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1 Chemistry 360 Dr. Jean M. Standard Problem Set 5 Solutions 1. Determine the amount of pressure-volume work performed by 1 mole of water freezing to ice at 0 C and 1 atm pressure. The density of liquid water at 0 C is g/ml and the density of ice is g/ml. Going from liquid water to ice involves a volume change, so PV work must be determined. The definition of work w is w = P ext dv. At constant pressure (standard, 1 atm), we can pull the pressure out of the integral and integrate the volume from the initial to the final value, w = P ext V1 V 2 dv w = P ext ( V 2 V 1 ). To determine the volumes, we must use the densities given. For 1 mole of liquid water, the volume is For 1 mole of ice, the volume is The work is therefore V 1 = ( 1 mol) ( g/mol) V 1 = ml = L. V 2 = ( 1 mol) ( g/mol) " ml % $ ' # g& V 2 = ml = L. w = P ext V 2 V 1 " ml % $ ' # g & ( L L) = 1atm # = L atm J & % ( $ 1L atm ' w = 0.165J.

2 2. One of the primary components of nuclear fuel rods is uranium(iv) oxide, UO 2. The melting point of UO 2 is 3140 K and the enthalpy of fusion is 70.0 kj/mol. Determine the enthalpy change per mole for the process 2 UO 2 ( s, 3100 K) UO 2 l, 3200 K. = J mol 1 K 1 and C p,m ( l) =134.2 J mol 1 K 1. Assume 1 bar The molar heat capacities are C p,m s pressure and that the molar heat capacities of solid and liquid UO 2 are independent of temperature. For a physical change involving heating/cooling and/or phase transitions, we can break the process up into steps and then calculate the enthalpy changes for each of the steps. The overall enthalpy change is then just the sum of the enthalpy changes for the steps since H is a state function. For this particular process, the steps are: ( 1) heating solid UO 2 from 3100 K to 3140 K, UO 2 ( s, 3100 K) UO 2 ( s, 3140 K). ( 2) phase transition from solid to liquid at 3140 K, UO 2 ( s, 3140 K) UO 2 ( l, 3140 K). ( 3) heating liquid UO 2 from 3140 K to 3200 K, UO 2 ( l, 3140 K) UO 2 ( l, 3200 K). The overall molar enthalpy change for the process is then just equal to the sum of the molar enthalpy changes for each of the steps, ΔH m = ΔH 1,m + ΔH 2,m + ΔH 3,m. In order to calculate the molar enthalpy changes for each of the steps, the molar heat capacities for the liquid and solid phases are required, along with the molar enthalpy of fusion; these quantities are provided in the problem. The steps listed above can be divided into two types: heating steps and phase transition steps. For the heating steps, we can begin with the exact differential for H, dh = C p dt + # H & % ( dp. $ P ' T Since the process occurs at constant pressure, dp = 0, the expression becomes dh = C p dt. Integrating in order to calculate the enthalpy change for the heating process from T 1 to T 2, ΔH = T 2 C p dt. T 1 If we assume that the heat capacity is independent of temperature, then we can integrate to get ΔH = C p ( T 2 T 1 ). Expressing this relation in terms of molar quantities, we divide both sides by the number of moles to give ΔH m = C p,m ( T 2 T 1 ).

3 2.) Continued 3 For the steps involving phase transitions, the enthalpy change is just the moles of substance multiplied by the standard molar enthalpy change (in this case, fusion), ΔH m = ΔH fus Using these expressions, the enthalpy changes for each of the steps may be calculated. Step 1 In this step, solid UO 2 is heated from 3100 K to the transition temperature of 3140 K. Substituting the appropriate heat capacity value and temperatures, we have ΔH 1,m = C p,m ( UO 2, s) T 2 T K 3100 K = J mol 1 K 1 ΔH 1 = 8440 J/mol.. Step 2 In this step, UO 2 undergoes a transition from solid to liquid at the transition temperature of 3140 K. Substituting the appropriate standard molar enthalpy change, we have ΔH 2,m = ΔH fus = ( 70.0 kj/mol) ΔH 2 = J/mol. Step 3 In this step, liquid UO 2 is heated from the transition temperature of 3140 K to 3200 K. Substituting the appropriate heat capacity value and temperatures, we have ΔH 3,m = C p,m ( UO 2,l) ( T 2 T 1 ) 3200 K 3140 K = J mol 1 K 1 ΔH 3 = 8052 J. The overall enthalpy change for the process is then just equal to the sum of the enthalpy changes for each of the steps, ΔH m = ΔH 1,m + ΔH 2,m + ΔH 3,m. Substituting, we have ΔH m = ΔH 1,m + ΔH 2,m + ΔH 3,m = J/mol ΔH = J/mol or kj/mol.

