CHAPTER 5 Problems:1, 12, 16, 18, 20, 28, 30, 36, 38, 40, 48, 50, 60, 62, 66, 87, 96, 108, 122

1) Define the following terms: system, surroundings, thermal energy, chemical energy, potential energy, kinetic energy, law of conservation of energy.

system - A part of the universe that we choose to study.

surroundings - The part of the universe not included in the system.

thermal energy - The kinetic energy due to the random motion of the particles making up a substance.

potential energy - Energy possessed by a system due to its position or composition.

kinetic energy - Energy due to the motion of an object in a particular direction. The kinetic energy of an object with mass m moving with a speed v is EK = 1/2 mv2.

law of conservation of energy - The total energy in an isolated system is constant. That is, we can convert energy from one form into another form, but the total energy present does not change.

12) Explain what is meant by a state function. Give two examples of quantities that are state functions and two that are not state functions.

A state function is a property of a system whose change in value is path independent. Two examples of state functions are internal energy (U) and enthalpy (H). Two examples of quantities that are not state functions are heat (q) and work (w).

16) Calculate q, and determine whether heat is absorbed or released when a system does work on the surroundings equal to 64. J and U = 213. J.

From the first law, U = q + w.

Therefore q = U - w

When a system does work on the surroundings this is considered negative work, and so w = - 64. J. Since U = 213. J, it follows that

q = 213. J - (- 64. J) = 277. J

Since q > 0, heat is absorbed by the system from the surroundings.

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18) The diagram below (left) shows a system before a process. Determine which of the diagrams ( i, ii, or iii ) could represent the system after it undergoes a process for each of the following: NOTE - It will help to recall that for mechanical work w = - p V.

a) work is done on the system and U is negative

If work is done on the system w > 0. Therefore V < 0 (since p cannot be negative). Diagram iii is the only one where volume decreases, and so the only one that could represent this process.

b) the system releases heat and U is positive

U = q + w, and so w = U - q. The system releases heat, and so q < 0. Since U > 0 and ( - q) > 0, it follows that w > 0, which means V < 0. So again the only diagram that could represent this process is iii, since it is the only one where volume decreases.

c) the system absorbs heat and U is positive

If the system absorbs heat then q > 0. From part b we know w = U - q. Since both U and q are positive, w could be positive ( if U > q), zero (if U = q), or negative (if U < q). So any of the diagrams could correspond to the process taking place.

20) Define these terms: enthalpy and enthalpy of reaction. Under what conditions is the heat of a reaction equal to the enthalpy change for the same reaction?

Enthalpy is defined by the equation H = U + pV. Its physical significance is that H = q for a process if it takes place under conditions of constant pressure.

The enthalpy of reaction is the value for H for a balanced chemical equation when one mole of reaction occurs under conditions of constant pressure.

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28) Determine the amount of heat (in kJ) given off when 1.26 x 104 g of NO2 are produced according to the equation

2 NO(g) + O2(g) 2 NO2(g) H = - 114.6 kJ/mol

M(NO2) = 46.01 g/mol

n(NO2) = 1.26 x 104 g 1 mol NO2 = 273.9 mol NO2. 46.01 g

H = 273.9 mol NO2 1 mol reaction ( - 114.6 kJ) = - 1.57 x 104 kJ 2 mol NO2 mol reaction

Assuming that the reaction is carried out under conditions of constant pressure, q = H = - 1.57 x 104 kJ. Since q < 0, heat is given off by the system.

29) Consider the reaction

2 H2(g) + 2 Cl2(g) 2 HCl(g) H = - 184.6 kJ/mol

If 3.00 moles of H2 reacts with 3.00 moles of Cl2 to form HCl, calculate the work done (in joules) against a pressure of 1.00 atm. What is U for the reaction? Assume the reaction goes to completion and that V = 0. Note 1 Latm = 101.3 J.

Since w = - p V, and V = 0, it follows that w = 0.

