crsch06-concept

Chapter 6

Thermochemistry

QUESTION

The combustion of a fuel is an exothermic process. This means…

1. the surroundings have lost exactly the amount of energy gained by the system.

2. the potential energy of the chemical bonds in the products should be less than the potential energy of the chemical bonds in the reactants.

3.

q must be positive; w must be negative

4. D E would have a + overall value because the surroundings have gained energy.

Copyright © Houghton Mifflin Company. All rights reserved.

CRS Question, 6 –2

ANSWER

Choice 2 correctly summarizes the connection between reactants and products for an exothermic process. The exothermic nature of the process means that the products must be less energetic than the reactants. The energy descriptions and comparisons are from the view point of the system.

Section 6.1: The Nature of Energy



QUESTION

The change in enthalpy, or heat of reaction, for chemical changes is equal to the heat flow when…

1. the pressure is constant and only PV work is allowed.

2. D H = – q (constant pressure)

3. D H = D E – P D V

4. the pressure is constant and enthalpy = internal energy ( E )



ANSWER

Choice 1 properly relates enthalpy to q . At constant pressure, any work is change in volume work. So, D H = D E + P D V and since q (at constant pressure) = D E + P D V it follows that D H = q (at constant P where most chemical reactions take place).

Section 6.2: Enthalpy and Calorimetry



QUESTION

Acetylene torches used for welding work so effectively because of the large release of energy ( D H ) when acetylene, C

2

H

2

, reacts with oxygen. For the combustion of acetylene, D H is approximately

–1,300 kJ/mol. What is the value for D H when 1.0 gram of acetylene reacts in this way?

1. Approximately +50. kJ

2. Approximately –50. kJ

3. Approximately +34,000 kJ

4. Approximately –34,000 kJ



ANSWER

Choice 2 indicates the proper sign, value, and label. The reported

–1,300 kJ is for the combustion of one mole. After 1.0 gram is properly converted to moles, (1.0/26.04) multiplying by –1,300 kJ/mol will yield the answer. The negative sign indicates that heat is released to the surroundings.




QUESTION

The instant cold packs associated with athletic injuries makes use of the heat change when NH

4 grams of NH

4

NO

3

NO

3 dissolves. If 8.0

(molar mass = 80.1) were placed in 100.0 mL of water such as shown in the constant pressure calorimeter, and the temperature change was – 6.0°C, what would you calculate as the D H for dissolving one mole of

NH

4

NO

3

?

1. –2,500 kJ/mol

2. +2,500 kJ/mol

3. –25 kJ/mol

4. +25 kJ/mol



ANSWER

Choice 4 shows both the proper sign for this endothermic process and the proper value. The constant 4.184 J/°C g was used to determine the heat absorbed from the 100.0 mL of water as its temperature dropped by 6.0°C, during dissolving. Once this was known for 8.0 grams, a ratio to 80.1 grams was calculated.




QUESTION



QUESTION (continued)

The heat capacity of a calorimeter of the type shown on the previous slide must be determined before other investigations, using the instrument, can be performed. When 0.5000 grams of pure carbon are combusted in the “bomb” the temperature rose by

1.842°C. (Note: use –393.5 kJ/mol as the heat of combustion for carbon.) What is the calorimeter’s heat capacity?

1. 30.20 kJ/°C

2. 106.8 kJ/K

3. 8.894 kJ/°C

4. 35.60 kJ/°C



ANSWER

Choice 3 is correct. The heat released per mole must be converted to heat released for 0.5000 gram (0.5000/12.01)  –393.5. This should be divided by the number of degrees the calorimeter registered and the sign changed to be positive.




QUESTION

A certain small piece of candy is made of 1.525 grams of sucrose

(C

12

H

22

O

11

; molar mass = 342.30 g/mol). When placed in a bomb calorimeter with a specific heat of 5.024 kJ/°C, how much should the temperature rise when the candy undergoes combustion? (Note: use

–5,640 kJ/mol as the heat released for the combustion of one mole of sucrose.)

