Chemistry Solutions PDF

b. 250.0 mL of 0.250 M Zn(NO3)2 solution b. 0.100 L c. 150.0 g of 2.25% (w/w) NaCl solution...

b. 250.0 mL of 0.250 M Zn(NO3)2 solution b. 0.100 L c. 150.0 g of 2.25% (w/w) NaCl solution c. 1.10 L d. 125.0 mL of 0.75% (w/v) KCl solution d. 350.0 mL 7.43 Explain how you would prepare the following solutions us- 7.50 What is the molarity of the solution prepared by diluting ing pure solute and water. Assume water has a density of 50.0 mL of 0.195 M KBr to each of the following volumes? 1.00 g/mL. a. 1.50 L a. 250. mL of a 2.00 M NaOH solution b. 200. mL b. 500. mL of a 40.0% (v/v) alcohol solution (C2H5OH) c. 500. mL c. 100. mL of a 15.0% (w/v) glycerol solution. Glycerol is a d. 700. mL liquid with a density of 1.26 g/mL. Describe two ways to measure out the amount of glycerol needed. Solution Stoichiometry (Section 7.6) d. Approximately 50. mL of a normal saline solution, 0.89% 7.51 How many milliliters of 6.00 M HCl solution would be (w/w) NaCl needed to react exactly with 20.0 g of pure solid NaOH? 7.44 * A saline solution used for intravenous injections is a 0.16 M HClsaqd 1 NaOHssd S NaClsaqd 1 H2Os/d solution of sodium chloride. Calculate the mass in grams of sodium chloride needed to prepare 250. mL of solution. 7.52 How many grams of solid Na2CO3 will react with 250. mL of 7.45 Calculate the following for these common solutions found in a 1.25 M HCl solution? clinics and hospitals: Na2CO3ssd 1 2HClsaqd S 2NaClsaqd 1 CO2sgd 1 H2Os/d a. The number of moles of dextrose (glucose) in 250. mL of a 0.278 M dextrose solution. 7.53 How many milliliters of 0.250 M HCl solution will exactly react with 10.5 g of solid NaHCO3? b. The number of grams of NaCl in 120. mL of a 0.154 M normal saline solution. NaHCO3ssd 1 HClsaqd S NaClsaqd 1 CO2sgd 1 H2Os/d c. The number of grams of iodine in 20.0 mL of a 3.00% 7.54 How many milliliters of 0.250 M AgNO3 solution will ex- (w/v) tincture of iodine solution. actly react with 25.0 mL of a 0.200 M NaCl solution? d. The number of milliliters of alcohol in 250. mL of a NaClsaqd 1 AgNO3saqd S NaNO3saqd 1 AgClssd 90.0% (v/v) disinfectant solution. 7.55 How many milliliters of 0.120 M Na2S solution will exactly 7.46 Calculate the following: ­react with 35.0 mL of a 0.150 M AgNO3 solution? a. The number of grams of Li2CO3 in 250. mL of 1.75 M Li2CO3 solution 2AgNO3saqd 1 Na2Ssaqd S Ag2Sssd 1 2NaNO3saqd b. The number of moles of NH3 in 200. mL of 3.50 M NH3 7.56 How many milliliters of 0.225 M NH3 solution will exactly solution ­react with 30.0 mL of a 0.190 M H2SO4 solution? c. The number of mL of alcohol in 250. mL of 12.5% (v/v) 2NH3(aq) 1 H2SO4(aq) S (NH4)2SO4(aq) solution d. The number of grams of CaCl2 in 50.0 mL of 4.20% 7.57 How many milliliters of 0.124 M NaOH solution will exactly (w/v) CaCl2 solution react with 35.0 mL of a 0.210 M H3PO4 solution? 7.47 Explain how you would prepare the following dilute solu- 3NaOH(aq) 1 H3PO4(aq) S Na3PO4(aq) 1 3H2O(O) tions from the more concentrated ones: 7.58 How many milliliters of 0.124 M NaOH solution will exactly a. 200. mL of 0.500 M HCl from a 6.00 M HCl solution react with 25.0 mL of a 0.210 M HCl solution? b. 50.0 mL of 2.00 M H2SO4 from a 6.00 M H2SO4 solution NaOH(aq) 1 HCl(aq) S NaCl(aq) 1 H2O(O) c. 100. mL of normal saline solution, 0.89% (w/v) NaCl, from a 5.0% (w/v) NaCl solution 7.59 How many milliliters of 0.115 M NaOH solution will exactly d. 250. mL of 5.00% (v/v) acetone from 20.5% (v/v) acetone react with 25.0 mL of a 0.210 M H2SO4 solution? 