# Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions

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1 PRACTICING SKILLS Energy Chapter 5 Principles of Chemical Reactivity: 1. To move the lever, one uses mechanical energy. The energy resulting is manifest in electrical energy (which produces light); thermal energy would be released as the bulb in the flashlight glows. Energy Units 3. Express the energy of a 100 Calories/day diet in joules: 100 Cal 1000 calorie = 5.0 x 10 6 oules/day 1 day 1 Cal 1 cal 5. Compare 170 kcal/serving and 80 k/serving. 170 kcal 1000 calorie i k i 1 serving 1 kcal 1 cal 1000 So 170 kcal/serving has a greater energy content. Specific Heat Capacity = 710 koules/serving 7. What is the specific heat capacity of mercury, if the molar heat capacity is 8.1 /mol K? Note that the difference in units of these two quantities is in the amount of substance. In one case, moles, while in the other grams. 8.1 mol K 1 mol g = g K 9. Heat energy to warm 168 g copper from -1. C to 5.6 C: Heat = mass x heat capacity x ΔT For copper = (168 g)( g K ) [5.6 C ( 1.) C] 1 K 1 C =.44 x 103 or.44 k 11. The final temperature of a 344 g sample of iron when.5 k of heat are added to a sample originally at 18. C. The energy added is: q Fe = (mass)(heat capacity)(δt).5 x 10 3 = (344 g)(0.449 g K )(x) and solving for x we get: K = x and since 1K = 1 C, T = C. The final temperature is ( ) C or 3.8 C.

2 13. Final T of copper-water mixture: We must assume that no energy will be transferred to or from the beaker containing the water. Then the magnitude of energy lost by the hot copper and the energy gained by the cold water will be equal (but opposite in sign). q copper = -q water Using the heat capacities of HO and copper, and expressing the temperatures in Kelvin (K = C ) we can write: (45.5 g)(0.385 g K )(T final K) = -(15. g)(4.184 g K )(T final K) Simplifying each side gives: 17.5 K T final = K T final + 185, K T final = Tfinal Don t forget: Round numbers only at the end. 15. Final temperature of water mixture: = K or ( ) or 0.7 C This problem is solved almost exactly like question 13. The difference is that both samples are samples of water. From a mechanical standpoint, the heat capacity of both samples will be identical and can be omitted from both sides of the equation: q water (at 95 C) = -q water (at C) (85. g)(4.184 g K )(T final K ) = -(156 g)(4.184 g K )(T final K) or (85. g)(tfinal K) = -(156 g)(tfinal K) or 85. /K Tfinal = -156 /K Tfinal rearranging: 41. /K Tfinal = so 31.0 K = Tfinal or 47.8 C 17. Here the warmer Zn is losing heat to the water: q metal = -q water Remembering that T = T final T initial, we can calculate the change in temperature for the water and the metal. Further, since we know the final and initial for both the metal and the water, we can calculate the temperature difference in units of Celsius degrees, since the change in temperature on the Kelvin scale would be numerically identical. For the metal: T = T final T initial = ( ) or C or 71.7 K. For the water: T = T final T initial = ( ) or.1 C or.1 K (recalling that a Celsius degrees and a Kelvin are the same size. (13.8 g)(cmetal)( 71.7 K) = -(45.0 g)( g K(Cmetal) = g K )(.1 K) 78

3 Cmetal = 0.40 g K (to significant figures) Changes of State 19. Quantity of energy evolved when 1.0 L of water at 0 C solidifies to ice: The mass of water involved: If we assume a density of liquid water of g/ml, 1.0 L of water (1000 ml) would have a mass of 1000 g. 333 To freeze 1000 g water: 1000 g ice g ice = 333. x 103 or 330 k (to sf) 1. Heat required to vaporize (convert liquid to gas) 15 g C 6 H 6 : The heat of vaporization of benzene is 30.8 k/mol. Convert mass of benzene to moles of benzene: 15 g g/mol = 1.60 mol Heat required: 1.60 mol C 6 H k/mol = 49.3 k NOTE: No sign has been attached to the amount of heat, since we wanted to know the amount. If we want to assign a direction of heat flow in this question, then we would add a (+) to 49.3 k to indicate that heat is being added to the liquid benzene. 3. To calculate the quantity of heat for the process described, think of the problem in two steps: 1) cool liquid from 3.0 C to liquid at 38.8 C ) freeze the liquid at its freezing point ( 38.8 C) Note that the specific heat capacity is expressed in units of mass, so convert the volume of liquid mercury to mass ml 13.6 g/ml = 13.6 g Hg ( Recall:1 cm 3 = 1 ml) 1) The energy to cool 13.6 g of Hg from 3.0 C to liquid at 38.8 C is: T = (34.35 K K) or K 13.6 g Hg K = g K ) To convert liquid mercury to solid Hg at this temperature: from the Hg /g 13.6 g = (The (-) sign indicates that heat is being removed The total energy released by the Hg is: [ ] = - 73 and since q mercury = - q surroundings the amount released to the surroundings is

