Chemical and Physical Changes

Chapter 4 Check for Understanding 4.1 1. Write the shorthand notation for the conversion of liquid water into water vapor. Is this process exothermic or endothermic? Explain. Answer: heat + H2O(l) ÷ H2O(g) (endothermic) Solution Energy must be added to cause molecules in the liquid phase to move fast enough to break away from each other and form a gas. Since heat is added this process is endothermic. 2. When water and sulfuric acid (H2SO4) are mixed together the mixture becomes very warm to the touch. Is this mixing process exothermic or endothermic? Explain. Answer: exothermic Solution Since the mixture becomes warm, heat must be released during the mixing. Such a process is exothermic. Check for Understanding 4.2 1. On an energy level diagram indicate the relative energies of H2O(s) and H2O(g). Is the conversion of the solid to the gas phase, a process called sublimation, exothermic or endothermic? Explain. Answer: H2O(g) Energy (arbitrary units) endo H2O(s) Solution Energy must be added to cause the molecules in the solid phase to move fast enough to break away from each other and form a gas so the energy of the gas phase is higher than that of the solid. Since heat is added the process is endothermic. Check for Understanding 4.3 1. How many joules of energy are needed to raise the temperature of a hot tub containing 1500 kg of water by 5 EC? Answer: 3.1 x 107 J 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 mass of water (m) in grams is 1500 x 103 g. The temperature change (Δt) is 5 EC. Substituting these values and calculating q gives: q  (1500 x103 g )(4.18 J/ g  o C )(5 o C ) = 3.1x107 J The answer is given to 2 significant figures because of the 2 significant figures in the mass. It is assumed that the temperature change is an exact number. 2. Calculate the specific heat of a substance in J/g @EC if 4478 J raises the temperature of 1168 g of the material from 22.9 EC to 28.3 EC. Answer: 0.71 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 4478 J. The mass of the substance (m) in grams is 1168 g and the temperature change (Δt) is 28.3 EC - 22.9 EC = 5.4 EC. Substituting these values and calculating c gives: c  4478 J J = 0.71 o o (1168g)(5.4 C) g C The answer is given to 2 significant figures because of the 2 significant figures in the temperature change. Check for Understanding 4.4 1. Convert each of the following descriptions of chemical reactions into a chemical equation and balance it using whole-number coefficients. Be sure to indicate the physical state of each reactant and product. a) hydrochloric acid reacts with an aqueous solution of sodium hydroxide to form water and an aqueous solution of sodium chloride b) elemental carbon reacts with iron(III) oxide to form elemental iron and carbon monoxide gas c) lithium metal reacts with oxygen gas to form lithium oxide Answers: a) HCl(aq) + NaOH(aq) b) 3C(s) + Fe2O3(s) c) 4Li(s) + O2(g) ÷ ÷ ÷ H2O(l) 2Fe(s) + + NaCl(aq) 3CO(g) 2Li2O(s) 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: hydrochloric acid - The hydro- prefix indicates that this is a binary acid and the stem chlor- tells you that the nonmetal chlorine is bonded to hydrogen. Thus, the binary acid formula will look like HnCl(aq). In order to determine the value for n, remember that in water this molecule will break up to form H+ ions and chloride anions. The charge on the cation must be balanced by the charge on the anion that is formed. Since chlorine is in Group 7A we expect it to form an anion with a minus one charge (Cl-). Thus, only one H+ is needed for each Cl- so n = 1 and the correct formula is HCl(aq). aqueous solution of sodium hydroxide - The metal cation is Na+. The polyatomic anion formula is OH-. Since the ion charges are the same, it takes one cation to balance the charge on the anion and the compound