7.48 Explain how you would prepare the following dilute solu- 2NaOHsaqd 1 H2SO4saqd S Na2SO4saqd 1 2H2Os/d tions from the more concentrated ones: 7.60 Stomach acid is essentially 0.10 M HCl. An active ingredient a. 5.00 L of 6.00 M H2SO4 from a 18.0 M H2SO4 solution found in a number of popular antacids is calcium carbonate, b. 250. mL of 0.500 M CaCl2 from a 3.00 M CaCl2 solution CaCO3. Calculate the number of grams of CaCO3 needed to c. 200. mL of 1.50% (w/v) KBr from a 10.0% (w/v) KBr exactly react with 250.0 mL of stomach acid. solution CaCO3ssd 1 2HClsaqd S CO2sgd 1 CaCl2saqd 1 H2Os/d d. 500. mL of 10.0% (v/v) alcohol from 50.0% (v/v) alcohol 7.61 An ingredient found in some antacids is magnesium hydrox- 7.49 What is the molarity of the solution prepared by diluting ide, Mg(OH)2. Calculate the number of grams of Mg(OH)2 25.0 mL of 0.412 M Mg(NO3)2 solution to each of the follow- needed to exactly react with 200. mL of stomach acid ing final volumes? (see Exercise 7.60). a. 40.0 mL MgsOHd2ssd 1 2HClsaqd S MgCl2saqd 1 2H2Os/d Even-numbered exercises answered in Appendix B.       233 Exercises with an asterisk (*) are more challenging. Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53384_ch07_ptg01.indd 233 22/06/21 4:42 PM  Solution Properties (Section 7.7) 7.71 Calculate the osmotic pressure of a 0.125 M solution of 7.62 Before it is frozen, ice cream is essentially a solution of Na2SO4, a strong electrolyte. sugar, flavorings, etc., dissolved in water. Use the idea of col- 7.72 Calculate the osmotic pressure of a 0.200 M solution of ligative solution properties and explain why a mixture of ice Na2SO4, a strong electrolyte. (and water) and salt is used to freeze homemade ice cream. 7.73 Calculate the osmotic pressure of a 0.250 M solution of Why won’t just ice work? methanol, a nonelectrolyte. 7.63 If you look at the labels of automotive products used to prevent 7.74 Calculate the osmotic pressure of a solution that contains 95.0 radiator freezing (antifreeze) and radiator boiling, you will find g of the nonelectrolyte urea, CH4N2O, per 500. mL of solution. the same ingredient listed, ethylene glycol. Use the idea of col- 7.75 Calculate the osmotic pressure of a solution that contains ligative properties to explain how the same material can prevent 1.20 mol of CaCl2, a strong electrolyte, in 1.50 L. an automobile cooling system from freezing and boiling. 7.76 * Calculate the osmotic pressure of a solution that has a freez- 7.64 Calculate the boiling and freezing points of water solutions ing point of 20.35ºC. that are 1.50 M in the following solutes: 7.77 * Calculate the osmotic pressure of a solution that is 0.122 M in a. KCl, a strong electrolyte solute and has a boiling point 0.19ºC above that of pure water. b. glycerol, a nonelectrolyte 7.78 * Calculate the osmotic pressure of a solution that contains c. (NH4)2SO4, a strong electrolyte 5.30 g of NaCl and 8.20 g of KCl per 750. mL. d. Al(NO3)3, a strong electrolyte 7.79 Calculate the osmotic pressure of a solution that contains 7.65 Calculate the boiling and freezing points of water solutions 220.0 g of ethylene glycol (C2H6O2), a nonelectrolyte, per that are 1.15 M in the following solutes: liter. a. KBr, a strong electrolyte 7.80 Suppose an osmotic membrane separates a 5.00% sugar so- b. ethylene glycol, a nonelectrolyte lution from a 10.0% sugar solution. In which direction will ­water flow? Which solution will become diluted as osmosis c. (NH4)2CO3, a strong electrolyte takes place? d. Al2(SO4)3, a strong electrolyte 7.66 Calculate the boiling and freezing points of the following so- Colloids (Section 7.8) lutions. Water is the solvent, unless otherwise indicated. 7.81 Explain how the following behave in a colloidal suspension: a. A 0.50 M solution of urea, a nonelectrolyte dispersing medium, dispersed phase, and colloid emulsifying b. A 0.250 M solution of CaCl2, a strong electrolyte agent. c. A solution containing 100. g of ethylene glycol (C2H6O2), a 7.82 Explain why detergents or soaps are needed if water is to be nonelectrolyte, per 250. mL used as a solvent for cleaning clothes and dishes. 