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

SOLUTIONS - CHAPTER 10 Problems NOTE: The final exam is Thursday, December 8 th, from 2:15pm to 4:45pm. 1) (Burdge, 10.1) Define the following terms: system, surroundings, thermal energy, chemical 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. 2) Define the following terms: potential energy, kinetic energy, law of conservation of energy. 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 mv 2. 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. 3) (Burdge, 10.9) 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). 4) Calculate q, and determine whether heat is absorbed or released when a system does work on the surroundings equal to 85. J and U = 240. 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 = - 85. J. Since U = 240. J, it follows that q = 240. J - (- 85. J) = 325. J Since q > 0, heat is absorbed by the system from the surroundings. 1

5) Consider a process involving a gas confined in a cylinder by a piston. During the process 635. J of heat is added to the gas, and the gas expands against a constant pressure pex = 2.00 atm from an initial volume Vi = 10.00 L to a final volume Vf =12.00 L. Find q, w, and U for the process. Heat is added to the system, and so q is positive. Therefore q = + 635. J The gas expands against a constant external pressure, and so w = - pex (Vf - Vi) = - (2.00 atm) (12.00 L - 10.00 L) = - 4.00 L atm (101.3 J/L atm) = -405. J From the first law, U = q + w = 635. J + (- 405. J) = + 230. J 6) (Burdge, 10.15) 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. 2

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. 7) (Burdge, 10.17) 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. 8) (Burdge, 10.24) Determine the amount of heat (in kj) given off when 1.26 x 10 4 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 10 4 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 10 4 kj 2 mol NO2 mol reaction Assuming that the reaction is carried out under conditions of constant pressure, q = H = - 1.57 x 10 4 kj. Since q < 0, heat is given off by the system. 9) (Burdge, 10.26) Consider the reaction H2(g) + 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 L atm = 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. 3

Finally, H = 3.00 mol H2 1 mol reaction ( - 184.6 kj) = - 553.8 kj 1 mol H2 mol reaction 10) (Burdge, 10.32) 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 = s m ( T). From Table 5.2, s(hg(l)) = 0.139 J/g C. So q = (0.139 J/g C)(366. g)(12.0 C - 77.0 C) = - 3.31 x 10 3 J = - 3.31 kj 11) (Burdge, 10.34) 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 g 24.31 g = - 602.0 kj/mol mol 12) (Burdge, 10.40) 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/g C. So m (4.184 J/g C)(39.4 C - 24.1 C) + (26.3 J/ C)(39.4 C - 24.1 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 = (50.75 g)(4.184 J/g C)(75.6 C - 39.4 C) 4

13) (Burdge, 10.46) Calculate the standard enthalpy change for the reaction given that 2 Al(s) + Fe2O3(s) 2 Fe(s) + Al2O3(s) 2 Al(s) + 3 /2 O2(g) Al2O3(s) H rxn = - 1601. kj/mol 2 Fe(s) + 3 /2 O2(g) Fe2O3(s) H rxn = - 821. kj/mol 2 Al(s) + 3 /2 O2(g) Al2O3(s) H rxn = - 1601. kj/mol Fe2O3(s) 2 Fe(s) + 3 /2 O2(g) H rxn = - (- 821. kj/mol) 2 Al(s) + Fe2O3(s) 2 Fe(s) + Al2O3(s) H rxn = - 780. kj/mol 14) (Burdge, 10.48) From the following data C(graphite) + O2(g) CO2(g) H 2 (g) + 1 / 2 O 2 (g) H 2 O( ) H rxn = - 393.5 kj/mol H rxn = - 285.8 kj/mol 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O( ) H rxn = - 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) H rxn = 2 (- 393.5 kj/mol) 2 CO2(g) + 3 H2O(l) C2H6(g) + 7 /2 O2(g) H rxn = (- 1 /2) (- 3119.6 kj/mol) 3 H2(g) + 3 /2 O2(g) 3 H2O( ) H rxn = 3 (- 285.8 kj/mol) 2 C(graphite) + 3 H2(g) C2H6(g) H rxn = - 84.6 kj/mol 15) Using the data in Appendix 2 find H rxn for the following processes, all of which take place at p = 1.00 atm and T = 25. C. Note that the general expression for the enthalpy change for a chemical reaction is H rxn = ( H f(products)) - ( H f(reactants)) 5

