Chapter 4 Solutions to Supplementary Check

Chapter 4 Solutions to Supplementary Check for Understanding Problems Energy Changes 1. Indicate whether each of the following involves primarily kinetic energy or potential energy. a) b) c) d) e) f) Answers: stretched rubber band running water stack of books person riding a bike car parked on a hill pitched baseball a) potential energy d) kinetic energy b) kinetic energy e) potential energy c) potential energy f) kinetic energy Solutions Kinetic energy is associated with motion as is found in (b), (d) and (f). Potential energy is stored energy associated with the position of an object as in (a), (c) and (e). 2. Which physical state involves constituent particles with the highest kinetic energy? Solution The gas phase is associated with constituent particles having the highest kinetic energy because energy must be added as a substance moves from the solid state to the liquid state to the gas phase. This added energy is causes the constituent particles to move faster. 3. Is the condensation of a gas an exothermic or endothermic process? Explain. Answer: exothermic Solution An exothermic process is one in which energy (heat) is released. The liquid state is at a lower energy level than the gas phase so energy is released as a gas condenses. It is helpful to note that the reverse process (liquid ö gas) is endothermic because it requires an input of heat. S.4.3 S.4.4 4. CHAPTER 4 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS How much energy (in kJ) is needed to heat 3.8 L of water from 22.3 EC to 100 EC? 1.2 x 103 kJ Answer: Solution The amount of heat energy (q) needed is calculated from q = m @ c @ Δt. For water, the specific heat (c) equals 4.18 J/g @EC. The temperature change (Δt) is 100 EC - 22.3 EC = 77.7 EC. The mass of water (m) in grams must be calculated from its volume using the density of water (in Table 2.3 the water density = 0.9978 g/mL at 22 EC). The solution map for this conversion is: L H2O ö mL H2O ö g H2O This yields: 3.8 L H 2 O x 1 mL H 2 O 0.9978g H 2 O x = 3.8 x103 g H 2 O -3 1 mL H 2 O 10 L H 2 O Substituting these values and calculating q gives: q = (3.8 x103 g H 2 O )(4.18 J/ g H 2 O × o C )(77.7 o C ) = 1.2 x106 J 1.2 106 J x 5. 1kJ = 1.2 x103 kJ 3 10 J When 25 g of water freezes 8.4 kJ of energy is released. How much energy is required to melt a pound of ice? Answer: 1.5 x 102 kJ Solution Note that is takes 8.4 kJ to melt 25 g of water so this serves as the conversion factor between grams of water and energy. The solution map is: lb H2O ö g H2O ö kJ 1 lb H 2 O x 454 g H 2 O 8.4 kJ x = 1.5 x102 kJ 25 g H 2 O 1 lb H 2 O S.4.5 6. CHAPTER 4 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Calculate the specific heat of a solid (J/g@EC) if 118 kcal of energy are needed to raise the temperature of 12.7 kg of the solid from 24 EC to 327 EC? Answer: 0.128 J/g@EC Solution The specific heat (c) can be calculated by rearranging equation 4.1 to give: c  q mΔt The amount of heat added (q) is 118 kcal which must be converted to joules. Recall that 1 cal equals exactly 4.184 J. The solution map is: kcal ö cal ö J 118 kcal x 103 cal 4.184 J = 4.94 x105 J x 1 kcal 1 cal The mass of the substance is 12.7 kg or 12.7 x 103 g, and the temperature change (Δt) is 327 EC 24 EC = 303 EC. Substituting these values and calculating c gives: c  7. 4.94 x 105 J J = 0.128 o 3 o (12.7 x10 g)(303 C) g C If a 10-g sample of copper and a 10-g sample of aluminum are placed in a large container of boiling water for several minutes than then removed and quickly transferred to a large container of ice water, which metal will lose the greater amount of energy? Explain. Answer: aluminum Solution The amount of heat energy lost is calculated from q = m @ c @ Δt. Since the masses are the same (10 g) and Δt is the same (100 EC - 0 EC = 100 EC), q will be larger for the metal with the larger specific heat which is aluminum. 