4 3. Determine the enthalpy change for one mole of water undergoing the following transformation, 4 H 2 O ( s, 30 C) H 2 O ( g, 300 C). The pressure is constant at 1 bar, and assume that the molar heat capacities are independent of temperature. For a physical change involving heating/cooling and/or phase transitions, we can break the process up into steps and then calculate the enthalpy changes for each of the steps. The overall enthalpy change is then just the sum of the enthalpy changes for the steps. For this particular process, the steps are: ( 1) heating solid water from 30 C to 0 C, H 2 O s, 30 C ( 2) phase transition from solid to liquid at 0 C, H 2 O s, 0 C ( 3) heating liquid water from 0 C to 100 C, H 2 O l, 0 C ( 4) phase transition from liquid to gas at 100 C, H 2 O l, 100 C ( 5) heating gaseous water from 100 C to 300 C, H 2 O g, 100 C H 2O ( s, 0 C). H 2 O ( l, 0 C). H 2 O ( l, 100 C). H 2O ( g, 100 C). H 2 O ( g, 300 C). The overall enthalpy change for the process is then just equal to the sum of the enthalpy changes for each of the steps, ΔH = ΔH 1 + ΔH 2 + ΔH 3 + ΔH 4 + ΔH 5. In order to calculate the enthalpy changes for each of the steps, the molar heat capacities for all three phases are required, along with the molar enthalpies of fusion and vaporization. These values are available from standard reference sources such as your textbook, the CRC, or the NIST Webbook, and are tabulated below. Substance C p,m (Jmol 1 K 1 ) Phase change for H 2 O ΔH R (kj/mol) H 2 O (s) 36.2 fusion 6.01 H 2 O (l) 75.3 vaporization 44.0 H 2 O (g) 33.6 The steps listed above can be divided into two types: heating steps and phase transition steps. For the heating steps, we can begin with the exact differential for H, dh = C p dt + # H & % ( dp. $ P ' T Since the process occurs at constant pressure, dp = 0, the expression becomes dh = C p dt. Integrating in order to calculate the enthalpy change for the heating process from T 1 to T 2, ΔH = T 2 C p dt. T 1

5 3.) Continued 5 If we assume that the heat capacity is independent of temperature, then we can integrate to get ΔH = C p ( T 2 T 1 ). Expressing this relation in terms of the molar heat capacity, we have ΔH = nc p,m ( T 2 T 1 ). For the steps involving phase transitions, the enthalpy change is just the moles of substance multiplied by the standard molar enthalpy change for the phase transition, ΔH = nδh phase change Using these expressions, the enthalpy changes for each of the steps may be calculated. Step 1 In this step, solid water is heated from 30ºC to the transition temperature of 0ºC. Substituting the appropriate heat capacity value and temperatures, we have ΔH 1 = nc p,m ( H 2 O,s) ( T 2 T 1 ) = ( 1 mol) 36.2 J mol 1 K 1 ΔH 1 = 1090 J.. ( K K ) Step 2 In this step, water undergoes a transition from solid to liquid at the transition temperature of 0ºC. Substituting the appropriate standard molar enthalpy change, we have ΔH 2 = nδh fus ( 6.01 kj/mol) = 1 mol ΔH 2 = 6010 J. Step 3 In this step, liquid water is heated from 0ºC to the transition temperature of 100ºC. Substituting the appropriate heat capacity value and temperatures, we have ΔH 3 = nc p,m ( H 2 O,l) ( T 2 T 1 ) = ( 1 mol) 75.3 J mol 1 K 1 ΔH 3 = 7530 J. ( K K ) Step 4 In this step, water undergoes a transition from liquid to vapor at the transition temperature of 100ºC. Substituting appropriate standard molar enthalpy change, we have ΔH 4 = nδh vap ( 44.0 kj/mol) = 1 mol ΔH 4 = J.

