr)

Question 12

Is the precipitation reaction exothermic or endothermic?

Select one: Explanation

The reaction(system) is releasing heat so the solution (surrounding) is getting warmer. Therefore the reaction is an exothermic reaction.

Question 13

A 161 mL solution of a dilute acid is added to 120 mL of a base solution in a coffee-cup calorimeter. The temperature of the solution increases from 23.45oC to 27.07 oC. Assuming the mixture has the same specific heat (4.184J/goC) and density (1.00 g/cm3) as water, calculate the heat (in J) transferred to the surroundings, qsurr.

Hint given in feedback Explanation

1 mL = 1 cm3. Now use the density of water to go from cm3 to g.

(281g)(4.184)(27.07-23.45)=q

1174.58(3.62)=q

4251.98=q

Question 14

Is the neutralization reaction exothermic or endothermic?

Select one: Explanation

The reaction(system) is releasing heat so the solution (surrounding) is getting warmer. Therefore the reaction is an exothermic reaction.

Question 15

Dissolving 6.51 g of CaCl2 in enough water to make 345 mL of solution causes the temperature of the solution to increase by 3.51 oC. Assume the specific heat of the solution and density of the solution are the same as water′s (about 4.18 J/goC and 1.00 g/cm3, respectively) Calculate ΔH per mole of CaCl2 (in kJ) for the reaction under the above conditions.

Hint given in feedback

Aside, the ΔH per mole for dilution depends on the process. For example, more energy is released when starting with a large volume of water (infinite dilution), than when starting with a small volume of water. Why do you think this happens? Question 16

In the process of isolating iron from its ores, carbon monoxide reacts with iron (III) oxide, as described by the equation:

Fe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g) ΔH = -24.8 kJ

The enthalpy change for the combustion of carbon monoxide is:

2CO(g) + O2(g) --> 2CO2(g) ΔH = -566 kJ

Use the preceding thermochemical equations to calculate the enthalpy change (in kJ) for the equation:

4Fe(s) + 3O2(g) --> 2Fe2O3(s)(Use 4 sig. fig.) Explanation

-2(Fe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g) ΔH = -24.8 kJ)

4Fe(s) + 6CO2(g)-->2Fe2O3(s) + 6CO(gΔH = 49.6 kJ

3(2CO(g) + O2(g) --> 2CO2(g) ΔH = -566 kJ)

6CO(g) + 3O2(g) --> 6CO2(g) ΔH = -1698 kJ

4Fe(s) + 3O2(g) --> 2Fe2O3(s) ΔH= 49.6-1698= -1648 kJ

Question 17

At 25oC, the following heats of reaction are known:

2C2H2(g)+5O2(g) → 4CO2(g) + 2H2O(l) ΔH = -2,600 kJ
C(s) + O2(g) → CO2(g) ΔH = -394 kJ
2H2(g) + O2(g) → 2H2O(l) ΔH = -572 kJ

At the same temperature, calculate ΔH for the reaction:

2C(s) + H2(g) → C2H2(g) Explanation

-1(2C2H2(g)+5O2(g) → 4CO2(g) + 2H2O(l) ΔH = -2,600 kJ)

4CO2(g) + 2H2O(l)→ 2C2H2(g)+5O2(g)

4(C(s) + O2(g) → CO2(g) ΔH = -394 kJ)

4C(s) + 4O2(g)4CO2(g)

2H2(g) + O2(g) 2H2O(l) ΔH = -572 kJ

4C(s) + 2H2(g) → 2C2H2(g)

2C(s) + H2(g) → C2H2(g)

(-1(2600) + 4(-394) -572) /2

226kJ

Question 18

Use the standard enthalpies of formation from Appendix II (A-7 to A-12) to calculate the enthalpy change (in kJ) for the photosynthesis of glucose at 298 K and 1 atm.

6CO2(g) + 6H2O(l) --> C6H12O6(s) + 6O2(g)

Use 4 SF. Explanation

Now Heat of formation of CO2 = -394 kJ

Heat of formation of H2O = -286

Heat of formation of Glucose = -1271 kJ

So Heat of Formation of Reaction = -1271 - 6 * (-394) - 6 * (286) = 2801J

Question 19

Is the preceding reaction endothermic or exothermic?