4 5. To accomplish the process, one must: 1) heat the ethanol from 0.0 C to 78.9 C (ΔT = 58.9 K) ) boil the ethanol (convert from liquid to gas) at 78.9 C Using the specific heat for ethanol, the energy for the first step is: (.44 g K )(1000 g)(58.9 K) = 14,7.6 ( 14,000 to 3 sf) To boil the ethanol at 78.9 C, we need: 855 g 1000 g = 855,000 The total heat energy needed (in ) is (14, ,000) = 997,000 or 9.97 x 10 5 Enthalpy Changes Note that in this chapter, I have left negative signs with the value for heat released (heat released = - ; heat absorbed = +) 7. For a process in which the ΔH is negative, that process is exothermic. To calculate heat released when 1.5 g NO react, note that the energy shown ( k) is released when moles of NO react, so we ll need to account for that: 1 mol NO k 1.5 g NO g NO mol NO = -.38 k 9. The combustion of isooctane (IO) is exothermic. The molar mass of IO is: 114. g/mol. The heat evolved is: 0.69 g IO 1.00 L of IO 1mL Calorimetry 1x103 ml 1 L 1 mol IO 114. g IO -109 k mol IO = -3.3 x 104 k ml of 0.00 M CsOH and 50.0 ml of M HCl each supply moles of base and acid respectively. If we assume the specific heat capacities of the solutions are 4. /g K, the heat evolved for 0.00 moles of CsOH is: q = (4. /g K)(150. g)(4.8 C -.50 C) [and since 1.78 C = 1.78 K] q = (4. /g K)(150. g)(1.78 K) q = 110 The molar enthalpy of neutralization is: or -56 k/mol 110 = /mol (to sf) mol CsOH 80

5 33. For the problem, we ll assume that the coffee-cup calorimeter absorbs no heat. Since q metal = -q water Remembering that T = T final T initial, we can calculate the change in temperature for the water and the metal. Further, since we know the final and initial for both the metal and the water, we can calculate the temperature difference in units of Celsius degrees, since the change in temperature on the Kelvin scale would be numerically identical. For the metal : T = T final T initial = ( ) or C or -75.K. For the water: T = T final T initial = ( ) or.6 C or.6 K (recalling that a Celsius degrees and a Kelvin are the same size. (0.8 g)(cmetal)( -75. K) = -(75.0 g)( g K(Cmetal) = Cmetal = 0.5 g K g K )(.6 K) (to significant figures) 35. Enthalpy change when 5.44 g of NH 4 NO 3 is dissolved in g water at 18.6 C.. Calculate the heat released by the solution: T = ( ) or -.4 C or -.4 K (155.4 g)(4. g K )( -.4 K) = or 1600 ( to sf) 1 mol NH Calculate the amount of NH 4 NO 3 : 5.44 g NH 4 NO 3 4 NO 3 = mol g NH 4 NO 3 Recall that the energy that was released by the solution is absorbed by the ammonium nitrate, so we change the sign from (-) to (+). The enthalpy change has been requested in units of k, so divide the energy (in ) by 1000: Enthalpy of dissolving = k mol = 3.0 k/mol or 3 k/mol (to sf) 37. Calculate the heat evolved (per mol SO ) for the reaction of sulfur with oxygen to form SO There are several steps: 1) Calculate the heat transferred to the water: 815 g g K ( ) C 1K/1 C = 18,700 ) Calculate the heat transferred to the bomb calorimeter 93 /K ( ) C 1K/1 C = 5, mol S 3) Amount of sulfur present:.56 g 8 = mol S g S 8 8 Note from the equation that 8 mol of SO form from each mole of S 8 4) Calculating the quantity of heat related per mol of SO yields: (18, , 050 ) = 97,000 /mol SO 0.08 mol SO or 97 k/mol SO 81