6 3.) Continued 6 Step 5 Finally, in this step, gaseous water is heated from 100ºC to the final temperature of 300ºC. Substituting the appropriate heat capacity value and temperatures, we have ΔH 5 = nc p,m ( H 2 O,g) ( T 2 T 1 ) = ( 1 mol) 33.6 J mol 1 K 1 ΔH 5 = 6720 J. ( K K ) The overall enthalpy change for the process is then just equal to the sum of the enthalpy changes for each of the steps, Substituting, we have ΔH = ΔH 1 + ΔH 2 + ΔH 3 + ΔH 4 + ΔH 5. ΔH = ΔH 1 + ΔH 2 + ΔH 3 + ΔH 4 + ΔH 5 = J ΔH = J or kj.

7 4. Metabolic activity in the human body releases about kj of heat per day. Since the body is made up of mostly water, approximate a body as consisting of 50 kg of water. 7 (a) How fast would the body temperature rise (in degrees K/day) if the body is considered to be an isolated system at constant pressure (1 bar)? The molar heat capacity of liquid water is C p,m = J mol 1 K 1. Since the process occurs at constant pressure, q p = ΔH = T 2 C p dt. T 1 If we assume that the heat capacity of the body (i. e., the 50 kg of water) is independent of temperature, then we can integrate to get in terms of the molar heat capacity. q p = ΔH = C p ( T 2 T 1 ) = nc p,m ( T 2 T 1 ), We are told how much heat is produced per day, and we know the number of moles of water in 50 kg is " ( 50 kg) 1000 g %" 1mol % $ ' $ ' = 2775 mol. # 1kg &# g & We need to figure out the temperature change. Solving the equation above for the temperature difference, q p = nc p,m ΔT or ΔT = q p nc p,m. Substituting, ΔT = q p nc p,m = ΔT = 48 K J ( 2775mol) J mol 1 K 1 Thus, the temperature rise per day is ΔT = 48 K per day. (b) How much water in kg per day must the body eliminate as perspiration to maintain normal body temperature of 98.6 F? Assume that the energy required to vaporize water is 2.41 kj/g. To maintain a constant body temperature of 98.6 F, the body would have to eliminate kj of heat per day. If this were done by using the heat generated to vaporize water, each g of water would require 2.41 kj of heat. Thus, the amount of water vaporized per day would be kj 2.41kJ/g = 4150 g. In other words, 4.15 kg of water would be vaporized as perspiration every day in order to maintain constant body temperature

8 5. Consider the following compounds: NaHCO 3 ( s), Na 2 CO 3 ( s), CO 2 ( g), and H 2 O( l). 8 (a) Write out the formation reactions for each of the compounds listed above. The formation reaction of NaHCO 3 (s) is The formation reaction of Na 2 CO 3 (s) is The formation reaction of CO 2 (g) is The formation reaction of H 2 O (l) is Na( s) H 2( g ) + C ( s) O 2( g ) NaHCO 3 ( s). 2 Na( s) + C ( s) O 2( g ) Na 2 CO 3 ( s). C ( s) + O 2 ( g) CO 2 ( g). H 2 ( g) O 2 ( g ) H 2 O ( l). (b) Show that the formation reactions for the compounds listed above may be combined to produce an overall reaction given by 2 NaHCO 3 ( s) Na 2 CO 3 ( s) + CO 2 ( g) + H 2 O( l). Now we have to take the formation reactions from part (a) in the appropriate combination to form the overall reaction given here. Taking twice the reverse of the first reaction and adding all the others yields [ Na( s) H 2( g) + C ( s) O 2( g) ] + C ( s) O 2( g) Na 2 CO 3 ( s) + O 2 ( g) CO 2 ( g) O 2 ( g ) H 2 O ( l) 2 NaHCO 3 s 2 Na s C s H 2 g 2 NaHCO 3 ( s) Na 2 CO 3 ( s) + CO 2 ( g) + H 2 O ( l)

