1A Exercise Key 6

Chapter 6 Exercise Key 1

Chapter 6 Exercise Key Exercise 6.1 Kinetic and Potential Energy: For each of the following situations, you are asked which pair has the higher energy. Explain your answer with reference to the capacity of each to do work and say whether the energy that distinguishes them is kinetic energy or potential energy. a. Nitric acid molecules, HNO3, in the upper atmosphere decompose to form HO

molecules and NO2 molecules by breaking a bond between the nitrogen atom and one of the oxygen atoms. Which has higher energy, a nitric acid molecule or the HO molecule and NO2 molecule the come from its decomposition?

HNO3(g) HO(g) + NO2(g) HO and NO2 have higher potential energy than HNO3. When a system shifts to decrease the strengths or number of attractions between the systems components, the potential energy in the system increases. Energy is required to separate the nitrogen and oxygen atoms being held together by mutual attraction in a chemical bond. The energy supplied goes to an increased potential energy of the separate HO and NO2 compared to HNO3. If the bond is reformed, the potential energy is converted into a form of energy that could be used to do work.

b. Nitrogen oxides, NO(g) and NO2(g), are released into the atmosphere in the exhaust of our cars. Which has higher energy, a NO2 molecule moving at 439 m/s or the same NO2 molecule moving at 399 m/s. (These are the average velocities of NO2 molecules at 80 C and 20 C.)

A nitrogen dioxide molecule with a velocity of 439 m/s has greater kinetic energy than the same molecule with a velocity of 399 m/s. Any object in motion can collide with another object and move it, so any object in motion has the capacity to do work. This capacity to do work resulting from the motion of an object is called kinetic energy, KE. The particle with the higher velocity will move another object (like another molecule) farther, so it can do more work. It must therefore have more energy.

c. Which has higher energy, a nitrogen monoxide molecule, NO, moving out your cars tailpipe at 450 m/s or a nitrogen dioxide molecule, NO2, moving at the same velocity?

The more massive nitrogen dioxide molecule has greater kinetic energy than the less massive nitrogen monoxide molecule with the same velocity. The moving particle with the higher mass can move another object (like another molecule) farther, so it can do more work. It must therefore have more energy.

Copyright 2004 Mark Bishop 1


d. Liquid nitrogen is used for a number of purposes, including the freezing of warts. Which has higher energy, liquid nitrogen or gaseous nitrogen? (Disregard the likely difference in temperature, and assume that the two systems are at the same temperature.)

Gaseous nitrogen has higher potential energy than liquid nitrogen. When a system shifts to decrease the forces of attractions between the systems components, the potential energy in the system increases. When nitrogen goes from liquid to gas, the attractions that link the N2 molecules together are broken. The energy that the nitrogen liquid must absorb to break these attractions goes to an increased potential energy of the nitrogen gas. If the nitrogen returns to the liquid form, attractions are reformed, and potential energy is converted into a form of energy that could be used to do work.

e. Halons, like Halon-1301 (CF3Br) and halon-1211 (CF2ClBr), which have been used as fire extinguishing agents, are a threat to our protective ozone layer. When released into the atmosphere, they can migrate into the upper atmosphere where bromine atoms are stripped from the molecules. These bromine atoms react with ozone molecules to form BrO molecules, which can react with NO2 molecules to form BrONO2. Which has higher energy, separate BrO and NO2 molecules or the BrONO2 that they form?

BrO(g) + NO2(g) BrONO2(g) Separate BrO and NO2 molecules have a higher potential energy than the BrONO2 molecule that they form. When a system shifts to increase the forces of attractions between the systems components, the potential energy in the system decreases. When BrO and NO2 are converted into BrONO2, a new bond is formed, and some of the potential energy of the BrO and NO2 is released. The energy could be used to do some work. For example, if some of the potential energy is converted into increased kinetic energy of a molecule like N2, the faster moving molecule could bump into something and move it and therefore do work.

f. Alpha particles, which are released in alpha decay of large radioactive elements, like uranium, are helium nuclei that contain two protons and two neutrons. Which has higher energy, alpha particles that are close together or alpha particles that are farther apart?

The positive charge of the alpha particles causes them to repel each other, and the closer the charges are, the more repulsion there is between them. Therefore, the alpha particles that are close together are less stable and higher potential energy than alpha particles that are farther apart.

g. Which has higher energy, an uncharged helium atom or an alpha particle and two separate electrons?

Decreasing the forces of attractions between particles in a system will increase the potential energy of the system, so an alpha particle and two separate electrons has higher potential energy than an uncharged helium atom, which has two electrons attracted to its nucleus. The attraction between the alpha particle and the electrons will pull them together, and as they move together, they could bump into something, move it, and do work.



Exercise 6.2 H as a Conversion Factor: When 1.245 x 104 kJ of heat are evolved from the combustion of ethane, what mass of water is formed?

