= (25.0 g)(0.137 J/g C)[61.2 C - (-31.4 C)] = 317 J (= kj)

Transcription

1 CHEM 101A ARMSTRONG SOLUTIONS TO TOPIC D PROBLEMS 1) For all problems involving energy, you may give your answer in either joules or kilojoules, unless the problem specifies a unit. (In general, though, energies that are more than 1000 J are expressed in kilojoules.) a) The heat required to change the temperature of a substance is given by q mcδt, where m is the mass of the substance, C is the specific heat capacity of that substance, and ΔT is the temperature change in Celsius degrees. q mcδt (25.0 g)(0.137 J/g C)[61.2 C - (-31.4 C)] 317 J ( kj) b) The heat required to boil a substance equals the mass of the substance times the heat of vaporization (in J/g). q (25.0 g)(296 J/g) 7400 J 7.40 kj c) The heat required to melt a substance equals the mass of the substance times the heat of fusion (in J/g). q (25.0 g)(11.7 J/g) 292 J ( kj) d) We use the same relationship here that we used in part b. Note that the heat should be converted to joules, because the heat of vaporization is in joules per gram J m 296 J/g Solving this equation gives m 5.07 g. This is the mass of Hg that will boil; the rest will remain in the liquid state. The original mass of Hg is irrelevant. e) For this part, we use q mcδt J (50.0 g)(0.137 J/g ºC) ΔT Solving this equation gives ΔT 219 C. Therefore, the final temperature of the mercury is 20.0 C C 239 C. f) The specific heat capacity (0.137 J/g ºC) tells us that it takes joules of heat to raise the temperature of one gram of Hg by 1 C. One mole of Hg weighs grams, so J/g ºC g/mol 27.5 J/mol ºC 2) The key concept here is conservation of energy: the overall energy change of the entire system must equal zero. ΔE Be + ΔE water 0 Note that one of these ΔE values will be positive and one will be negative (in fact, the ΔE for the beryllium will be negative, since the beryllium gets colder). Then we recognize that no work is done, so each ΔE equals the heat. q Be + q water 0 Next, we make the substitution q mcδt twice, being sure to label the masses and ΔT s:

2 m Be C Be ΔT Be + m water C water ΔT water 0 Then we rewrite each ΔT as T final T initial : m Be C Be (T final T initial ) Be + m water C water (T final T initial ) water 0 The final temperatures must be equal, since the two substances reach thermal equilibrium. We have values for all of the other variables in this equation, so we just substitute and solve. [(22.31 g)(1.82 J/g ºC)(T f 58.7ºC)] + [(41.33 g)(4.18 J/g ºC)(T f 16.1ºC)] 0 Solving this equation gives T f 24.2ºC. 3) This problem is similar to #2, but now we have three substances to deal with. Our conservation of energy statement becomes: ΔE Al + ΔE Ti + ΔE water 0 Since no work is done, we rewrite this statement as q Al + q Ti + q water 0 Making the substitution q mcδt gives: m Al C Al ΔT Al + m Ti C Ti ΔT Ti + m water C water ΔT water 0 In this case, we know all of the temperatures, so we can calculate the ΔT values right away: ΔT Al 14.7ºC 68.0ºC -53.3ºC ΔT Ti 14.7ºC 75.0ºC -60.3ºC ΔT water 14.7ºC 6.3ºC 8.4ºC Now we insert all the values we know into our equation: [(50.0 g)(0.902 J/g ºC)(-53.3ºC)] + [m Ti (0.522 J/g ºC)(-60.3ºC)] + [(125.0 g)(4.18 J/g ºC)(8.4ºC)] 0 Solving this equation gives us m Ti 63.1 g. 4) Converting water at 25.0ºC into steam at 150.0ºC requires three steps: Raise the temperature of the water from 25.0ºC to 100.0ºC. Convert the liquid water into steam. Raise the temperature of the steam from 100.0ºC to 150.0ºC. Each of these processes requires heat. We need to calculate the heat needed for each step, then add these values to get the total heat requirement. Raising the temperature of the water from 25.0ºC to 100.0ºC: q (50.0 g)(4.18 J/g ºC)(100.0ºC 25.0ºC) J kj Converting the liquid water into steam: q (50.0 g)(2257 J/g) J kj Raising the temperature of the steam from 100.0ºC to 150.0ºC: q (50.0 g)(2.04 J/g ºC)(150.0ºC 100.0ºC) 5100 J 5.10 kj The total heat needed is therefore: kj kj kj kj 134 kj