9 5.) Continued 9 (c) Use values of standard enthalpies of formation from the CRC or NIST to calculate ΔH R 25 C for the reaction given in part (b). in kj/mol at The standard enthalpies of formation at 25 C as obtained from the CRC are: Compound ΔH f (kj/mol) NaHCO 3 (s) Na 2 CO 3 (s) CO 2 (g) H 2 O (l) The overall enthalpy of reaction is therefore given by ΔH R = ΔH f ( Na 2 CO 3 ) + ΔH f ( CO 2 ) + ΔH f ( H 2 O) 2ΔH f ( NaHCO 3 ) + ( kJ/mol) + ( kJ/mol) 2( kJ/mol) = kJ/mol ΔH R = 91.51kJ/mol.

10 6. The following may be considered as reactions used to power rockets, 10 ( a) H 2 ( g) O 2 ( g) H 2 O ( g) ( b) CH 3 OH( l) O 2 ( g) CO 2 ( g) + 2 H 2 O ( g) ( c) H 2 ( g) + F 2 ( g) 2 HF ( g). (a) Using values from the tables in the Appendix of your textbook, calculate the molar enthalpies of reaction at 25 C for each of these reactions. Convert your results to units of kj per total mass of reactants (in kg). For reaction (a), H 2 ( g) O 2 ( g ) H 2 O ( g), the enthalpy of reaction is ΔH R = ΔH f ( H 2 O,g) ΔH f = 241.8kJ/mol 0 0 ΔH R = 241.8kJ/mol. ( H 2 ) 1 ΔH 2 f ( O 2 ) Expressing this per kg of reactants, the weight of 1 mole of H 2 and 1/2 mole of O 2 is g, or kg. Thus, # ΔH R = ( 241.8kJ/mol) 1molreactants & % ( $ kg ' ΔH R = 13420kJ/kgreactants. For reaction (b), CH 3 OH ( l) O 2 ( g ) CO 2 ( g) + 2 H 2 O ( g), the enthalpy of reaction is ΔH R = ΔH f ( CO 2 ) + 2ΔH f ( H 2 O,g) ΔH f ( CH 3 OH) 2 3 ΔH f ( O 2 ) ( 239.2kJ/mol) 0 = 393.5kJ/mol kJ/mol ΔH R = 637.9kJ/mol. Expressing this per kg of reactants, the weight of 1 mole of CH 3 OH and 3/2 mole of O 2 is g, or kg. Thus, # ΔH R = ( 637.9kJ/mol) 1molreactants & % ( $ kg ' ΔH R = 7970kJ/kgreactants. For reaction (c), H 2 ( g) + F 2 ( g) 2 HF ( g), the enthalpy of reaction is ΔH R = 2ΔH f ( HF) ΔH f 0 0 = kJ/mol ΔH R = 546.6kJ/mol. ( H 2 ) ΔH f ( F 2 )

11 6 a). Continued 11 Expressing this per kg of reactants, the weight of 1 mole of H 2 and 1 mole of F 2 is g, or kg. Thus, # ΔH R = ( 546.6kJ/mol) 1molreactants & % ( $ kg ' ΔH R = 13661kJ/kgreactants. (b) Since the thrust in a rocket is greater when the molar mass of the exhaust gas is lower, divide the absolute enthalpy per kilogram by the average molar mass of the products. Arrange the reactions in order of effectiveness on the basis of thrust. What is the most effective reaction? The thrust T is the absolute enthalpy per kilogram divided by the average molecular weight of the products. For reaction (a), H 2 ( g) O 2 ( g ) H 2 O ( g), the thrust T is T = ΔH R ( kj/kg) M products kj/kg = kg/mol T = kj kg 2 mol. For reaction (b), CH 3 OH ( l) O 2 ( g ) CO 2 ( g) + 2 H 2 O ( g), the thrust T is T = ΔH R ( kj/kg) M products 7970 kj/kg = kg/mol kg/mol 3 T = kj kg 2 mol. For reaction (c), H 2 ( g) + F 2 ( g) 2 HF ( g), the thrust T is T = ΔH R ( kj/kg) M products = 13661kJ/kg kg/mol T = kj kg 2 mol. Therefore, the reactions in order of largest magnitude of thrust are: a > c > b. Of the three, reaction (a) is the most effective for providing rocket thrust by this measure.