Since H = U + pV, H = U + (pV) = U + (pfVf - piVi), where i = initial value and f = final value. But pressure is constant (at 1.00 atm) and V is constant (since V = 0), and so for this process H = U.

Finally, H = 3.00 mol H2 1 mol reaction ( - 184.6 kJ) = - 276.9 kJ 2 mol H2 mol reaction

36) Calculate the amount of heat liberated (in kJ) from 366. g of mercury when it cools from 77.0 C to 12.0 C.

q = sm(T). From Table 5.2, s(Hg(l)) = 0.139 J/gC.

So q = (0.139 J/gC)(366. g)(12.0 C - 77.0 C) = - 3.31 x 103 J = - 3.31 kJ

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38) A 0.1375 g sample of solid magnesium is burned in a constant volume bomb calorimeter that has a heat capacity of 3024. J/C. The temperature increases by 1.126 C. Calculate the heat given off by the burning Mg in kJ/g and kJ/mol.

q (per gram) = - C ( T) = - (3024. J/ C) (1.126 C) = - 24.76 kJ/g m 0.1375 g

q (per mol) = - 24.76 kJ 24.31 g = - 602.0 kJ/mol g mol

40) A 50.75 g sample of water at 75.6 C is added to a sample of water at 24.1 C in a constant pressure calorimeter. If the final temperature of the combined water is 39.4 C and the heat capacity of the calorimeter is 26.3 J/C, calculate the mass of water originally in the calorimeter.

We know the heat gained by the cold water and calorimeter must be equal to the heat lost by the hot water.

Let m = mass of water originally in the calorimeter. Also recall that for water s = 4.184 J/gC. So

m (4.184 J/gC)(39.4 C - 24.1 C) + (26.3 J/C)(39.4 C - 24.1 C)

= (50.75 g)(4.184 J/gC)(75.6 C - 39.4 C)

If we multiply the above out (and do not include the units) we get

64.015 m + 402.39 = 7686.64

m = 7686.64 - 402.39 = 114. g water 64.015

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48) From the following data

C(graphite) + O2(g) CO2(g) Hrxn = - 393.5 kJ/mol

H2(g) + 1/2 O2(g) H2O(l) Hrxn = - 285.8 kJ/mol

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l) Hrxn = - 3119.6 kJ/mol

calculate the enthalpy change for the reaction

2 C(graphite) + 3 H2(g) C2H6(g)

2 C(graphite) + 2 O2(g) 2 CO2(g) Hrxn = 2 (- 393.5 kJ/mol)

2 CO2(g) + 3 H2O(l) C2H6(g) + 7/2 O2(g) Hrxn = (-1/2) (- 3119.6 kJ/mol)

3 H2(g) + 3/2 O2(g) 3 H2O(l) Hrxn = 3 (- 285.8 kJ/mol)____________________________________________________________________

2 C(graphite) + 3 H2(g) C2H6(g) Hrxn = - 84.6 kJ/mol

50) Calculate the standard enthalpy change for the reaction

2 Al(s) + Fe2O3(s) 2 Fe(s) + Al2O3(s)

given that

2 Al(s) + 3/2 O2(g) Al2O3(s) Hrxn = - 1669.8 kJ/mol

2 Fe(s) + 3/2 O2(g) Fe2O3(s) Hrxn = - 822.2 kJ/mol

2 Al(s) + 3/2 O2(g) Al2O3(s) Hrxn = - 1669.8 kJ/mol

Fe2O3(s) 2 Fe(s) + 3/2 O2(g) Hrxn = - (- 822.2 kJ/mol)___________________________________________________________________ 2 Al(s) + Fe2O3(s) 2 Fe(s) + Al2O3(s) Hrxn = - 847.6 kJ/mol

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60) The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to the enthalpy of formation of H+ ion; that is, by saying Hf(H+(aq)) = 0.

a) For the reaction

HCl(g) H+(aq) + Cl-(aq) Hrxn = - 74.9 kJ/mol

find Hf for the Cl-(aq) ion. Note Hf(HCl(g)) = - 92.3 kJ/mol.