6 39. Quantity of heat evolved in the combustion of benzoic acid: As in study question 37, we can approach this in several steps: 1) Calculate the heat transferred to the water: 775 g g K ( ) C 1K/1 C = 9,800 ) Calculate the heat transferred to the bomb calorimeter 893 /K ( ) C 1K/1 C = 8,10 3) Amount of benzoic acid: g benzoic acid 1 mol benzoic acid 1.1 g benzoic acid =1.9 x 10- mol benzoic acid 4) Heat evolved per mol of benzoic acid is: ( 9, ,10 ) = 3.09 x x 10 - mol 6 /mol or 3.09 x 10 3 k/mol 41. Heat absorbed by the ice : g ice 3.54 g ice = 1,180 (to 3 sf) Since this energy (1180 ) is released by the metal, we can calculate the heat capacity of the metal: heat = heat capacity x mass x ΔT = C x 50.0 g x (73. K K) [Note that ΔT is negative!] 0.36 g K = C Note that the heat released (left side of equation) has a negative sign to indicate the directional flow of the energy. Hess s Law 43. (a) Hess Law allows us to calculate the overall enthalpy change by the appropriate combination of several equations. In this case we add the two equations, reversing the second one (with the concomitant reversal of sign) CH 4 (g) + O (g) CO (g) + H O (g) ΔH = k CO (g) + H O (g) CH 3 OH(g) + 3/ O (g) ΔH = k CH 4 (g) + 1/ O (g) CH 3 OH(g) ΔH = - 16 k 8

7 (b) A graphic description of the energy change: CH 4 (g) + 1/ O (g) + 3/ O (g) ΔH = k CH OH(g) 3 ΔH = k ΔH = k CO (g) + H O (Ú) 45. The overall enthalpy change for 1/ N (g) + 1/ O (g) NO (g) For the overall equation, note that elemental nitrogen and oxygen are on the left side of the equation, and NO on the right side of the equation. Noting that equation has 4 ammonia molecules consumed, let s multiply equation 1 by : N (g) + 6 H (g) 4 NH 3 (g) ΔH = ()( k) The second equation has NO on the right side : 4 NH 3 (g) + 5 O (g) 4 ΝΟ (g ) + 6 HO (g) ΔH = k The third equation has water as a product, and we need to consume the water formed in equation two, so let s reverse equation 3 changing the sign AND multiply it by 6 6 HO (g) 6 H (g) + 3 O (g) ΔH = (+41.8)(6) k Adding these 3 equations gives N (g) + O (g) 4 ΝΟ (g ) ΔH = 361 k So, dividing all the coefficients by 4 provides the desired equation with a ΔH = k or 90.3 k Standard Enthalpies of Formation 47. The equation requested requires that we form one mol of product liquid CH 3 OH from its elements each in their standard state. Begin by writing a balanced equation: C (s,graphite) + O(g) + 4 H(g) CH3OH(l) 83

8 Now express the reaction so that you form one mole of CH3OH divide coefficients by. C (s) + 1/ O(g) + H(g) CH3OH(l) And from Appendix L, the f H is reported as 38.4 k/mol 49. (a) The equation of the formation of Cr O 3 (s) from the elements: Cr (s) + 3/ O (g) Cr O 3 (s) from Appendix L f H is reported as: k/mol for the oxide. (b) The enthalpy change if.4 g of Cr is oxidized to Cr O 3 (g) is:.4 g Cr 1 mol Cr 5.0 g Cr k = - 6 k (to sf) mol Cr 51. Calculate r H for the following processes: (a) 1.0 g of white phosphorus burns: P 4 (s) + 5 O (g) P 4 O 10 (s) from Appendix L: f H k/mol 1.0 mol P 1.0 g P k = - 4 k g P 4 1 mol P 4 (b) 0.0 mol NO (g) decomposes to N (g) and O (g): From Appendix L: f H for NO = 90.9 k/mol Since the reaction requested is the reverse of f H, we change the sign to 90.9 k/mol k The enthalpy change is then 0.0 mol = -18 k 1 mol (c).40 g NaCl is formed from elemental Na and elemental Cl : From Appendix L: f H for NaCl (s) = k/mol 1 mol NaCl The amount of NaCl is:.40 g NaCl = mol g NaCl The overall energy change is: k/mol mol = k (d) 50 g of Fe oxidized to Fe O 3 (s): From Appendix L: f H for Fe O 3 (s) = k/mol The overall energy change is: 1 mol Fe 50 g Fe g Fe 1 mol Fe O 3 mol Fe k = x 10 3 k 1 mol Fe O (a) The enthalpy change for the reaction: 4 NH3(g) + 5 O(g) 4 NO(g) + 6 HO(g) Δ f H (k/mol)