12 7. Consider the reactions 12 ( a) C( graphite) + O 2 ( g) CO 2 ( g) ΔH R = 393.5kJ/mol ( b) H 2 ( g) + 1 O 2 2 ( g ) H 2 O( l) ΔH R = 285.5kJ/mol ( c) 2C 2 H 6 ( g) + 7O 2 ( g) 4CO 2 ( g) + 6H 2 O( l) ΔH R = kJ/mol. Using these reactions in conjunction with Hess' Law to determine the enthalpy of formation of ethane. The formation reaction of ethane is 2C ( s, graphite) + 3H 2 ( g) C 2 H 6 ( g). Note that reaction (a) above contains graphite as a reactant; the reaction must be doubled, though, since the formation reaction involves 2 moles of graphite. Also, reaction (b) contains hydrogen gas as a reactant; this reaction must be multiplied by a factor of 3 to match the molar ratio needed in the formation reaction. Finally, reaction (c) involves ethane, but as a reactant; thus, the reverse of reaction (c) is required, multiplied by 1/2 to get the correct stoichiometry: [ + O 2 ( g) CO 2 ( g) ] [ O 2 ( g ) H 2 O ( l) ] [ + 6 H 2 O ( l) 2C 2 H 6 ( g) + 7O 2 ( g) ] 2 C graphite 3 H 2 g CO 2 g + 3H 2 ( g) C 2 H 6 ( g) 2C graphite From the combination of reactions, the enthalpy of formation of ethane can be calculated from Hess' Law, ΔH f = 2ΔH R ( a) + 3ΔH R ( b) 1 ΔH 2 R c + 3( 285.5kJ/mol) kj/mol 2 = kJ/mol ΔH f = 83.7kJ/mol.

13 8. The combustion of g naphthalene in a constant volume bomb calorimeter caused a temperature rise in the water bath of C. The final temperature was K. The heat capacity of the calorimeter was J/K. 13 (a) Determine the molar internal energy of combustion of naphthalene and its molar enthalpy of combustion. Using the bomb calorimeter equation developed in class, ΔU comb = C v,cal ΔT ( K) = J/K ΔU comb = 17565J. To get the molar internal energy of combustion, we divide by the number of moles of naphthalene. ΔU comb,m = ΔU comb n J = $ g ' & ) % g/mol( = J/mol or ΔU comb,m = kj/mol. Finally, to get the molar enthalpy of combustion, we use ΔH comb,m = ΔU comb,m + Δn gas RT. For this conversion, the balanced reaction for the combustion of naphthalene is required, C 10 H 8 ( s) + 12 O 2 ( g) 10 CO 2 ( g) + 4 H 2 O ( l). From the balanced reaction, the change in number of moles of gas is Substituting, Δn gas = Δn gas = 2. ΔH comb,m = ΔU comb,m + Δn gas RT K = J mol 1 + ( 2) J mol 1 K 1 ΔH comb,m = J mol 1 ΔH comb,m = kj mol 1.

14 8. Continued 14 (b) Using the results from part (a), calculate the enthalpy of formation of naphthalene. The formation reaction of naphthalene is 10C ( graphite) + 4 H 2 ( g) C 10 H 8 ( s). Since naphthalene appears on the product side of the formation reaction, we will have to somehow use the reverse of the combustion reaction. The reverse of the combustion reaction is 10 CO 2 ( g) + 4 H 2 O ( l) C 10 H 8 ( s) + 12 O 2 ( g). We can use this to construct the formation reaction, along with the formation reactions for water and carbon dioxide, 10 CO 2 ( g) + 4 H 2 O ( l) C 10 H 8 ( s) + 12 O 2 ( g) 4 H 2 ( g) + 1 [ 2 O 2 ( g) H 2 O ( l) ] 10 [ C ( graphite) + O 2 ( g) CO 2 ( g) ] 10 C ( graphite) + 4 H 2 ( g) C 10 H 8 ( s) Thus, we see that the enthalpy of formation of naphthalene is equal to 4 times the enthalpy of formation of water, plus 10 times the enthalpy of formation of carbon dioxide, minus the enthalpy of combustion of naphthalene, = 4 ΔH f ( H2 O) +10 ΔH f ( CO2 ) ΔH comb,m ( C 10 H 8 ) = 4( kj/mol) + 10( kj/mol) ( kj/mol) ΔH f C10 H 8 ΔH f ( C10 H 8 ) = kj/mol.