a) Ca(s) + 2 H2O( ) Ca(OH)2(s) + H2(g) H rxn = [ H f(ca(oh)2(s)) + H f(h2(g))] - [ H f(ca(s)) + 2 H f(h2o( ))] = [(- 986.6 kj/mol) + (0.0 kj/mol)] - [(0.0 kj/mol) + 2(- 285.8 kj/mol)] = - 415.0 kj/mol b) 4 FeO(s) + O2(g) 2 Fe2O3(s) H rxn = [2 H f(fe2o3(s))] - [4 H f(feo(s)) + H f(o2(g)] = [2 (- 822.2 kj/mol)] - [4 (- 272.0 kj/mol) + (0.0 kj/mol)] = - 556.4 kj/mol 16) (Burdge, 10.56) 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 H f(h + (aq)) = 0. a) For the reaction HCl(g) H + (aq) + Cl - (aq) H rxn = - 74.9 kj/mol find H f for the Cl - (aq) ion. Note H f(hcl(g)) = - 92.3 kj/mol. H rxn = [ H f(h + (aq)) + H f(cl - (aq))] - [ H f(hcl(g))] So H f(cl - (aq)) = H rxn + H f(hcl(g)) - H f(h + (aq)) = (- 74.9 kj/mol) + (- 92.3 kj/mol) + 0.0 kj/mol = - 167.2 kj/mol b) Given that H f(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( ) Note H f(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( ) So H rxn = [ H f(h2o( ))] - [ H f(h + (aq)) + H f(oh - (aq))] = (- 285.8 kj/mol) - [(0.0 kj/mol) + - 229.6 kj/mol)] = - 56.2 kj/mol 6

17) (Burdge, 10.57) 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.) H rxn = [ H f(cao(s)) + H f(co2(g))] - [ H f(caco3(s))] = [(-635.6 kj/mol) + (-393.5 kj/mol)] - (-1206.9 kj/mol) = 177.8 kj/mol 18) (Burdge, 10.58) 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( ) H rxn = [2 H f(co2(g)) + 2 H f(h2o( ))] - [ H f(c2h4(g)) + 3 H f(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). b) 2 H2S(g) + 3 O2(g) 2 H2O( ) +2 SO2(g) H rxn = [2 H f(h2o(l)) + 2 H f(so2(g))] - [2 H f(h2s(g)) + 3 H f(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 H rxn 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. 19) (Burdge, 10.74) From the following data find the average bond enthalpy for an N-H bond NH3(g) NH2(g) + H(g) NH2(g) NH(g) + H(g) NH(g) N(g) + H(g) H = 435. kj/mol H = 381. kj/mol H = 360. kj/mol 7

From Hess law, the combination of the above three processes is equivalent to NH3(g) N(g) + 3 H(g) So the average N-H bond enthalpy is 1 /3 [ 435. kj/mol + 381. kj/mol + 360. kj/mol ] ; or 392. kj/mol (close to the value in Table 10.4). 20) Using the table of average bond enthalpies estimate the value for H rxn for the following process CH3SCH3(g) + O2(g) C2H6(g) + SO2(g) A general expression for the approximate change in enthalpy for a gas phase reaction is H rxn [ (bonds broken)] - [ (bonds formed)] bonds broken bonds formed 6 (C-H) = 6 (414) = 2484. 6 (C-H) = 6 (414) = 2484. 2 (C-S) = 2 (255) = 510. 1 (C-C) = 1 (347) = 347. 1 (O=O) = 1 (498.7) = 498.7 2 (S=O) = 2 (469) = 938. = 3493. = 3769. And so H rxn [ 3493. kj/mol ] - [ 3769. kj/mol ] = - 276. kj/mol 21) (Burdge, 10.93) Calculate the value for H for the reaction H2(g) + I2(g) 2 HI(g) using eq 10.18, and using eq 10.19. Note that H f(i2(g)) = 61.0 kj/mol. From 10.18 From 10.19 H = [ (H-H) + (I-I) ] - [ 2 (H-I) ] = [ (436.4 kj/mol) + (151.0 kj/mol) ] - [ 2(298.3 kj/mol) ] = - 9.2 kj/mol H = [ 2 H f(hi(g)) ] - [ H f(h2(g)) + H f(i2(g)) ] = [ 2(25.9 kj/mol) ] - [ (0.0 kj/mol) + (61.0 kj/mol) ] = - 9.2 kj/mol 8