9 Δ rh = [(4 mol)(+90.9 k k mol ) + (6 mol)( mol )] - [(4 mol)( k mol ) + (5 mol)(0)] = ( k) - ( k) = k The reaction is exothermic. (b) Heat evolved when 10.0 g NH3 react: The balanced equation shows that 4 mol NH3 result in the release of 906. k. 1 mol NH g NH g NH k = -133 k 4 mol NH (a) The enthalpy change for the reaction: BaO (s) BaO (s) + O (g) Given Δ f H for BaO is: k/mol and Δ f H for BaO is: k/mol This equation can be seen as the summation of the two equations: (1) Ba (s) + O (g) BaO (s) () BaO (s) Ba (s) + O (g) Equation (1) corresponds to the formation of BaO x while equation() corresponds to (x) the reverse of the formation of BaO Δ rh = (Δ fh for BaO) + -(ΔfH for BaO) = Δ rh = k and the reaction is endothermic. (b) Energy level diagram for the equations in question: Ba (s) + 1/ O Ba (s) + O (g) (g) ΔH = k ΔH = k BaO (s) + 1/ O (g) ΔH rxn = k BaO (s) 85

10 57. The molar enthalpy of formation of naphthalene can be calculated since we re given the enthalpic change for the reaction: C10H8 (s) + 1 O(g) 10 CO(g) + 4 HO(Ú) ΔH f(k/mol)? ΔH rxn = ΔH f products - ΔH f reactants k = [(10 mol)( k k mol ) + (4 mol)( mol )] - [ΔH f C10H8] k = ( k) - ΔH f C10H k = - ΔH f C10H k = ΔH f C10H8 GENERAL QUESTIONS ON THERMOCHEMISTRY 59. Define and give an example of: (a) Exothermic and Endothermic the suffix thermic talks about heat, and the prefixes exo and endo tell us whether heat is ADDED to the surroundings from the system (exo) or REMOVED from the surroundings to the system (endo). Combustion reactions (e.g. gasoline burning in your automobile) are EXOthermic reactions, while ice melting is an ENDOthermic reaction. (b) System and Surroundings The system is the reactant(s) and product(s) of a reaction, while the surroundings is EVERYTHING else. Suppose we burn gasoline in an internal combustion engine. The gasoline (and air) in the cylinder(s) composes the system, while the engine, and the air contacting the engine is the surroundings. Together the system and surroundings compose the universe at least in thermodynamics. (c) Specific heat capacity is the quantity of heat required to change the temperature of 1g of a substance by 1 degree Celsius. Water has a specific heat capacity of about 4. /g K, meaning that 1 gram of water at 15 degrees C, to which 4. of energy is added, will have a temperature of 16 degrees C. (or 14 degrees C if 4. of energy is removed). (d) State function Any parameter which is dependent ONLY on the initial and final states. Chemists typically use CAPITAL letters to indicate state functions (e.g. H, S, ) while nonstate functions are indicated with LOWER CASE letters (e.g. q, w). Your checking account balance is a state function! (e) Standard state Defined as the MOST STABLE (PHYSICAL) STATE for a substance at at a pressure of 1 bar and at a specified temperature. (Typically 98K) The standard state for elemental nitrogen at 5 C (98K) is gas. (f) Enthalpy change the heat transferred in a process that is carried out under constant pressure conditions is the enthalpy change, H. 86

11 The enthalpy change upon the formation of 1 mol of water(ú) is 85.8 k, meaning that 85.8 k is released upon the formation of 1 mol of liquid water from 1 mol of hydrogen (g) and 1/ mol oxygen (g). (g) Standard Enthalpy of Formation the enthalpy change for the formation of 1 mol of a compound directly from its component elements, each in their standard states. The standard enthalpy of formation of nitrogen gas (N ) = 0 k/mol. 61. Define system and surroundings for each of the following, and give direction of heat transfer: (a) Methane is burning in a gas furnace in your home: (System) methane + oxygen (Surroundings) components of furnace and the air in your home. The heat flows from the methane + oxygen to the furnace and air. (b) Water drops on your skin evaporate: (System) water droplets (Surroundings) your skin and the surrounding air. The heat flows from your skin and the air to the water droplet. (c) Liquid water at 5 C is placed in freezer: (System) water (Surroundings) freezer. The heat flows from the water to the freezer. (d) Aluminum and FeO react in a flask on a lab bench: (System) Al and FeO (Surroundings) flask, lab bench, and air around flask. The heat flows from the reaction of Al and FeO into the surroundings. 63. Standard Enthalpies of Formation for O(g), O (g), O 3 (g). Substance f H (at 98K) k/mol O(g) O (g) 0 O 3 (g) What is the standard state of O? The standard state of oxygen (O ) is as a gas. Is the formation of O from O exothermic? r H = Σ f H products - Σ f H reactants (O O) r H = ( mol)(49.170) (1 mol)(0) = k (endothermic) What is the H for 3/ O (g) O 3 (g) r H = Σ f H products - Σ f H reactants r H = (1 mol)(14.67 k/mol) (3/ mol)(0) = k 87