formula is NaOH(aq). Product formulas: water - H2O(l) aqueous solution of sodium chloride - The metal cation is Na+. The anion formula is Cl-. Since the ion charges are the same, it takes one cation to balance the charge on the anion and the compound formula is NaCl(aq). This results in the equation: HCl(aq) + NaOH(aq) ÷ H2O(l) + NaCl(aq) In this case, all of the atoms are balanced so no adjustment of the coefficients is necessary. b) Reactant formulas: elemental carbon - This nonmetal is a solid at room temperature with the formula C(s). iron(III) oxide - The Roman numeral indicates that the cation is Fe3+. The oxide ion formula is O2-. Using the value of the anion charge as the subscript for the cation and the value of the cation charge as the subscript for the anion results in the formula Fe2O3. This ionic compound is a solid at room temperature. Product formulas: elemental iron - This metal is a solid at room temperature with the formula Fe(s). carbon monoxide gas - The name suggests a binary molecular compound with the formula CO(g). This results in the unbalanced equation: C(s) + Fe2O3(s) ÷ Fe(s) + CO(g) To balance this equation, place a coefficient of 3 in front of CO to balanced oxygen atoms. This requires a coefficient of 3 in front of C to balanced carbon atoms. Finally, a coefficient of 2 in front of Fe will balance iron atoms. The balanced equation is: 3C(s) + Fe2O3(s) ÷ 2Fe(s) + 3CO(g) c) Reactant formulas: lithium metal - This metal is a solid at room temperature with the formula Li(s). oxygen gas - This element exists as a diatomic molecule with the formula O2(g). Product formulas: lithium oxide - The metal cation is Li+. The oxide ion formula is O2-. Since it takes 2 cations to balance the charge on each anion, the formula is Li2O. This ionic compound is a solid at room temperature. This results in the unbalanced equation: Li(s) + O2(g) ÷ Li2O(s) To balance this equation, place a coefficient of 2 in front of Li2O to balanced oxygen atoms. This requires a coefficient of 4 in front of Li to balanced lithium atoms. The balanced equation is: 4Li(s) + O2(g) ÷ 2Li2O(s) Check for Understanding 4.5 1. 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) CH3OH(l) + O2(g) ÷ b) Pb(s) + AgNO3(aq) ÷ c) Bi(NO3)3(aq) + H2S(aq) ÷ Answers: a) combustion 2CH3OH(l) + 3O2(g) ÷ 2CO2(g) + 4H2O(l) b) single displacement Pb(s) + 2AgNO3(aq) ÷ 2Ag(s) + Pb(NO3)2(aq) c) double displacement 2Bi(NO3)3(aq) + 3H2S(aq) ÷ Bi2S3(s) + 6HNO3(aq) Solutions a) 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. CH3OH(l) + O2(g) ÷ CO2(g) + H2O(l) Notice that the carbon atoms are already balanced, so balance the hydrogen atoms. Since there are 4 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. CH3OH(g) + O2(g) ÷ CO2(g) + 2H2O(l) (C and H atoms balanced) Finally, balance the oxygen atoms. Since there are 3 O atoms on the reactant side and 4 O atoms on the product side, multiplying O2 by 3/2 will balance oxygen atoms. CH3OH(g) + 3/2O2(g) ÷ CO2(g) + 2H2O(l) (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: 2CH3OH(g) + 3O2(g) ÷ 2CO2(g) + 4H2O(l) (all atoms balanced using whole-number coefficients) b) This reaction between an elemental form and an ionic compound 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 lead (Pb) displaces the silver ion in the compound to form elemental silver, Ag(s), and an ionic compound consisting of lead(II) ions and nitrate (NO3-) ions. Recall that the common ion of lead is Pb2+. This means that the formula of the compound formed between Pb2+ and NO3- is Pb(NO3)2. Since this ionic compound contains a -1 anion that is not OH-, this compound should be soluble in water. The unbalanced equation is: Pb(s) + AgNO3(aq) ÷ Ag(s) + Pb(NO3)2(aq) At this stage Pb and Ag are balanced. Notice that the polyatomic nitrate ion does not undergo any change in composition in this