7.67 Calculate the boiling and freezing points of the following so- lutions. Water is the solvent, unless otherwise indicated. Dialysis (Section 7.9) 7.83 Suppose you have a bag made of a membrane like that in a. A solution containing 50.0 g of H2SO4, a strong electro- Figure 7.26. Inside the bag is a solution containing water lyte (both Hs dissociate), per 250. mL and dissolved small molecules. Describe the behavior of the b. A solution containing 200. g of table sugar (C12H22O11), system when the bag functions as an osmotic membrane and a nonelectrolyte, per 250. mL when it functions as a dialyzing membrane. c. A solution containing 75.0 g of octanoic acid (C8H16O2), 7.84 Each of the following mixtures was placed in a dialysis bag a nonelectrolyte, in enough benzene to give 250. mL of like the one shown in Figure 7.26. The bag was immersed in solution pure water. Which substances will pass through the bag into 7.68 Calculate the osmolarity for the following solutions: the surrounding water? a. A 0.15 M solution of glycerol, a nonelectrolyte a. NaCl solution and starch (colloid) b. A 0.15 M solution of (NH4)2SO4, a strong electrolyte b. Urea (small organic molecule) and starch (colloid) c. A solution containing 25.3 g of LiCl (a strong electrolyte) c. Albumin (colloid), KCl solution, and glucose solution per liter (small organic molecule) 7.69 Calculate the osmolarity for the following solutions: Additional Exercises a. A 0.25 M solution of KCl, a strong electrolyte 7.85 When 5.0 mL of water and 5.0 mL of rubbing alcohol are b. A solution containing 15.0 g of urea (CH4N2O), a non- mixed together, the volume of the resulting solution is electrolyte, per 500. mL 9.7 mL. Explain in terms of molecules why the final solution c. A solution containing 50.0 mL of ethylene glycol volume is not 10.0 mL. (C2H6O2), a nonelectrolyte with a density of 1.11 g/mL, 7.86 A student made a 7.00% (w/v) solution by dissolving per 250. mL 7.00 grams of sodium chloride (salt) in an appropriate volume Note: In Exercises 7.70–7.79, assume the temperature is 25.0°C, of water. ­Assume the density of the water was 1.00 g/mL, and express your answer in torr, mmHg, and atm. and that there was no increase in volume when the salt was 7.70 Calculate the osmotic pressure of any solution with an osmo- added to the w ­ ater. Calculate the density of the resulting larity of 0.250. ­solution in g/mL. Even-numbered exercises answered in Appendix B.     234   Exercises with an asterisk (*) are more challenging. Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53384_ch07_ptg01.indd 234 22/06/21 4:42 PM  7.87 Ethylene glycol is mixed with water and used as an anti- 7.91 Can the terms saturated and supersaturated be used to de- freeze in automobile cooling systems. If a 50% (v/v) solu- scribe solutions made of liquids that are soluble in all pro- tion is made using pure ethylene glycol and pure water, it portions? Explain. will have a freezing point of about 2658C. What will hap- 7.92 Refer to Figure 7.4 and propose a reason why fish some- pen to the freezing point of the mixture if ethylene glycol is times die when the temperature of the water in which they added until the concentration is 70% (v/v) ethylene glycol? live increases. Explain your answer. 7.93 Small souvenir salt-covered objects are made by forming the 7.88 How many mL of 1.50 M HCl solution contains enough object out of wire mesh and suspending the mesh object in a HCl to react completely with 0.500 g of zinc metal? The container of water from a salt lake such as the Dead Sea or reaction is Znssd 1 2HClsaqd S ZnCl2saqd 1 H2sgd. the Great Salt Lake. As the water evaporates, the wire mesh 7.89 At 208C ethyl alcohol has a vapor pressure of 43.9 torr, and becomes coated with salt crystals. Describe this process us- the vapor pressure of water at the same temperature is 17.5 ing the key terms introduced in Section 7.2. torr. Suppose a 50% (v/v) solution of water and ethyl alcohol 7.94 Refer to Figure 7.9 and answer the question. How would was brought to a boil and the first vapors given off were col- the solubility of sugar compare in equal amounts of hot and lected and condensed to a liquid. How would the percentage iced tea? of alcohol in the condensed liquid compare to the percent- 7.95 Refer to Figure 7.19 and explain the process as requested. age of alcohol in the original 50% solution (greater than, less Draw simple diagrams showing the initial appearance and than, equal to)? Explain your reasoning. appearance after some time for a similar experiment in which Chemistry for Thought the two liquids are 0.20 M copper sulfate solution and 2.0 M copper sulfate solution. Explain your reasoning. 