12 65. To calculate the formation of SnCl 4 (Ú) and TiBr (s) from SnBr (s) + TiCl 4 (Ú) we can use Hess Law to add a series of appropriate equations. A screen capture from the CD-ROM shows the overall process: Given the integral values for Reaction, H = 74 k. 67. If 187 raises the temperature of g of Ag from 18.5 to 7.0 C, what is the specific heat capacity of silver? Recall that q = m c t; so 187 = g c ( ) C. and 187 C Ag = = 0.4 /g K g ( ) C 69. Addition of g of water at 60 C to g of ice at 0.00 C. The water cools to 0.00 C. How much ice has melted? As the ice absorbs heat from the water, two processes occur: (1) the ice melts and () the water cools. We can express this with the equation q water = -q ice The melting of ice can be expressed with the heat of fusion of ice, 333 /g, as q = m 333 /g. The cooling of the water may be expressed: q = m c T. Setting these quantities equal gives: m c T = m 333 /g or g (4.184 /g K) (0 60)K = -x g 333 /g [x = quantity of ice that melts. Note that since Celsius degrees and kelvin are the same size, t is 60 C or 60 K] g (4.184 /g K) (0 60)K = -x 333 /g = - x 333 /g or /g = x or 75.4 g of ice. 88

13 g (two 45 g cubes)of ice cubes (at 0 C) are dropped into 500. ml tea at 0.0 C (Assume a density of 1.00 g/ml for tea). What is the final temperature of the mixture if all the ice melts? Since q water = -q ice and we can set up the expression. m c T = - m 333 /g NOTE however, that not only does all the ice melt, but the melted ice warms to a temperature above 0 C. We add a term to account for that: m tea c T tea = - [m ice 333 /g + m ice c T ice ] 500. g (4.184 /g K) (F 93. K) = -[(90 g 333 /g) + where F is the final temperature of the tea and melted ice. 09F - 613,374 = - [ F - 10,876 ] Simplifying: 09F - 613, F - 10,876 = 0 (09F + 377F ) + (- 613, ,876 ) = 0 (90 g (4.184 /g K) (F -73. K))] 469 F ,80 = 0 or 469 F = 686,80 and F = (686,80/469) = 78 K and noting that 45 g of ice cube has sf, we report a final temperature of 80 K. 73. One can arrive at the desired answer if you recall the definition of ΔH f. The definition is the enthalpy change associated with the formation of one mole of the substance (in this case BH6) from its elements each in their standard state (s for boron and g for hydrogen). (a) Note that the 1 st equation given uses four moles of B as a reactant and we ll need only, so divide the first equation by to give: B (s) + 3/ O (g) BO3 (s) ΔH = 1/( k) = k The formation of 1 mole of BH6 will require the use of 6 moles of H (or 3 moles of H), so multiply the second equation by 3 to give: 3 H (g) + 3/ O (g) 3 HO (g) ΔH = 3(-41.8 k) = k Finally the third equation given has BH6 as a reactant and not a product. So reverse the third equation to give: BO3 (s) + 3 HO (g) BH6 (g) + 3 O (g) ΔH = -(-03.9 k)= +03.9k (b) Adding the three equations gives the equation: B (s) +3 H (g) BH6 (g) with a ΔH = ( )k (c ) Energy level diagram for the reactions: or a ΔH for BH6 (g) of k 89

14 B H 6 (g) ΔH = k B (s) + 3 H (g) - 3 O (g) ΔH = +03.9k + 3/ O (g) ΔH = k + 3/ O (g) ΔH = k B O 3 (s) + 3 H O (g) B O 3 (s) + 3 H O (g) (d) Formation of B H 6 (g) is reactant-favored 75. (a) Enthalpy change for: C(s) + HO(g) CO(g) + H(g) ΔH f(k/mol) ΔH rxn = [(1 mol)( k k mol ) + 0] - [ 0 + (1 mol)( mol )] = k (b) The process is endothermic, so the reaction is reactant-favored. (c) Heat involved when 1.0 metric ton ( kg) of C is converted to coal gas: kg C 1000 g C 1 kg C 1 mol C k g C 1 mol C = x 107 k 77. For the combustion of C8H18: C8H18(l) + 5/ O(g) 8 CO(g) + 9 HO(l) rh = [(8 mol)( k k mol ) + (9 mol)( mol rh = k Expressed on a gram basis: k mol 1 mol C 8 H g C 8 H 18 = k/g )] - [(1 mol)(-59. k mol ) + 0] For the combustion of CH 3 OH: CH3OH(l) + 3 O (g) CO (g) + 4 HO(l) rh = [(mol)( k/mol) + (4 mol)( k/mol)] - [(mol)( k/mol) + 0] 90