15 9. A sample of solid KNO 3 (1.668 g) is dissolved in 100 ml of water initially at C in a constant pressure solution calorimeter. The final temperature was C. Assume the constant pressure heat capacity of the calorimeter is cal/k. Determine the molar enthalpy of solution of KNO Using the calorimeter equation for solution calorimetry, ΔH soln = C p,cal ΔT ( K) = 101.3cal/K ΔH soln = cal. To get the molar enthalpy of solution, we divide by the number of moles of KNO 3, ΔH soln,m = ΔH soln n cal = # g & % ( $ g/mol' = 6692 cal/mol or ΔH soln,m = 6.69 kcal/mol.

16 10. Using the values in the table below, reported at 25 C, determine ΔH R at 500ºC for the reaction 16 1 CH 3 OH ( l) CH 4 ( g) + 2 O 2 ( g). Assume that C p,m is independent of temperature for all species in the reaction. CH 3 OH l ΔH f (kj/mol) C p,m (Jmol 1 K 1 ) CH 4 (g) O 2 (g) The first step is to determine the enthalpy of reaction at 298 K: ΔH R = ΔH f ( CH 4 ) + 1 ΔH 2 f = 74.81kJ/mol ( O 2 ) ΔH f ( CH 3 OH) ( kj/mol) ΔH R = kJ/mol. The temperature dependence of the molar enthalpy of reaction is given by ΔH R ( T 2 ) = ΔH R ( T 1 ) + ΔC p,m ( T 2 T 1 ). This equation assumes that the molar heat capacities are independent of temperature. In the equation, ΔC p,m is the difference in molar heat capacities between products and reactants with the appropriate signed stoichiometric coefficients included, ΔC p,m = ν i C p,m ( i). i For this reaction, the heat capacity difference is ΔC p,m = i ν i C p,m i = C p,m CH C p,m ( O 2 ) C p,m ( CH 3 OH) = ( ) 81.6 Jmol 1 K 1 ΔC p,m = Jmol 1 K 1. Substituting, the molar enthalpy of reaction at 500 C (773 K) is ΔH R ΔH R ( 773 K) = ΔH R ( 298 K) + ΔC p,m ( K) = kj/mol + ( Jmol 1 K 1 )( K) = kj/mol ( 773 K) = kj/mol. $ 1 kj ' & ) % 1000 J (

17 11. Calculate the standard molar enthalpy change at 600 K for the reaction 17 N 2 ( g) + 3H 2 ( g) 2 NH 3 ( g). The standard molar enthalpies of formation and molar heat capacities at 25 C are listed below. Assume that the molar heat capacities are independent of temperature. ΔH f (kj/mol) C p,m (Jmol 1 K 1 ) N 2 (g) H 2 (g) NH 3 (g) First, the enthalpy of reaction at 298 K is calculated, ΔH R = 2ΔH f ( NH 3 ) ΔH f ( N 2 ) 3ΔH f ( H 2 ) 0 3( 0) = kJ/mol ΔH R = 91.8kJ/mol. The temperature dependence of the molar enthalpy of reaction is given by ΔH R ( T 2 ) = ΔH R ( T 1 ) + ΔC p,m ( T 2 T 1 ). For this reaction, the heat capacity difference is ΔC p,m = ν i C p,m i i = 2C p,m ( NH 3 ) C p,m ( N 2 ) 3C p,m ( H 2 ) = 2( 35.1) ( 28.8) Jmol 1 K 1 ΔC p,m = 45.3 Jmol 1 K 1. Substituting, the molar enthalpy of reaction at 600 K is ΔH R ΔH R ( 600 K) = ΔH R ( 298 K) + ΔC p,m ( K) = 91.8 kj/mol + ( 45.3 Jmol 1 K 1 )( K) = kj/mol ( 773 K) = kj/mol. $ 1 kj ' & ) % 1000 J (