8. When you enter a room having a uniform temperature and touch a metal object it feels cooler than when you touch a wooden object in the room. What is your explanation (hypothesis) for this difference? What simple experiments could you do to test your hypothesis? S.4.6 CHAPTER 4 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Solution Hypothesis: Assuming both objects have been in the room for some time they should both be at room temperature. Because you are warmer than room temperature, heat leaves your body when you touch either object. Heat is transferred at a faster rate (your skin cools faster) when you touch the metal object than it does when you touch the wooden object because metals conduct heat much better (faster) than objects made of wood or plastic (mostly composed of nonmetals) so it feels like the metal object is cooler. Experiment: One way to compare the ability of different materials to conduct heat is to have two containers at different temperatures connected by a sample of each material you wish to test. You could use a pot of boiling water for the higher temperature and an ice cube (or other ice mold) for the lower temperature. The samples of different materials should be exactly the same in size and shape (see figure below). boiling water ice sample When this arrangement is created, heat will be transferred from the boiling water to the ice. By measuring the time it takes to melt the ice, or how much ice melts in a given time, you can determine which materials conduct heat more effectively. Chemical Reactions 1. Convert each of the following word equations into a chemical equation and then balance it. Indicate the physical state of each reactant and product. a) phosphoric acid + calcium hydroxide ÷ water + calcium phosphate b) ammonium nitrate ÷ nitrogen gas + water + oxygen gas Answers: a) 2H3PO4(aq) b) 2NH4NO3(s) + ÷ 3Ca(OH)2(s) 2N2(g) + ÷ 6H2O(l) 4H2O(l) + + Ca3(PO4)2(s) O2(g) S.4.7 CHAPTER 4 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Solutions First, write the chemical formula for each reactant and product, along with its physical state. Then use whole-number coefficients in front of the reactant/product formulas to balance atoms on both sides of the equation. a) Reactant formulas: phosphoric acid - The name indicates an oxoacid (no hydro- prefix) based on the anion phosphate (PO43-). Thus, three H+ ions are needed to balance by the charge on the anion and the correct formula is H3PO4(aq). Note that oxoacids are generally used as an aqueous solution and are rarely used in pure form. calcium hydroxide - The metal cation is Ca2+. The polyatomic anion formula is OH-. It takes two hydroxides to balance the charge on the cation so the compound formula is Ca(OH)2(s). The solid state is used since there is no information to suggest that this is not the pure compound. Product formulas: water - H2O(l) calcium phosphate - The metal cation is Ca2+. The anion formula is PO43-. Use the value of the anion charge as the subscript for the cation and use the value of the cation charge as the subscript for the anion. The compound formula is Ca3(PO4)2. This ionic compound is formed in the presence of water so you must consider if this compound dissolves readily in water. The combination of +2 and -3 ion charges suggest this will be an insoluble compound so the solid state is used. This results in the equation: H3PO4(aq) + Ca(OH)2(s) ÷ H2O(l) + Ca3(PO4)2(s) Notice that the polyatomic phosphate ion does not undergo any change in composition in this reaction. This means that you can balance PO43- units just like an individual atom. Since there are 2 phosphate ions on the product side, a coefficient of 2 is needed in front of H3PO4 on the reactant side. To balance calcium atoms, place a coefficient of 3 in front of Ca(OH)2 on the reactant side. There are now 12 hydrogen atoms on the reactant side and only 2 on the product side so a cooefficient of 6 is needed in front of water on the product side. Notice that this also balances the oxygen atoms, a total of 14 on each side of the equation. The balanced equation is: 2H3PO4(aq) + 3Ca(OH)2(s) ÷ 6H2O(l) + Ca3(PO4)2(s) S.4.8 b) CHAPTER 4 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Reactant formula: ammonium nitrate - The metal cation is NH4+. The polyatomic anion formula is NO3-. Since the ion charges are the same, it takes one cation to balance the charge on the anion and the compound formula is NH4NO3. A pure ionic compound is expected to be a solid. Product formulas: nitrogen gas - This element exists as a diatomic molecule with the formula N2(g). water - H2O(l) oxygen gas - This element exists as a diatomic molecule with the formula O2(g). This results in the unbalanced equation NH4NO3(s) ÷ N2(g) + H2O(l) + O2(g) (N and O atoms balanced) Notice that the nitrogen atoms and oxygen atoms are balanced, so balance the hydrogen atoms. Place a coefficient of 2 in front of H2O to balanced hydrogen atoms, but now there are 3 oxygen atoms on the reactant side and four on the product side. NH4NO3(s) ÷ N2(g) + 2H2O(l) + O2(g) (N and H atoms balanced) A coefficient of ½ in front of O2 will balance oxygen atoms. NH4NO3(s) ÷ N2(g) + 2H2O(l) + ½O2(g) (all atoms balanced) The fractional coefficient in front of O2 can be converted to the smallest whole number by multiplying by 2. This requires that all other coefficients be multiplied by 2 in order to retain the atom balance. The resulting balanced equation is: 2NH4NO3(s) ÷ 2N2(g) + 4H2O(l) + O2(g) S.4.9 2. CHAPTER 4 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Balance each of the following chemical equations. a) CaC2(s) + H2O(l) ÷ Ca(OH)2(s) + C2H2(g) b) NO2(g) + H2O(l) ÷ HNO3(aq) + NO(g) c) Fe2O3(s) + CO(s) ÷ Fe(s) + CO2(g) d) H2O2(aq) ÷ H2O(l) + O2(g) Answers: a) CaC2(s) + 2H2O(l) ÷ Ca(OH)2(s) + C2H2(g) b) 3NO2(g) + H2O(l) ÷ 2HNO3(aq) + NO(g) c) Fe2O3(s) + 3CO(s) ÷ 2Fe(s) + 3CO2(g) d) 2H2O2(aq) ÷ 2H2O(l) + O2(g) Solutions a) Initially the carbon atoms and calcium atoms are balanced but the hydrogen and oxygen atoms are not. CaC2(s) + H2O(l) ÷ Ca(OH)2(s) + C2H2(g) 2 atoms ² C ÷ 2 atoms 1 atom ² Ca ÷ 1 atom 2 atoms ² H ÷ 4 atoms 1 atom ² O ÷ 2 atoms Placing a coefficient of 2 in front of water will balance both the hydrogen and oxygen atoms. CaC2(s) + 2H2O(l) ÷ Ca(OH)2(s) + C2H2(g) S.4.10 b) CHAPTER 4 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Initially no atoms are balanced. NO2(g) + H2O(l) ÷ HNO3(aq) + NO(g) 1 atom ² N ÷ 2 atoms 2 atoms ² H ÷ 1 atom 3 atoms ² O ÷ 4 atoms Also note that any new coefficient will change the number of two or three different atoms. Start by placing a coefficient of 2 in front of HNO3 to balance the hydrogen atoms. This results in the following. NO2(g) + H2O(l) ÷ 2HNO3(aq) + NO(g) 1 atom ² N ÷ 3 atoms 2 atoms ² H ÷ 2 atoms 3 atoms ² O ÷ 7 atoms Next place a coefficient of 3 in front of NO2 to balance the nitrogen atoms. This also balances the oxygen atoms and results in a balanced equation. 