7.90 When a patient has blood cleansed by hemodialysis, the blood is circulated through dialysis tubing submerged in a 7.96 Strips of fresh meat can be preserved by drying. In one pro- bath that contains the following solutes in water: 0.6% NaCl, cess, the strips are coated with table salt and exposed to the 0.04% KCl, 0.2% NaHCO3, and 0.72% glucose (all percent- air. Use a process discussed in this chapter and one discussed ages are w/v). Suggest one or more reasons why the dialysis in Chapter 6 to explain how the drying takes place. tubing is not submerged in pure water. Even-numbered exercises answered in Appendix B.       235 Exercises with an asterisk (*) are more challenging. Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53384_ch07_ptg01.indd 235 22/06/21 4:42 PM  8 Reaction Rates and Equilibrium Health Career Focus Physician’s Assistant The hardest part is getting your patient experience. I’m lucky that I went to a high school that allowed students release time to at- tend a local technical college. I took advantage of that and earned a certificate in Medical Assisting by the time I graduated, so I worked at a doctor’s office throughout college. The more inten- sive Physician Assistant program I completed later required 2000 hours of patient experience, but some require twice that! I didn’t decide to become a PA right away. I didn’t even decide to go to college right away. I got married and started a family. I liked the Medical Assisting job but working with PAs helped me see more possibilities. I started dreaming about it, then I started planning for it. My Physician Assistant program accepted online credit, and since I have kids, that made it much easier for me to meet the re- quirements. I could do my schoolwork for an hour or two after my children went to bed. It wasn’t easy, but I’m so glad I did it. It’s a lot of hard work, but if you want it, you’ll find a way. Follow-up to this Career Focus appears at the end of the chapter age fotostock/SuperStock before the Concept Summary. Learning Objectives When you have completed your study of this chapter, you 5 Explain how factors such as reactant concentrations, tempera- should be able to: ture, and catalysts influence reaction rates. (Section 8.5) 1 Use the concepts of energy and ­entropy to predict the spon- 6 Relate experimental observations to the establishment taneity of processes and reactions. (Section 8.1) of equilibrium. (Section 8.6) 2 Calculate reaction rates from experimental data. 7 Write equilibrium expressions based on reaction equations, (Section 8.2) and do calculations based on equilibrium expressions. 3 Use the concept of molecular collisions to explain reaction (Section 8.7) characteristics. (Section 8.3) 8 Use Le Châtelier’s principle to predict the influence of 4 Represent and interpret the energy relationships for reactions changes in concentration and reaction temperature on the by using energy diagrams. (Section 8.4) position of equilibrium for reactions. (Section 8.8) 236   Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53384_ch08_ptg01.indd 236 22/06/21 12:46 PM  According to calculations, carbon in the form of a diamond will spon- taneously change into graphite. Owners of diamonds seem unconcerned by this fact; they know from experience that their investments in gems are secure. In real- ity, the change does take place, but occurs so slowly that it is not detectable over many human lifetimes. The low rate of the reaction makes the difference. The element barium is very poisonous when it is taken into the body as barium ions (Ba21). When barium sulfate (BaSO4) dissolves, Ba21 and SO422 ions are formed. However, suspensions of solid BaSO4 are routinely swallowed by p ­ atients undergoing Skitterphoto/2307 images/Pixabay diagnostic X-ray photography of the stomach and intestinal tract. The patients are not affected because very little BaSO4 dissolves in the body. This lack of solubility is an equilibrium property of BaSO4. 