3 5) Three processes occur in this experiment: The liquid gallium freezes (remaining at 29.8 C as it does). The solid gallium cools from 29.8 C to 21.4 C. The water warms from 5.2 C to 21.4 C. The law of conservation of energy tells us that the energies involved in these three processes must add up to zero: (q of Ga freezing) + (q of Ga cooling) + (q of water warming) 0 Now we write an expression for each of these heat terms. q of Ga freezing m Ga ( H fusion ) g H fusion (Note that the heat of freezing is the same as the heat of fusion, but with the opposite sign, so we need the minus sign here.) q of Ga cooling m Ga C Ga T Ga g J/g ºC (21.4 C 29.8 C) q of water warming m H2O C H2O T H2O g 4.18 J/g ºC (21.4 C 5.2 C) So we have the following equation: [ g H fusion ] + [54.72 g J/g ºC (21.4 C 29.8 C)] + [67.09 g 4.18 J/g ºC (21.4 C 5.2 C)] 0 Solving this equation gives H fusion 79.9 J/g. To convert to kj/mol 79.9 J/g g/mol (1 kj/1000 J) 5.57 kj/mol. 6) a) Since we are given ΔH, we can calculate ΔE using the relationship: ΔH ΔE + RTΔn gases The ΔH we are given is the standard ΔH, corresponding to the decomposition of 2 moles of H 2 (as specified in the balanced equation). Therefore, we can calculate Δn gases using the balanced equation. The reaction produces one mole of gas (the ) and does not consume any gases, so: Δn gases 1 mol 0 mol 1 mol Now we calculate ΔE: kj ΔE + (8.314 J/mol K)( K)(1 mol) kj ΔE J kj ΔE kj ΔE kj. b) Both ΔE and ΔH are negative, telling us that this reaction is exothermic. Therefore, the reaction produces thermal energy, making the mixture hotter. c) The negative sign of ΔE tells us that the internal energy of the atoms decreases. d) In an open container, the pressure remains constant, so the heat produced by the reaction equals ΔH. However, the correct answer is not kj, because that is the amount of heat produced when 2 moles of H 2 break down. We must set up a ratio to calculate the amount of heat we get when 7.50 g of H 2 breaks down. To set up the ratio, we must convert 7.50 g to moles: you should get mole of H 2. Our ratio is: kj ΔH 2 mol H mol H 2

4 Solving this equation gives ΔH kj. We can therefore conclude that the reaction will release 20.9 kj of heat. e) In a closed, sealed container, the volume remains constant, so no PV work is done and the heat produced by the reaction equals ΔE. As in part d, we must set up a ratio to calculate the heat kj ΔE 2 mol H mol H 2 Solving this equation gives ΔE kj. The reaction will release 21.1 kj of heat. f) Since the reaction occurs in an open container, we use ΔH to relate heat to moles of reactant. Again, we must set up a ratio, but in this case the moles of H 2 is our unknown. Calling this number n, we have: kj 2 mol H kj n Solving this equation gives n mol H 2, which equals 7.19 g of H 2 (aq). g) In an open container, the PV work is given by w PV -RTΔn gases First, we should calculate the standard value of w PV, corresponding to the reaction of 2 moles of H 2 : w PV (8.314 J/mol K)( K)(1 mol) J Now we use ratios to calculate the PV work when we use 7.50 g of H grams of H 2 equals moles, so: J w PV 2 mol H mol H 2 Solving this equation gives w PV J. Therefore, the reaction does 273 J of PV work. The value of w PV is negative, telling us that the work is done by the system (the reacting chemicals). h) In a sealed rigid container, no PV work is done. i) We found in part g that when the standard value of w PV is J. As before, we can set up a ratio between moles of H 2 and PV work, but now our unknown is the number of moles of H 2 : J -150 J 2 mol H 2 n Solving this equation gives n mol H 2, which equals 4.12 g of H 2. 7) If the reaction produces more heat in an open container than it does in a closed container, the number of moles of gases must decrease during the reaction. When the number of moles of gas decreases, the volume of the gases decreases; the surrounding air does PV work as it forces the volume to shrink. This means that the surroundings transfer energy to the system (the reacting chemicals). The system releases this energy back to the surroundings in the form of extra heat.