15 = [( ) + (- 967.)] k = k or k/mol Express this on a per mol and per gram basis: k mol CH 3 OH 1 mol CH 3 OH 3.04 g CH 3 OH = -.68 k/g On a per gram basis, octane liberates the greater amount of heat energy. 79. (a) Enthalpy change for formation of 1.00 mol of SrCO 3 Sr (s) + C (graphite) + 3/ O (g) SrCO 3 (s) using the data: Sr(s) + 1/ O (g) SrO(s) SrO(s) + CO (g) SrCO 3 (s) C (graphite) + O (g) CO (g) f H = 59 k/mol-rxn r H = 34 k/mol-rxn f H = 394 k/mol-rxn Let s add the equations to give our desired overall equation. Sr(s) + 1/ O (g) SrO(s) f H = 59 k/mol-rxn SrO(s) + CO (g) SrCO 3 (s) C (graphite) + O (g) CO (g) Sr (s) + C (graphite) + 3/ O (g) SrCO 3 (s) (b) Energy diagram relating the energy quantities: Sr (s) + 1/O (g) + C (graphite) + O (g) H f = 394 k H f = 59 k CO r H = 34 k/mol-rxn f H = 394 k/mol-rxn r H = 10. k/mol-rxn SrO (s) H rxn = 34 k H rxn = 10. k SrCO 3 (s) 91

16 81. The desired equation is: CH4 (g) + 3 Cl (g) 3 HCl (g) + CHCl3 (g) Begin with equation 1 (the combustion of methane) CH4 (g) + O (g) HO (l) + CO (g) ΔH= k = k/mol-rxn Noting that we form HCl as one of the products, using the second equation, we need to reverse it and (to adjust the coefficient of HCl to 3), multiply by 3/ to give: 3/ H (g) + 3/ Cl (g) 3 HCl ΔH = -3/(+184.6) k/mol-rxn = k Note that CO formed in equation 1 doesn t appear in the overall equation so let s use the equation for the formation of CO (reversed) to consume the CO: CO (g) C (graphite) + O (g) ΔH = -1(-393.5) k = k Noting also that equation 1 produces water molecules, let s consume them by using the equation for the formation of water (reversed) multiplied by : HO (l) H (g) + O (g) ΔH = -(-85.8) k = k and finally we need to produce CHCl3 which we can do with the equation that represents the ΔHf for CHCl3: C(graphite) + 1/ H (g) + 3/ Cl (g) CHCl3 (g) ΔH = k The overall enthalpy change would then be: ΔH = k k k k k = k IN THE LABORATORY 83. qmetal = heat capacity x mass x ΔT qmetal = Cmetal 7.3 g (99.47 K K) Note that ΔΤ is negative, since Tfinal of the metal is LESS THAN Tinitial and qwater = 15.0 g g K (99.47 K K) = 39.7 Setting qmetal = - qwater Cmetal 7.3 g (-7.58 K) = and solving for C gives: Cmetal = 0.11 g K 85. Calculate the enthalpy change for the precipitation of AgCl (in k/mol): 1) How much AgCl is being formed? 50. ml of 0.16 M AgNO 3 will contain (0.50L 0.16 mol/l) mol of AgNO 3 15 ml of 0.3 M NaCl will contain (0.15L 0.3 mol/l) mol of NaCl. Given the stoichiometry of the process, we anticipate the formation of mol of AgCl. 9

17 ) How much energy is evolved? [(50 +15ml) 1g/mL = 375g of water 375 g 4. g K (96.05 K ) K =,800 (to sf) The enthalpy change is then (since the reaction releases heat). The change in k/mol is mol 1 k = - 69 k/mol Heat evolved when ammonium nitrate is decomposed: T = ( ) = 1.8 C (or 1.8 K). Heat absorbed by the calorimeter: 155 /K 1.8K = 8 Heat absorbed by the water: 415 g K = 3160 g K Total heat released by the decomposition: = 3,440 (to 3 sf) With g NH 4 NO 3 = mol, heat released = 3440 = 36.0 k/mol mol 89. The enthalpy change for the reaction: Mg(s) + HO(l) Mg(OH) (s) + H(g) Δ fh (k/mol) ΔrH = (1 mol)( k k mol ) - ( mol)( mol ) = k or x 10 5 Each mole of magnesium releases k of heat energy. Calculate the heat required to warm 5 ml of water from 5 to 85 C. heat Magnesium required: 6.3 k = heat capacity x mass x ΔT k g = (4.184 mol )(5 ml)( ml )(60 K) = 676 or 6300 or 6.3 k (to sf) 1 mol Mg k 4.3 g Mg 1mol Mg = 0.43 g Mg 93