Hrxn = [Hf(H+(aq)) + Hf(Cl-(aq))] - [Hf(HCl(g))]

So Hf(Cl-(aq)) = Hrxn + Hf(HCl(g)) - Hf(H+(aq))

= (- 74.9 kJ/mol) + (- 92.3 kJ/mol) + 0.0 kJ/mol = - 167.2 kJ/mol

b) Given that Hf(OH-(aq)) = -229.6 kJ/mol, find the enthalpy change for the neutralization reaction of a strong acid and a strong base, such as

HCl(aq) + KOH(aq) KCl(aq) + H2O(l)

Note Hf(H2O(l)) = - 285.8 kJ/mol. HINT: Start by writing the net ionic equation corresponding to the above neutralization reaction.

The net ionic equation for the neutralization reaction is

H+(aq) + OH-(aq) H2O(l)

So Hrxn = [Hf(H2O(l))] - [Hf(H+(aq)) + Hf(OH-(aq))]

= (- 285.8 kJ/mol) - [(0.0 kJ/mol) + - 229.6 kJ/mol)] = - 56.2 kJ/mol

62) Calculate the heat of combustion for the following reactions from the standard enthalpies of formation given in Appendix 2 of Burdge. Note that reaction b assumes the combustion product for sulfur is SO2(g).

a) C2H4(g) + 3 O2(g) 2 CO2(g) + 2 H2O(l)

Hrxn = [2 Hf(CO2(g)) + 2 Hf(H2O(l))] - [Hf(C2H4(g)) + 3 Hf(O2(g))]

= [2(-393.5 kJ/mol) + 2(-285.8 kJ/mol)] - [(52.3 kJ/mol) = 3(0.0 kJ/mol)]

= - 1410.9 kJ/mol The reaction meets the definition of a combustion reaction, so this value is the heat of combustion for C2H4(g).

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b) 2 H2S(g) + 3 O2(g) 2 H2O(l) +2 SO2(g)

Hrxn = [2 Hf(H2O(l)) + 2 Hf(SO2(g))] - [2 Hf(H2S(g)) + 3 Hf(O2(g))]

= [2(-285.8 kJ/mol) + 2(-296.4 kJ/mol)] - [2(-20.15 kJ/mol) = 3(0.0 kJ/mol)]

= - 1124.1 kJ/mol Since the reaction involves 2 moles of H2S reacting with oxygen to form combustion products, the value for Hrxn is twice the value for the heat of combustion for H2S(g). So the heat of combustion is 1/2 (- 1124.1 kJ/mol) = - 562.0 kJ/mol.

66) Calculate the heat of decomposition for the process below at constant pressure and T = 25.0 C.

CaCO3(s) CaO(s) + CO2(g)

(Data are given in Appendix 2 of Burdge.)

Hrxn = [Hf(CaO(s)) + Hf(CO2(g))] - [Hf(CaCO3(s))]

= [(-635.6 kJ/mol) + (-393.5 kJ/mol)] - (-1206.9 kJ/mol) = 177.8 kJ/mol

87) For which of the following reactions does Hrxn =Hf ?a) H2(g) + S(rhombic) H2S(g) Yes (the rhombic form of solid S is the

thermodynamically most stable form.

b) C(diamond) + O2(g) CO2(g) No (the graphite form of solid C, not the diamond form, is the more stable form of the solid.

c) H2(g) + CuO(s) H2O(l) + Cu(s) No. Not a formation reaction.

d) O(g) + O2(g) O3(g) No. Not a formation reaction. The formationreaction for O3 is (3/2 O2(g) O3(g))

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96) Acetylene (C2H2) and benzene (C6H6) have the same empirical formula. In fact, benzene can be made from acetylene by the process

3 C2H2(g) C6H6(l)

The enthalpies of combustion for acetylene and benzene are

Hcom(C2H2(g)) = - 1299.4 kJ/mol Hcom(C6H6(l)) = - 3267.4 kJ/mol

Using this information and the enthalpies of formation for carbon dioxide and water

Hf(CO2(g)) = - 393.5 kJ/mol Hf(H2O(l)) = - 285.8 kJ/mol

find the enthalpies of formation for C2H2(g) and C6H6(l). From those values find the enthalpy change for the reaction that converts acetylene into benzene.