5 Another way to think of this is by considering the First Law of Thermodynamics, ΔE q + w. In an open container, the only work done is PV work. If q is more negative than ΔE, then w PV must be a positive number, which means that the work must be done by the surroundings. When the surroundings do the PV work, the system gets smaller, which can only happen if the number of moles of gases decreases during the reaction. 8) This reaction produces a gas, so the PV work is negative (it is done by the system) when the reaction is run in an open container. In a sealed, rigid container, no PV work occurs. Let s now write the First Law: ΔE q + w The reaction is endothermic, so ΔE and q are both positive numbers. If the reaction is carried out in a sealed rigid container, w 0, so ΔE q. In an open container, ΔE q + w, and since w is negative, q must be larger than ΔE. Therefore, the reaction absorbs more heat when it is run in an open container, so it produces a lower temperature in an open container. A physical interpretation of this is that in an open container, the reaction absorbs some energy from its surroundings to rearrange the positions of the atoms (that s ΔE), and it absorbs some more energy from its surroundings to do the PV work. In a closed container, it still absorbs the energy to change the positions of the atoms, but it doesn t need to do the PV work. 9) Note that throughout this problem, we are looking for the heat produced at constant pressure. Therefore, the heat will equal H in all parts of this problem. However, we need to compensate for the fact that we are not using standard amounts of the chemicals. a) The amount of heat that is produced is determined by the limiting reactant, just like everything else about a reaction, so we start by finding the limiting reactant. The most straightforward approach is to calculate the amount of heat that is produced based on each reactant; the reactant that gives the smallest amount of heat is the limiting reactant (and the corresponding heat is our answer) For C 3 H 9 N (molar mass g/mol), g moles kj 4 mol C 3 H 9 N ΔH mol C 3 H 9 N Solving this equation gives H kj. For (molar mass g/mol), g moles kj ΔH 21 mol mol Solving this equation gives H kj. The corresponds to the smaller amount of heat, so is the limiting reactant and the reaction will produce kj of heat. b) First, we calculate the number of moles of C 3 H 9 N that we need to get kj of heat, using the fact that we get 9684 kj when we burn 4 moles of C 3 H 9 N.

6 -9684 kj kj 4 mol C 3 H 9 N n Solving this equation gives n mol C 3 H 9 N. Next, we use stoichiometry to calculate the corresponding number of moles of N 2 : mol C 3 H 9 N 2 mol N 2 4 mol C 3 H 9 N mol N 2 Finally, we use the ideal gas law to calculate the volume of the N 2. PV nrt, so V nrt mol P atm Doing the arithmetic gives us V L ( ml) ( )( L atm/mol K) ( K) c) For C (molar mass g/mol), g moles. We know from the balanced equation that we get 9684 kj of heat when we make 12 moles of C (the standard values), so kj ΔH 4 mol C mol C Solving this equation gives H kj, so we get 1746 kj of heat. 10) a) We are told that kj of heat is produced (q kj) when g of Al(s) reacts at constant pressure. When the pressure is constant, the heat equals ΔH, so ΔH kj when g of Al(s) reacts. However, this is not our final answer Whenever you are asked for ΔH for a reaction, and you are given the balanced equation, you must calculate the standard ΔH value, corresponding to the number of moles in the balanced equation. The kj is a value, corresponding to g of Al ( moles of Al). We can calculate the standard ΔH by setting up a ratio. ΔH standard 2 mol Al kj mol Al Solving this equation gives ΔH kj. Now that we know the standard value of ΔH, we can calculate the standard value of ΔE. The balanced equation tells us that the reaction consumes three moles of gases (the Cl 2 ) and does not produce any gases, so Δn gases 0 mol 3 mol -3 mol. Using a value of ΔH with a few extra digits to avoid rounding errors ( kj), we have: kj ΔE + (8.314 J/mol K)( K)(-3 mol) This gives ΔE kj (rounded from kj) b) In a sealed, rigid container, the heat equals ΔE. In part a, we calculated the standard value of ΔE; now we need to calculate a value that corresponds to the masses of Al