18 SUMMARY AND CONCEPTUAL QUESTIONS 91. Without doing calculations, decide whether each is product- or reactant- favored: (a) combustion of natural gas oxidation reactions of carbon and hydrogen are typically product-favored. This process is exothermic as well. (b) Decomposition of sugar to carbon and water- Sugar does not naturally decompose into carbon and water, hence we predict the reaction to be reactant-favored. 93. Determine the value of H for the reaction: Ca(s) + 1/8 S 8 (s) + O (g) CaSO 4 (s) Imagine this as the sum of several processes: 1) Ca(s) + 1/ O (g) CaO(s) ) 1/8 S 8 (s) + 3/ O (g) SO 3 (g) 3) CaO(s) + SO 3 (g) CaSO 4 (s) H = k Note that the SUM of the three processes is the DESIRED equation (the formation of CaSO 4 (s)). The OVERALL H is then the SUM of the H for process (1) and H for process (). We know that the r H for (3) = k or r H = ΔfH CaSO 4 (s) [Δ fh CaO(s) + Δ fh SO 3 (g)]. Since we know the ΔrH and BOTH the ΔfH for CaO(s) and SO 3 (g), we can calculate the ΔfH CaSO 4 (s). From Appendix L we find, ΔfH for CaO(s) = k/mol and ΔfH for SO 3 (g) = k/mol r H = ΔfH CaSO 4 (s) [ΔfH CaO(s) + ΔfH SO 3 (g)] k = ΔfH CaSO 4 (s) [ k/mol k/mol] - 1,433.6 k = ΔfH CaSO 4 (s) 95. The molar heat capacities for Al, Fe, Cu, and Au are: g K 6.98 g Al = 4. 1 mol Al mol K g K gfe 1 mol Fe = 5.1 mol K g Cu g K 1 mol Cu = 4.5 mol K 0.19 g K g Au 1 mol Au = 5.4 mol K 94

19 1.0 Specific heat capacity Ag Atomic Weight The graph shown is a plot of specific heat capacity versus atomic weight. As you can see, no simple linear relationship exists for these metals. The plot of the specific heat of Cu (atomic weight 63.55) and Au (atomic weight 197) does show a decreasing value of specific heat capacity as the atomic weight of the element increases. If you estimate the atomic weight to be about 100 (exact value is about 108), one could estimate a value of approximately 0.8 as the specific heat (compared to the experimental value of 0.36). Alternatively, a quick examination of the values for the four metals above indicates that they are quite similar, with an average of 4.8 /mol K. This translates into: 4.8 mol K 1 mol Au = 0.30 /g K g Au 97. Mass of methane needed to heat the air from 15.0 to.0 C: Calculate the volume of air, then with the density and average molar mass, the moles of air present: 75 m.50 m 1000 L 1 m 3 1. g air 1 L air 1 mol air 8.9 g air =.90 x 104 mol air The energy needed to change the temperature of that amount of air by ( ) C:.90 x 10 4 mol air 9.1 mol K 7.0 K = 5.9 x 106 What quantity of energy does the combustion of methane provide? The reaction may be written: CH 4 (g) + O (g) H O (g) + CO (g) 95

20 Using data from Appendix L: r H = [( mol)( k/mol) + (1mol)( k/mol)] The amount of methane necessary is: 1 k 5.9 x mol CH k [(1 mol)( k/mol) + ( mol)(0)]= k 16.0 g CH 4 1 mol CH 4 = 10 g CH 4 ( sf) 99. Calculate the quantity of heat transferred to the surroundings from the water vapor condensation as rain falls. Calculate the volume of water that falls, and then the mass of that water: From the conversion factors listed in your textbook, calculate the area of 1mi in cm [1 km = mi and 1 km = 10 5 cm. ] ( ) ( ) 10 5 cm = 1010 cm ( 1 km) 1 mi 1 km mi ( ) =.59 x 1010 cm 1 in =.54 cm so the VOLUME of water is.59 x cm x.54 cm = 6.6 x cm 3. The mass of water is: 6.6 x cm 3 x 1.0 g/cm 3 or 6.6 x g of water. The amount of heat: 6.6 x 1010 g water 1 1 mol water 18.0 g water 44.0 k 1 mol water = 1.6 x 1011 k Note the much larger energy for this process than for the detonation of a ton of dynamite (a) The diagram is: 1-butene + 6 O c H cis--butene + 6 O Energy c H trans--butene + 6 O c H 4 CO + 4 H O (b)for the combustion reaction: C 4 H 8 (g) + 6 O (g) 4 CO (g) + 4 H O(g) Δ c H = 4 Δ f H (CO ) + 4 Δ f H (H O) - Δ f H (C 4 H 8 ) + 6 Δ f H (O ) Note that the last term will be 0 in all cases. Substitute the thermodynamic data for each of the three isomers: 96