3NO2(g) + H2O(l) ÷ 2HNO3(aq) + NO(g) 3 atoms ² N ÷ 3 atoms 2 atoms ² H ÷ 2 atoms 7 atoms ² O ÷ 7 atoms c) Initially only the carbon atoms are balanced. Fe2O3(s) + CO(s) ÷ Fe(s) + CO2(g) 1 atom ² C ÷ 1 atom 2 atoms ² Fe ÷ 1 atom 4 atoms ² O ÷ 2 atoms S.4.11 CHAPTER 4 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Placing a coefficient of 2 in front of Fe will balance the iron atoms. Fe2O3(s) + CO(s) ÷ 2Fe(s) + CO2(g) 1 atom ² C ÷ 1 atom 2 atoms ² Fe ÷ 2 atoms 4 atoms ² O ÷ 2 atoms Any attempt to add more oxygen atoms to the product side by changing the coefficient in front of CO2 will upset the balance of carbon atoms so the same coefficient must be placed in front of CO as well. A coefficient of 3 will balance both the oxygen atoms and carbon atoms. Fe2O3(s) + 3CO(s) ÷ 2Fe(s) + 3CO2(g) 3 atoms ² C ÷ 3 atoms 2 atoms ² Fe ÷ 2 atoms 6 atoms ² O ÷ 6 atoms d) Initially the hydrogen atoms are balanced but the oxygen atoms are not. H2O2(aq) ÷ H2O(l) + O2(g) 2 atoms ² H ÷ 2 atoms 2 atoms ² O ÷ 3 atoms Placing a coefficient of ½ in front of O2 will balance oxygen atoms. H2O2(aq) ÷ H2O(l) + ½O2(g) 2 atoms ² H ÷ 2 atoms 2 atoms ² O ÷ 2 atoms S.4.12 CHAPTER 4 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The fractional coefficient in front of O2 can be converted to the smallest whole number by multiplying by 2. This requires that all other coefficients be multiplied by 2 in order to retain the atom balance. The resulting balanced equation is: 2H2O2(aq) ÷ 2H2O(l) + O2(g) 3. If ammonia gas and 0.128 g of hydrogen chloride gas react to form 0.188 g of ammonium chloride, what mass of ammonia reacted? Answer: 0.060 g Solution The given information is: NH4(g) + HCl(g) ÷ HNH4Cl(s) 0.128 g 0.188 g Since one expects the mass of the products in a chemical reaction to equal the mass of the reactants, the mass of ammonia equals 0.188 g - 0.128 g = 0.060 g. 4. What does the notation ‘(aq)’ signify about a reactant or product in a chemical equation? Solution The (aq) designation stands for aqueous solution and indicates that the substance is dissolved in water. 5. Classify each of the following reactions as combination, decomposition, combustion, single displacement or double displacement, then complete and balance each equation. Indicate the physical state of each product. a) K3PO4(aq) + BaCl2(aq) ÷ b) Al(s) + H2SO4(aq) ÷ c) C4H9OH(l) + O2(g) ÷ d) CuO(s) + HNO3(aq) ÷ e) Al(s) + Br2(l) ÷ S.4.13 CHAPTER 4 Answers: SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS a) double displacement 2K3PO4(aq) + 3BaCl2(aq) ÷ 6KCl(aq) + Ba3(PO4)2(s) b) single displacement 2Al(s) + 3H2SO4(aq) ÷ 3H2(g) + Al2(SO4)3(aq) c) combustion C4H9OH(l) + 6O2(g) ÷ 4CO2(g) + 5H2O(l) d) double displacement e) combination CuO(s) + 2HNO3(aq) ÷ Cu(NO3)2(aq) + H2O(l) 2Al(s) + 3Br2(l) ÷ 2AlBr3(aq) Solutions a) This reaction between two dissolved ionic compounds should suggest a double displacement reaction of the general form AB + CD ÷ AD + CB. In this case the two metal ions, potassium (K+) and barium (Ba2+), in the reactants switch places to form an ionic compound between potassium ion and chloride ion (Cl-) and an ionic compound between barium ion and phosphate ion (PO43- ). Therefore, the product formulas are KCl and Ba3(PO4)2. The combination of a +1 cation and a -1 anion is expected to soluble while the combination of a +2 cation and a -3 anion is expected to insoluble . This results in the unbalanced equation: K3PO4(aq) + BaCl2(aq) ÷ KCl(aq) + Ba3(PO4)2(s) To balance this equation, place a coefficient of 2 in front of K3PO4 to balance phosphate units. This requires a coefficient of 6 in front of KCl to balanced potassium atoms. Finally, a coefficient of 3 in front of BaCl2 will balance both the barium atoms and the chlorine atoms. The balanced