7 and Cl 2 given in the problem. To do this, we must start by calculating the moles of each chemical and then determining the limiting reactant. This is topic A material and should be familiar by now; you should get the following: mol Al mol Cl 2 Cl 2 is the limiting reactant Now we set up a ratio to calculate ΔE corresponding to moles of Cl 2 : kj ΔE 3 mol Cl mol Cl 2 Solving this equation gives ΔE kj, so the reaction produces kj of heat. 11) In a calorimetry problem, always start by thinking about kinetic energy (thermal energy). The temperature of the calorimeter rose because it absorbed thermal energy (in the form of heat). We can calculate the change in the kinetic energy of the calorimeter, using the temperature change and the heat capacity of the calorimeter. ΔT ºC ºC 5.798ºC ΔKE calorimeter 5.798ºC x 4287 J/ºC 24,856 J kj Now, energy cannot be created from nothing. It can only be transferred from one place to another. Since the calorimeter gained kj of energy, we can conclude that the chemicals must have lost kj of energy, in the form of heat q chemicals kj Since a bomb calorimeter is a sealed, rigid container, the heat that was produced by the chemicals equals ΔE for the reaction: ΔE kj This is the energy lost when g of C 6 H 14 reacts, but the problem is asking us for the standard value, corresponding to the reaction of 2 moles of C 6 H 14 (note the balanced equation). To get ΔE for two moles of C 6 H 14, we use our normal ratio method, after converting the mass of C 6 H 14 to moles (the molar mass of C 6 H 14 is g/mol, so g mol). ΔE standard kj 2 mol C 6 H mol C 6 H 14 Solving this equation gives ΔE standard kj (rounded from kj). To get the standard value of ΔH, we use ΔH ΔE + RTΔn gases : Δn gases 12 mol 19 mol -7 mol ΔH kj + (8.314 J/mol K)(298 K)(-7 mol) Solving this equation gives ΔH kj. 12) For any reaction, the overall ΔH equals the ΔH f values of the products minus the ΔH f values of the reactants. The values in the textbook give us: Products: 2 mol Na 2 O(s) x (-416 kj/mol) -832 kj 2 mol N 2 (g) x (0 kj/mol) 0 kj 5 mol C (g) x ( kj/mol) kj Total kj

8 Reactants: 4 mol NaNO 3 (s) x (-467 kj/mol) kj 5 mol C(s, graphite) x (0 kj/mol) 0 kj Total kj Overall ΔH for the reaction ( kj) (-1868 kj) kj This is the ΔH value that corresponds to the balanced equation, which calls for 4 moles of NaNO 3. In the problem, we are asked for the heat released when 5.00 g of NaNO 3 reacts. 1 mol 5.00 g NaNO mol NaNO g NaNO 3 Now we can set up a ratio to calculate the correct value of ΔH: kj ΔH 4 mol NaNO mol NaNO 3 Solving this equation gives ΔH kj. Since the reaction occurs in an open container, ΔH equals the heat, so the reaction releases 13.7 kj of heat. 13) a) This problem is similar to the previous one, except that in this case we know the overall ΔH, but we are missing one of the ΔH f values. Products: 8 mol C (g) x ( kj/mol) kj 6 mol H 2 O(l) x (-286 kj/mol) kj 2 mol N 2 (g) x (0 kj/mol) 0 kj Total kj Reactants: 4 mol C 2 H 3 N(l) x x kj/mol 4x kj 11 mol (g) x (0 kj/mol) 0 kj Total 4x kj Overall ΔH for the reaction kj (-4864 kj) (4x kj) Solving this equation gives x So ΔH f of C 2 H 3 N(l) 53 kj/mol (rounded to the nearest whole number, since all of the other numbers were only known to the nearest whole number.) b) To solve this part, think of ΔH vap as the ΔH for a reaction: C 2 H 3 N(l) C 2 H 3 N(g) ΔH 825 J/g Just as with any reaction, ΔH equals products minus reactants. Since we calculated the ΔH f for liquid C 2 H 3 N in part a, we can now calculate the ΔH f for solid C 2 H 3 N. However, to make the units work, we must convert the heat of vaporization from 825 J/g into kj/mol: 825 J 1 kj g 1000 J g kj/mol 1 mol So ΔH for the vaporization of 1 mol of C 2 H 3 N(l) is kj. Now we can apply products minus reactants :