8.1 Spontaneous and Nonspontaneous Processes Figure 8.1 The rusting of iron is a spontaneous process. Learning Objective 1 Use the concepts of energy and entropy to predict the spontaneity of processes and reactions. Years ago, tourists in Salt Lake City visited “gravity hill,” where an optical illusion made a stream of water appear to flow uphill. The fascination with the scene came from the appar- ent violation of natural laws—everyone knew that water does not flow uphill. Processes that take place naturally with no apparent cause or stimulus are called spontaneous pro- spontaneous process A process cesses (see Figure 8.1). Nonspontaneous processes take place only as the result of some that takes place naturally with no apparent cause or stimulus. cause or stimulus. For example, imagine you are in the positions depicted in Figure 8.2 and want to move a boulder. Figure 8.2 The problem of ­moving a boulder. a Spontaneous process b Spontaneous process c Nonspontaneous once started process In part (a), the boulder will roll down the hill as soon as you release it; the process begins and takes place spontaneously. In part (b), you must push the boulder a little to get it over the hump, but once started, the boulder spontaneously rolls downhill. The situation in part (c) is very different. You must continually push on the boulder, or it will not move up the hill; at no time can you stop pushing and expect the boulder to continue moving up. The process is non- spontaneous; it takes place only because of the continuous application of an external stimulus. As the boulder rolls downhill, it gives up energy; it moves from a state of high poten- tial energy to a lower one. As you push the boulder uphill, it gains energy (from you). Reaction Rates and Equilibrium  237 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53384_ch08_ptg01.indd 237 22/06/21 12:46 PM  exergonic process A process that Processes that give up energy are called exergonic (energy out), whereas those that gain gives up energy as it takes place. energy are called endergonic (energy in). Very often, energy changes in chemical pro- endergonic process A process cesses involve heat. Those changes that do are referred to as either exothermic (heat out) or that gains or accepts energy as it endothermic (heat in), two terms you have encountered before (see Sections 5.8 and 6.11). takes place. Many spontaneous processes give up energy. A piece of wood, once ignited, spontane- ously burns and liberates energy as heat and light. At normal room temperature, steam spontaneously condenses to water and releases heat. However, some spontaneous pro- cesses take place and either give up no energy or actually gain energy. These spontaneous entropy A measurement or processes are always accompanied by another change called an entropy increase. Entropy indication of the disorder or describes the disorder or mixed-up character (randomness) of a system (see Figure 8.3). randomness of a system. The more Thus, an entropy increase accompanies all spontaneous processes in which energy re- disorderly a system is, the higher its entropy. mains constant or is gained. A drop of ink placed in a glass of water will eventually be- come uniformly distributed throughout the water, even though the water is not stirred. No energy change accompanies the diffusion of the ink, but the distribution of ink throughout the ­water is a more disorderly (higher-entropy) state than when the ink is all together in a single drop. Ice melts spontaneously at 20°C and in the process absorbs heat. The process takes place because the random distribution of water molecules in the liquid is in a higher- entropy state than the orderly molecular arrangement found in crystalline ice. Figure 8.3 The mixture on the left has higher entropy (more randomness, or mixed-up character) than the mixture on the right. © Spencer L. Seager © Spencer L. Seager Energy and entropy changes influence the spontaneity of processes in several ways: 1. A process will always be spontaneous if the energy decreases and the entropy in- creases. An example of such a process is the burning of a piece of wood, in which heat is given up and the gaseous products of combustion are distributed throughout the surroundings, enabling the entropy to increase. 