4 22 3

2

6 mol H O 18.0153 g H O? g H O = 1.245 10 kJ 3.08 10 kJ 1 mol H O

2 = 437 g H2O

Exercise 6.3 H and Changing Coefficients: What is the H for the following equation?

4C2H6(g) + 14O2(g) 8CO2(g) + 12H2O(l)

2 (3.08 x 103 kJ) = 6.16 x 103 kJ Exercise 6.4 H and reverse Reactions: What is the H for the following reaction?

4CO2(g) + 6H2O(l) 2C2H6(g) + 7O2(g)

1 (3.08 x 103 kJ) = 3.08 x 103 kJ

Exercise 6.5 H and E: Nitrogen dioxide gas reacts with liquid water to yield liquid nitric acid and nitrogen monoxide gas. 23.84 kJ of heat is evolved when one mole of NO2 reacts at constant volume and 25 C and 1 atm. What is the H for this reaction?

NO2(g) + 1/3H2O(l) 2/3HNO3(l) + 1/3NO(g)

H = E + (n)RT

0.008314 kJH = 23.84 kJ + (1/3 1) mol 298.15 KK mol

= 25.49 kJ

Exercise 6.6 H and E: When 3.000 g of ethyl alcohol, C2H5OH(l), are burned in a bomb calorimeter at 25.00 C, 88.90 kJ of heat are evolved. Calculate the molar E and H for this reaction. (The reactions in a bomb calorimeter are run at constant volume.)

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)

2 5

2 5 2 5

46.069 g C H OH88.90 kJE = 3.000 g C H OH 1 mol C H OH

= 1365 kJ/mol

H = E + (n)RT

0.008314 kJH = 1365 kJ + (2 3) mol 298.15 K

K mol = 1367 kJ/mol



Exercise 6.7 Calculating H from Bomb Calorimeter Data: 3.200 g of ethyl alcohol, C2H5OH(l), is burned in a bomb calorimeter. It contains 504.5 g of water. The heat capacity of the calorimeter is 0.415 kJ/C. This is determined in an experiment described in Exercise 6.8. The temperature rises from 20.15 C to 56.49 C. What is the heat of combustion of ethyl alcohol?

v cal w0.00418 kJq = C + m T

g C

( )0.415 kJ 0.00418 kJ = + 504.5 g 56.49 20.15 C = 91.7 kJC g C

32 5

2 2 5 2 5 2

46.069 g C H OH? kJ 91.7 kJ kJE = = = 1.32 x 10 mol H 3.200 g C H OH 1 mol C H OH mol C H OH

5

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)

H = E + (n)RT

3 0.008314 kJH = 1.32 x 10 kJ + (2 3) mol 298.15 KK mol

= 1.32 x 103 kJ/mol C2H5OH

Exercise 6.8 Calculating Heat capacity from Bomb Calorimeter Data: The heat of combustion of benzene, C6H6(l), is 2066 kJ/mole. When 0.200 g of benzene is burned in the bomb calorimeter mentioned in Exercise 6.7 which now contains 526.0 g of water, the temperature rises from 22.56 C to 24.58 C. What is the heat capacity of the calorimeter?

C6H6(l) + 15/2O2(g) 6CO2(g) + 3H2O(l)

E = H - (n)RT

0.008314 kJE = 2066 kJ + (6 7.5) mol 298.15 K = 2062 kJK mol

6 6v 6 6

6 6 6 6

1 mol C H 2062 kJq = 0.200 g C H = 5.28 kJ78.114 g C H mol C H

v cal w0.00418 kJq = C + m T

g C

vcal w

q 0.00418 kJC = mT g C

( )

5.28 kJ 0.00418 kJ= 526.0 g24.58 22.56 C g C

= 0.415 kJ/C



Exercise 6.9 Law of Hess: Given the following data,

Hcombustion C2H2(g) = 1301 kJ/mole

Hcombustion C2H6(g) = 1562 kJ/mole

H2O(l) H2(g) + 1/2O2(g) H = 286 kJ

calculate the H for the following reaction.

1/2 C2H2(g) + H2(g) 1/2 C2H6(g)

1/2C2H2(g) + 5/4O2(g) CO2(g) + 1/2H2O(l) H2 = 1/2 (1301 kJ)

H2(g) + 1/2O2(g) H2O(l) H2 = 1 (286 kJ)

CO2(g) + 3/2H2O(l) 1/2C2H6(g) + 7/4O2(g) H3 = 1/2 (1562 kJ)

Hnet = H1 + H2 + H3 = 156 kJ

Exercise 6.10 Heats of Formation: Calculate the H for the following reaction.

2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g)

Hrxn = 2 Hf ZnO(s) + 2 Hf SO2(g) 2 Hf ZnS(s)

= 2(347.98 kJ) + 2(296.06 kJ) 2(202.9 kJ) = 882.3 kJ


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