It is a good idea to start by writing the formation and combustion reactions

2 C(s, graphite) + H2(g) C2H2(g) formation rxn for C2H2(g)6 C(s, graphite) + 3 H2(g) C6H6(l) formation rxn for C6H6(l)

C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O(l) Hcom = - 1299.4 kJ/molC6H6(l) + 15/2 O2(g) 6 CO2(g) + 3 H2O(l) Hcom = - 3267.4 kJ/mol

C(s,graphite) + O2(g) CO2(g) Hf = - 393.5 kJ/mol H2(g) + 1/2 O2(g) H2O(l) Hf = - 285.8 kJ/mol

So 2 CO2(g) + H2O(l) C2H2(g) + 5/2 O2(g) H = - (-1299.4 kJ/mol)

2 C(s,graphite) + 2 O2(g) 2 CO2(g) H = 2 (- 393.5 kJ/mol)

H2(g) + 1/2 O2(g) H2O(l) H = - 285.8 kJ/mol ________________________________________________________________2 C(s,graphite) + H2(g) C2H2(g) Hf = 226.6 kJ/mol

6 CO2(g) + 3 H2O(l) C6H6(l) + 15/2 O2(g) H = - (-3267.4 kJ/mol)

6 C(s,graphite) + 6 O2(g) 6 CO2(g) H = 6 (- 393.5 kJ/mol)

3 H2(g) + 3/2 O2(g) 3 H2O(l) H = 3 (- 285.8 kJ/mol) ________________________________________________________________6 C(s,graphite) + 3 H2(g) C6H6(l) Hf = 49.0 kJ/mol

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We can now find the enthalpy change for the reaction that convert acetylene into benzene.

3 C2H2(g) C6H6(l)

Hrxn = Hf(C6H6(l)) - 3 Hf(C2H2(g))

= (49.0 kJ/mol) - 3(226.6 kJ/mol) = - 630.8 kJ/mol

108) A 3.52 g sample of ammonium nitrate (NH4NO3) was added to 80.0 mL of water in a constant pressure calorimeter of negligible heat capacity. As a result the temperature of the solution decreased from 21.6 C to 18.1 C. Calculate the heat of solution for ammonium nitrate, the enthalpy change corresponding to the process

NH4NO3(s) NH4+(aq) + NO3

-(aq)

Assume that the specific heat of the solution is the same as the specific heat of water.

M(NH4NO3) = 80.04 g/mol

If we assume the specific heat of the solution is the specific heat of the water, and that the density of water is 1.00 g/mL, then

q = - ms(T) = - (80.0 g)(4.184 J/gC)(18.1 C - 21.6 C) = 1172. J

H = 1172. J = 333.0 J/g 3.52 g

The molar enthalpy of solution for ammonium nitrate is then

H = 333.0 J 80.04 g = 26.6 kJ/mol g mol

122) Calculate the standard enthalpy change for the fermentation process, in which glucose (C6H12O6(s)) is converted to ethanol (C2H5OH(l)) and carbon dioxide. Data are given in Appendix 2 of Burdge.

The reaction is

C6H12O6(s) 2 C2H5OH(l) + 2 CO2(g)

Hrxn = [2 Hf(C2H5OH(l)) + 2 Hf(CO2(g))] - [Hf(C6H12O6(s))]

= [2(-276.98 kJ/mol) + 2(-393.5 kJ/mol)] - (-1274.5 kJ/mol) = - 66.5 kJ/mol

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