9 Products: Reactants: 1 mol C 2 H 3 N(g) x x kj/mol x kj 1 mol C 2 H 3 N(l) x (53.25 kj/mol) kj kj x kj kj x So ΔH f of C 2 H 3 N(g) 87 kj/mol (again rounding to the nearest whole number) 14) a) To get this chemical equation, we need to multiply the original equation by 7. Therefore, we must also multiply H by 7: H 7(-722 kj) kj. b) To get this chemical equation, we need to multiply the original equation by 1 / 4. Therefore, we must also multiply H by 1 / 4 : H 1 / 4 (-722 kj) kj. c) To get this chemical equation, we need to reverse the original equation and multiply it by 2. Therefore, we must change the sign of H and multiply by 2: H 2(722 kj) 1444 kj. 15) This question is a typical application of Hess s Law of Heat Summation. We must find a way to manipulate the first three equations so that when we add them up, we produce the fourth equation. This usually ends up being a process of trial and error. The correct combination is: Keep reaction 1 as is: P 4 O 10 (s) + 6 H 2 O(l) 4 H 3 PO 4 (s) H kj Multiply reaction 2 by 4: 4 H 3 PO 4 (s) + 12 NaOH(s) 4 Na 3 PO 4 (s) + 12 H 2 O(l) H 4(-52.4 kj) kj Reverse reaction 3 and multiply it by 6: 6 Na 2 O(s) + 6 H 2 O(l) 12 NaOH(s) H -6(36.1 kj) kj Adding these three equations gives us our target reaction : we can cancel all of the H 3 PO 4, all of the NaOH, and all of the H 2 O from both sides of the equation. Therefore, our desired H is the sum of the three H values above: H (-99.7 kj) + ( kj) + ( kj) kj 16) a) The ΔH f value for a substance is H for the reaction that makes one mole of the substance from its constituent elements (in their standard states): elements 1 mol compound In this case, the ΔH f values for NaHCO 3 is H for the following reaction: Na(s) + 1 / 2 H 2 (g) + C (s, graphite) + 3 / 2 (g) NaHCO 3 (s) Of the four reactions we are given in the problem, only the third reaction can be related to this equation. To get the third equation, we can take the above reaction, reverse it, and then multiply it by 2. Therefore we can calculate ΔH for the reaction 2 NaHCO 3 (s) 2 Na(s) + H 2 (g) + 2 C(s, graphite) + 3 (g)

10 b) H for this reaction is -2(-948 kj) 1896 kj. 17) Start by writing the reactions that correspond to the enthalpies of combustion. For C 6 H 12 (l): C 6 H 12 (l) + 9 (g) 6 C (g) + 6 H 2 O(l) H kj For C 6 H 6 (l): C 6 H 6 (l) + 15 / 2 (g) 6 C (g) + 3 H 2 O(l) H kj Now treat this as a Hess s Law problem Multiply the first reaction by 2: 2 C 6 H 12 (l) + 18 (g) 12 C (g) + 12 H 2 O(l) H 2(-3916 kj) kj Reverse the second reaction and multiply it by 2: 12 C (g) + 6 H 2 O(l) 2 C 6 H 6 (l) + 15 (g) H -2(-3266 kj) 6532 kj Adding these two equations gives us the desired reaction, after we cancel 9 (g) and 6 H 2 O(l) from both sides. Therefore H for the target reaction (-7832 kj) kj kj 18) First, let s calculate the amount of heat we will need. Converting ice at 0ºC into steam at 100ºC involves three processes: Melting the ice (converting 0ºC ice into 0ºC water): 1 mol kj q melting g kj g 1 mol Warming the water from 0ºC to 100ºC: q warming g J/g C C J kj Boiling the water (converting 100ºC water into 100ºC steam): 1 mol kj q boiling g kj g 1 mol Total heat required kj kj kj kj So the question becomes how much butane would you need to burn in order to obtain kj of heat? Since the butane will be burned in a sealed, rigid container, the heat will equal E for the reaction. However, we are not given E for the combustion of butane. Instead, we are given some ΔH f values. Therefore, we will need to do the following Write a balanced equation for the combustion of butane. Use the ΔH f values to calculate the standard H for the combustion reaction. Convert the standard H into the standard E, using H E + RT n gases Use a ratio to determine the amount of butane that must be burned to get kj. Finally, we ll need to use the ideal gas law to translate moles of gaseous butane into a volume. The balanced equation for the combustion of C 4 H 10 is:

11 2 C 4 H 10 (g) + 13 (g) 8 C (g) + 10 H 2 O(l) For this reaction, products minus reactants gives us: H [8 mol( kj/mol) + 10 mol( kj/mol)] [2 mol( kj/mol) + 13 mol(0 kj/mol) H kj This is the standard value of H, corresponding to the burning of 2 moles of C 4 H 10 (g). Now, let s convert this into the standard E value. For this reaction, n gases -7 mol, and we assume a temperature of 25ºC ( K): kj E + (8.314 J/mol K)( K)(-7 mol) Solving this equation gives E kj. This is the standard value of E, corresponding to the burning of 2 moles of C 4 H 10 (g). Now, how much C 4 H 10 (g) must we burn in order to get kj of heat? kj 2 mol C 4 H kj n Solving this equation gives n , so we must burn moles of gaeous C 4 H 10. Finally, we can calculate the volume of the C 4 H 10 using PV nrt: (722.0 torr)v ( mol)(62.36 L torr/mol K)( K) V L (or ml)