reaction. This means that you can balance NO3- units just like an individual atom. Since there are 2 nitrate ions on the product side, a coefficient of 2 is needed in front of AgNO3 on the reactant side. This upsets the Ag balance. This can be restored by putting a coefficient of 2 in front of Ag on the product side. The balanced equation is: Pb(s) + 2AgNO3(aq) ÷ 2Ag(s) + Pb(NO3)2(aq) c) This reaction between an ionic compound and a binary acid should suggest a double displacement reaction of the general form AB + CD ÷ AD + CB. In this case the metal bismuth (Bi) and hydrogen in the reactants switch places to form an ionic compound between bismuth and sulfur and an oxoacid. Notice that the ionic compound reactant (Bi(NO3)3) involves Bi3+ and NO3- ions. Since sulfur is expected to form a S2- ion, the correct formula for the ionic compound formed is Bi2S3. This combination of a +3 cation and a -2 anion is expected to insoluble. The oxoacid formed from H+ and NO3- has the formula HNO3 and forms an aqueous solution. This results in the unbalanced equation: Bi(NO3)3(aq) + H2S(aq) ÷ Bi2S3(s) + HNO3(aq) To balance this equation, place a coefficient of 3 in front of H2S to balanced sulfur atoms. This requires a coefficient of 6 in front of HNO3 to balance hydrogen atoms. Finally, a coefficient of 2 in front of Bi(NO3)3 will balance both the bismuth atoms and the nitrate units. The balanced equation is: 2Bi(NO3)3(aq) + 3H2S(aq) ÷ Bi2S3(s) + 6HNO3(aq) 2. Indicate whether each of the following compounds is expected to be soluble or insoluble in water. a) Cu(NO3)2 b) Fe2S3 c) Zn(OH)2 d) K3PO4 Answers: a) soluble b) insoluble c) insoluble d) soluble Solutions Recall the solubility guidelines: (I) soluble ionic compounds contain (+1) and/or (-1) ions (main exception: compounds containing (+2) or (+3) cations and OH-), and (ii) insoluble ionic compounds contain combinations of (+2) or (+3) cations with either (-2) or (-3) anions (main exception: most compounds containing SO42-). a) The -1 anion, since it is not OH-, suggests this compound is soluble in water. b) The combination of a +3 cation and -2 anion suggests this compound is insoluble in water. c) The combination of a +2 cation with OH- suggests this compound is insoluble in water. d) The +1 cation suggests this compound is soluble in water. Chapter 4 1. It is converted into heat. 3. C 4. -45.2 EC 5. E 6. silver 7. The amount of thermal energy (heat) is the same since the same sample (same mass and specific heat) is undergoing the same temperature change (Δt = 26 EC). Recall q = m x c x Δt. 8. 0.24 J/g @EC 9. As water molecules in the gas phase are cooled they begin to slow down and eventually interact strongly enough so that they can only move around each other as the liquid state forms. Further cooling causes the molecules to move even more slowly, eventually forming a 3-dimensional arrangement (lattice) in which each molecule is locked into place by its strong attractions to other molecules as the liquid freezes. 10. Atoms are neither created nor destroyed in chemical reactions. 11. a) 4Na(s) + O2(g) ÷ 2Na2O(s) b) 2NaHCO3(s) ÷ Na2CO3(s) + H2O(l) + CO2(g) c) 2HCl(aq) + Ni(s) ÷ H2(g) + NiCl2(aq) 12. a) 2Na3PO4(aq) + 3BaCl2(aq) ÷ 6NaCl(aq) + Ba3(PO4)2(s) (double displacement) b) 2SO2(g) + O2(g) ÷ 2SO3(g) (combination) c) H2CO3(aq) + 2KOH(aq) ÷ K2CO3(aq) + 2H2O(l) (double displacement) d) Fe2O3(s) + 2Al(s) ÷ 2Fe(s) + Al2O3(s) (single displacement) e) 4C3H5(NO3)3(l) ÷ 12CO2 + 10H2O + 6N2 + O2 (decomposition) 13. C 14. C 15. D 16. a) double displacement 3Mg(OH)2(s) + 2H3PO4(aq) ÷ Mg3(PO4)2(s) + 6H2O(l) b) combustion 2C2H6O2(l) + 5O2(g) ÷ 4CO2(g) + 6H2O(l) c) single displacement Cl2(aq) + 2NaI(aq) ÷ 2NaCl(aq) + I2(s) d) combination 4Ag(s) + O2(g) ÷ 2Ag2O(s) 17. iron(III) perchlorate and potassium hydroxide nickel acetate and sodium sulfide NiS Fe(OH)3