21 cis--butene: Δ c H = 4 Δ f H (CO ) + 4 Δ f H (H O) - Δ f H (C 4 H 8 ) 1 mol k/mol = 4 mol k/mol+ 4 mol kmol 1 mol Δ f H (C 4 H 8 ) = k trans--butene: 1 mol -684.k/mol = 4 mol k/mol+ 4 mol kmol 1 mol Δ f H (C 4 H 8 ) = 14.8 k 1-butene: 1 mol k/mol = 4 mol k/mol+ 4 mol kmol 1 mol Δ f H (C 4 H 8 ) = k (c)relation of Enthalpies of Isomers to the elements: 1-butene f H o cis--butene Energy f H o trans--butene f H o 4 C+ 4 H (d)enthalpy change of cis--butene to trans--butene: Δ r H = Δ f H (trans--butene) - Δ f H (cis--butene) = 14.8 k k = -3.3 k 103. (a)a sample of 0.850g Mg corresponds to mol Mg. The amount of heat (evolved) is -5.4 k, corresponding to 5.4k/0.0350mol Mg = -76 k/mol (b) Final temperature of water and bomb calorimeter: Heat evolved = - Heat absorbed = -(80 /K) T + (750.g)(4.184 /g K) T = -(80 /K) T + (3138 /K) T and 5400 = /K T /-3958 /K = 6.41 K (or 6.41 C since a K and a C are the same size ) The new temperature of water will be 18.6 C C = 5.0 C 97

22 105. (a)the energy diagram shown here indicates that methane liberates k/mol while methanol liberates only k/mol. (b) Energy per gram: k For methane: 1 mol = k/g 1 mol g k For methanol: 1 mol 1 mol 3.04 g = k/g (c) Enthalpy conversion from methane to methanol: The diagram indicates that the difference in enthalpy for these two substances is the difference between the two top boxes. Hence H = k ( k) = - 79 k/mol (d) The equation for conversion of methane to methanol: CH 4 (g) + 1/ O (g) CH 3 OH(l) 107. (a) Piece of metal to heat? To cool to achieve a maximum T? Final temperature of water? To convey maximum heat per gram, one needs a metal with the greatest specific heat so of these 3 metals, Al, is the best candidate, and the larger piece of metal (1) would convey MORE heat than the smaller piece() of Al. To minimize the heat absorbed by the cooler metal, one needs a metal with the lesser specific heat and the smaller the better so the smaller piece of Au (4)is a prime candidate. As to final T: Heat loss (by warm metal) = Heat gain (by cool metal and water) Note that the SIGNS of the two will be opposite, so let s (arbitrarily) place a (-) sign in front of the heat loss side. -(100.0 g)(0.900 /g K)(T f 373) = (50.0 g)(0.189 /g K)(T f -63) + (300.g)(4.184 /g K)(T f -94) -90.0T f =6.445T f T f Collecting T f terms: -90.0T f T f T f = or T f = and a T f = 99.1K or (99-73) = 6 C 98

23 (b) Process is similar to that in (a) but we want minimal T change: Consider the following table of data and calculations: Note that the amount of heat lost by cooling 50.0g Zn and the heat gained by warming 50.0g of Al is approximately equal. Obviously you could do these calculations for all the combinations of metals (both type and mass). Let s see how these two compute! - Heat lost = Heat gained - (50.0 g)( /g K)(T f 373) = (50.0 g)(0.900 /g K)(T f -63) + (300.g)(4.184 /g K)( (T f -94) -19.3T f = 45.01T f T f Collecting T f terms: T for metal heated to 100 C T for metal cooled to - 10 C -19.3T f T f T f = or T f = and a T f = 94K or (94-73) = 1 C Heat lost upon cooling to 1 C Heat gained upon warming to 1 C Specific heat Mass Heat Capacity Metal Al Zn

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