Select one: Explanation

The reaction is endothermic because the heat of formation is positive

Question 20

Use the standard enthalpies of formation from Appendix II (A-7 to A-12) to calculate the enthalpy change (in kJ) for the reduction of iron (III) oixde to iron at 298 K and 1 atm. (Calculate it for the reaction as written, namely 2 moles of iron (III) oxide and 3 moles of carbon.)

2Fe2O3(s) + 3C(s) --> 4Fe(s) + 3CO2(g) Explanation Question 21

Is the preceding reaction endothermic or exothermic?

Select one: Explanation

The reaction is endothermic because the heat of formation is positive

Question 22

Frederick Trouton noted that the enthalpy of vaporization of 1 mole of a pure liquid, expressed in joules, is approximately 88 times the boiling point of the liquid on the Kelvin scale. This relationship is called Trouton's rule and is represented by the thermochemical equation: liquid → gas, ΔHvap = 88 Tbjoules/mol.

Combined with an empirical formula from chemical analysis, Trouton's rule can be used to find the molecular formula of a compound, as illustrated here. A compound that contains only carbon and hydrogen is 85.6% C and 14.4% H. Its enthalpy of vaporization is 389 J/g, and it boils at a temperature of 322 K.

First, determine the empirical formula of this compound using the percent composition of H and C. What is the empirical molar mass (in g/mol)? Question 23

Use Trouton's rule to calculate the approximate enthalpy of vaporization (in J) of one mole of the compound. Question 24

Using the result from the preceding question and the experimental ΔHvap of 389 J/g, find the approximate molar mass (in g/mol).

Hint given in feedback Explanation

This is a simple multiplication. Using units, you can see how to multiple the information to get the approximate molar mass or g/mol.

Question 25

What is the molar mass correct to 4 SIG. FIG.?

Hint given in feedback Explanation

You know the approximate molar mass. Use the empirical formula to get the actual molar mass.

Question 26

The price of gold is \$625 per troy ounce at this writing. How much heat (in J) is needed to raise the temperature of \$5000.00 worth of gold from 14.2 oC to 80.6 oC?

(1 troy oz = 31.10 g and the specific heat of Au is 25.42 J/(moloC).)

Review of multiplication by 1. Note, periodic table has additional needed information. Question 27

Problems27-31. The law of Dulong and Petit states that the approximate molar heat capacity of a metal is 25 J/(mol.K).

Experimentally, the specific heat of a metal is found to be 0.460 J/(goC). Use the law of Dulong and Petit to calculate the approximate molar mass of the metal.

Hint given in feedback Explanation

This is a simple multiplication. Using units, you can see how to multiple the information to get the approximate molar mass or g/mol.

Question 28

A chloride of this element is 67.2% chlorine by mass. In a 100 g sample how many moles of chlorine are there? (This info will be used to determine the empirical formula. Because the accuracy of the result in the last question for this set of questions depends on the accuracy of this answer, use the atomic mass of Cl accurate to 4 SIG. FIG.). Question 29

In the same 100 g sample approximately how many moles of metal are there? (This info will be used to determine the empirical formula.)

Hint given in feedback Explanation

Hint, use the approximate molar mass found earlier.

Question 30

Determine the subscripts x and y in the empirical formula of the chloride. MxCly (The mole ratio of the elements you find will not be exactly whole numbers, so considerable rounding is needed to obtain whole numbers in the formula.) What is y? Explanation

Answer must be a whole number.

Question 31

You know how many g of metal are in the 100 g sample. You can find the actual (not just approximate) moles of metal because you know the moles of chlorine in a 100 g sample and you know the relationship between moles of metal and moles of Cl (preceding question). Determine an accurate molar mass for the metal. ANSWER NEEDS TO BE RIGHT FOR 3 SIG. FIG.

Note, your approximate molar mass for the metal will indicate whether you are doing the calculation correctly. By the way, what is the metal? (Don't include in answer.) 