1. 5.00°C

2. 126°C

3. 25.1°C

4. Couldn’t I just eat the candy, I am not sure what to do with the calculations.



ANSWER

Choice 1 (although choice 4 might be tastier) is correct. Since it is given that one mole of sucrose releases 5,640 kJ, the determination of the number of moles combusted would enable a ratio to find the kJ released from 1.525 g (1.525/342.30)  5,640). Then the heat released by the combustion would be divided by the kJ/°C for the calorimeter; the result would be the change in temperature.




QUESTION

The following reaction shows the conversion of ozone to oxygen and provides the standard enthalpy change for the process:

2 O

3

 3 O

2

; D H = –427 kJ

What is the D H value for the conversion of one mole of O

2 to O

3

?

1. +427 kJ

2. +641 kJ

3. +142 kJ

4. +285 kJ



ANSWER

Choice 3 provides both the proper sign and value for the new reaction. The original reaction must be reversed in order to show oxygen forming ozone. This means the given sign for D H must also be reversed. Since the reversed reaction shows the process for

3 moles of oxygen, the D H value must be reduced by 1/3 to determine the value for only one mole.

Section 6.3: Hess’s Law



QUESTION

The following reaction depicts one way in which nitric acid could be formed:

N

2

O

5

+ H

2

O  2 HNO

3

D H = ?

Use the following information, and apply Hess’s l aw, to determine the missing value.

H

2

+ 1/2 O

½ N

2

2N

2

+ 3/2 O

2

+ 5 O

2

2

 H

2

+ ½ H

2

 2 N

O;

2

O

5

D H



= –285.8 kJ

HNO

3

; D H = –174.1 kJ

; D H = + 28.4 kJ




1. –362.4 kJ

2. –69.5 kJ

3. –76.6 kJ

4. I am having difficulty rearranging the given equations to produce the “target” equation.



ANSWER

Choice 3 is the answer when the three given equations are properly arranged. The arrangement that, when added, yields the desired equation is: reverse the first given equation, double the next one, and finally take ½ and reverse the third equation. Each manipulation should be done to the respective D H values before tallying those to yield the final D H value.

Section 6.3: Hess’s Law



QUESTION

Using information from Appendix Four, determine the standard enthalpy change at 25°C for the combustion of one mole of heptane (C

7

H

16

) sometimes used in gasoline fuel mixtures. The balanced equation is shown here:

C

7

H

16

( g ) + 11 O

2

( g )  7 CO

2

( g ) + 8 H

2

O ( l )

(Note: the standard molar enthalpy change of formation of C

7

( g ) is –264 kJ/mol)

H

16

1. –4430 kJ

2. –4780 kJ

3. –5310 kJ

4. This can’t be solved without the heat of formation of oxygen.



ANSWER

Choice 2 provides the correct sign and value. The moles of each reactant must be multiplied by the heat of formation of each component (noting their physical state), then the sum of the enthalpy for reactants is subtracted from the sum of the enthalpies of the products (noting that oxygen has a zero for enthalpy of formation).

Section 6.4: Standard Enthalpies of Formation



QUESTION




After examining the graph shown here, which energy source would you like to see increase and which would you like to see decrease? Be prepared to explain your choice.

1. Wood increase; nuclear decrease because…

2. Hydro and nuclear increase, petroleum decrease because…

3. Coal increase, hydro and nuclear decrease because…

4. I have another choice besides these, it is…



ANSWER

While there are a variety of opinions, the decrease or increase of energy sources will depend on a complex combination of politics, life style choices, economics and technology developments.

Section 6.5: Present Sources of Energy



QUESTION

One source of energy being considered as a replacement for petroleum fuels is the combustion of hydrogen. There are several technical considerations to bring this about, but the heat of combustion reaction is part of the determination. How much heat, under standard conditions and 25°C, could be obtained from the combustion of 2.00 kg of H

2

?

H

2

( g ) + ½ O

2

( g )  H

2

O ( l )

1. –572,000 kJ

2. –484,000 kJ

3. –284,000 kJ

4. –242,000 kJ



ANSWER

Choice 3 has taken into account the kg to g to mole conversion for 2.00 kg of H

2 mole of H

2

H

2 and the stoichiometry. Since combustion of one in the equation yields 286 kJ, the number of moles of multiplied by the D H value for the equation (using H will yield the correct value.

2

O ( l ))

Section 6.6: New Energy Sources



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Chapter 6

Thermochemistry

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