The agreement is not as impressive as it looks. If you examine Table 10.4, you will see all of the bond enthalpies used in the calculation using eq 10.18 are for diatomic molecules, and are therefore exact (and not average) values. So this is not a true test of the accuracy of eq 10.18 for estimating enthalpies of reaction. 22) (Burdge, 10.82) Calculate the lattice energy for CaCl2. Use data from Fig. 7.8 and Fig. 7.10, and data from Appendix 2. Note that the second ionization energy for Ca is IE2 = 1145. kj/mol. To do this problem we need a Born-Haber cycle. Ca(s) Ca(g) H f(ca(g)) Cl2 (g) 2 Cl(g) 2 H f(cl(g)) Ca(g) Ca + (g) + e - IE1(Ca) Ca + (g) Ca 2+ (g) + e - IE2(Ca) 2 Cl(g) + 2 e - 2 Cl - (g) 2 EA(Cl) Ca 2+ (g) + 2 Cl - (g) CaCl2(s) - H lattice(cacl2) Ca(s) + Cl2(g) CaCl2(s) H f(cacl2(s)) From Hess law H f(cacl2(s)) = H f(ca(g)) + 2 H f(cl(g)) + IE1(Ca) + IE2(Ca) and so + 2 EA(Cl) - H lattice(cacl2(s)) H lattice(cacl2(s)) = H f(ca(g)) + 2 H f(cl(g)) + IE1(Ca) + IE2(Ca) + 2 EA(Cl) - H f(cacl2(s)) = 179.3 kj/mol + 2 (121.7 kj/mol) + 590. kj/mol + 1145. kj/mol = + 2255. kj/mol + 2 ( - 349. kj/mol) - (- 794.96 kj/mol) 23) (Burdge, 10.107) For which of the following reactions does H rxn = H f? 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) H 2 (g) + CuO(s) H 2 O(l) + Cu(s) No. Not a formation reaction. 9

d) O(g) + O2(g) O3(g) No. Not a formation reaction. The formation reaction for O3 is ( 3 /2 O2(g) O3(g)) 24) (Burdge, 10.131) 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( ) The enthalpies of combustion for acetylene and benzene are H com(c2h2(g)) = - 1299.4 kj/mol H com(c6h6( )) = - 3267.4 kj/mol Using this information and the enthalpies of formation for carbon dioxide and water H f(co2(g)) = - 393.5 kj/mol H f(h2o( )) = - 285.8 kj/mol find the enthalpies of formation for C2H2(g) and C6H6( ). 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( ) formation rxn for C6H6( ) C2H2(g) + 5 /2 O2(g) 2 CO2(g) + H2O( ) C6H6(l) + 15 /2 O2(g) 6 CO2(g) + 3 H2O( ) C(s,graphite) + O 2 (g) CO 2 (g) H2(g) + 1 /2 O2(g) H2O( ) H com = - 1299.4 kj/mol H com = - 3267.4 kj/mol H f = - 393.5 kj/mol H f = - 285.8 kj/mol So 2 CO2(g) + H2O( ) 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( ) H = - 285.8 kj/mol 2 C(s,graphite) + H2(g) C2H2(g) H f = 226.6 kj/mol 6 CO2(g) + 3 H2O( ) C6H6( ) + 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( ) H = 3 (- 285.8 kj/mol) 6 C(s,graphite) + 3 H2(g) C6H6( ) H f = 49.0 kj/mol 10

We can now find the enthalpy change for the reaction that converts acetylene into benzene. 3 C2H2(g) C6H6( ) H rxn = H f(c6h6( )) - 3 H f(c2h2(g)) = (49.0 kj/mol) - 3(226.6 kj/mol) = - 630.8 kj/mol 25) 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 = - m s ( T) = - (80.0 g)(4.184 J/g C)(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 26) (Burdge, 10.112) 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( ) + 2 CO2(g) H rxn = [2 H f(c2h5oh( )) + 2 H f(co2(g))] - [ H f(c6h12o6(s))] = [2(-276.98 kj/mol) + 2(-393.5 kj/mol)] - (-1274.5 kj/mol) = - 66.5 kj/mol 11