equation is: 2K3PO4(aq) + 3BaCl2(aq) ÷ 6KCl(aq) + Ba3(PO4)2(s) b) This reaction between an elemental form and an oxoacid should suggest a single displacement reaction of the general form A + BC ÷ AC + B where A and B are elemental forms and BC and AC are compounds. The metal aluminum (Al) displaces hydrogen in the oxoacid to form elemental hydrogen, H2(g), and an ionic compound consisting of aluminum ions (Al3+) ions and sulfate (SO42-) ions. This means that the formula of the compound formed between Al3+ and SO42- is Al2(SO4)3. Although this ionic compound contains a +3 cation and a -2 anion, suggesting it is insoluble, compounds containing sulfate tend to be soluble. The unbalanced equation is: Al(s) + H2SO4(aq) ÷ H2(g) + Al2(SO4)3(aq) S.4.14 CHAPTER 4 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS To balance this equation, place a coefficient of 2 in front of Al to balance aluminum atoms and place a coefficient of 3 in front of H2SO4 to balance sulfate units. Finally, a coefficient of 3 in front of H2 will balance both the hydrogen atoms. The balanced equation is: 2Al(s) + 3H2SO4(aq) ÷ 3H2(g) + Al2(SO4)3(aq) c) This is a reaction between a CxHyOz compound and oxygen so it is a combustion reaction that forms CO2(g) and H2O(l) as products. C4H9OH(l) + O2(g) ÷ CO2(g) + H2O(l) (unbalanced) Start by balancing the carbon atoms by putting a coefficient of 4 in front of CO2. Since there are 10 H atoms on the reactant side and only 2 H atoms on the product side, a coefficient of 2 in front of H2O will balance hydrogen atoms. C4H9OH(g) + O2(g) ÷ 4CO2(g) + 5H2O(l) (C and H atoms balanced) Finally, balance the oxygen atoms. Since there are 3 O atoms on the reactant side and 13 O atoms on the product side, multiplying O2 by 6 will balance oxygen atoms. C4H9OH(g) + 6O2(g) ÷ 4CO2(g) + 5H2O(l) d) (all atoms balanced) This reaction between an ionic compound and an oxoacid should suggest a double displacement reaction of the general form AB + CD ÷ AD + CB. In this case the metal copper (Bi) and hydrogen in the reactants switch places to form water and an ionic compound between copper and nitrate. Notice that the ionic compound reactant (CuO) involves Cu2+. Since nitrate ion has a -1 charge, the correct formula for the ionic compound formed is Cu(NO3)2. This combination of a +2 cation and a -1 anion is expected to soluble. This results in the unbalanced equation: CuO(s) + HNO3(aq) ÷ Cu(NO3)2(aq) + H2O(l) Notice that the copper atoms and oxygen atoms are balanced. To balance this equation, place a coefficient of 2 in front of HNO3 to balance hydrogen atoms and the nitrate units. The balanced equation is: CuO(s) + 2HNO3(aq) ÷ Cu(NO3)2(aq) + H2O(l) S.4.15 e) CHAPTER 4 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The reaction of two elemental forms should suggest a combination reaction. Since a metal (Al) is reacting with a nonmetal (Br2) an ionic compound will form. Aluminum is expected to form an Al3+ ion and bromine is expected to form a Br- ion. The correct formula for the ionic compound formed is AlBr3. This combination of a +3 cation and a -1 anion is expected to soluble. This results in the unbalanced equation: Al(s) + Br2(l) ÷ AlBr3(aq) The aluminum atoms are balanced but there are 2 bromine atoms on the reactant side and 3 bromine atoms on the product side. Multiplying Br2 by 3/2 will balance bromine atoms. Al(s) + 3/2Br2(l) ÷ AlBr3(aq) The fractional coefficient in front of Br2 can be converted to the smallest whole number by multiplying by 2. This requires that all other coefficients be multiplied by 2 in order to retain the atom balance. The resulting balanced equation is: 2Al(s) + 3Br2(l) ÷ 2AlBr3(aq)