2. When a spontaneous process is accompanied by an energy increase, a large entropy increase must also occur. Thus, when ice spontaneously melts at 20°C, the increase in entropy is large enough to compensate for the increase in energy that also takes place. 3. A spontaneous process accompanied by an entropy decrease must also be accompa- nied by a compensating energy decrease. Thus, when water spontaneously freezes at 0°C, the energy loss compensates for the entropy decrease that occurs as the water molecules assume the well-ordered arrangement in ice. stable substance A substance It is useful to think of energy and entropy changes in terms of the directions favoring that does not undergo spontaneous spontaneity and the directions not favoring spontaneity. It is apparent from the preceding changes under the surrounding discussion that spontaneity is favored by an energy decrease and/or an entropy increase. conditions. Processes in which one of these factors changes in a nonspontaneous direction will be spontaneous only if the other factor changes in a spontaneous direction by a large enough amount to compensate for the nonspontaneous change. The influences listed under 2 and 3 Ukki Studio/Shutterstock.com above are examples of this. A substance that does not undergo spontaneous changes is said to be a stable substance (see Figure 8.4). However, stability depends on the surrounding conditions, and a change in those conditions might cause a nonspontaneous process to become spontaneous. Ice is a stable solid at 210°C and 760 torr pressure but spontaneously melts to a liquid at 5°C and 760 torr. Figure 8.4 Both gold and Wood is stable at room temperature but spontaneously burns when its temperature equals or diamonds are valuable because of exceeds the ignition temperature. In the following discussions, the surrounding conditions their beauty and their stability. are assumed to be those normally encountered. Otherwise, the differences are specified. 238   Chapter 8 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53384_ch08_ptg01.indd 238 22/06/21 12:46 PM  8.2 Reaction Rates Learning Objective 2 Calculate reaction rates from experimental data. The speed of a reaction is called the reaction rate. It is determined experimentally as reaction rate The speed of a the change in concentration of a reactant or product divided by the time required for the reaction. change to occur. This average rate is represented for changes in product concentration in Equation 8.1: DC Ct 2 C0 Rate 5 5 (8.1) Dt Dt where the delta symbol (D) stands for change. Ct and C0 are the concentrations of a prod- uct at the end and beginning of the reaction, respectively, of the measured time change, Dt. The time can be measured in any convenient unit. Example 8.1  Calculating Reaction Rates Nitric oxide, NO, is used in hospitals for lung vasodilation of preterm newborns and pa- tients with pulmonary distress. The gases NO2 and CO react as follows: NO2sgd 1 COsgd S NOsgd 1 CO2sgd Calculate the average rate of the reaction if pure NO2 and CO are mixed and after 50.0 seconds the concentration of CO2 is found to be 0.0160 mol/L. Solution STEP 1 Examine the question for what is given and what is unknown. Given: the balanced reaction; concentration of CO2 after 50.0 seconds 5 0.0160 mol/L Unknown: average rate of reaction 5 ? STEP 2 Identify a formula that links the known and unknown terms. DC Ct 2 C0 Equation 8.1: 5 DT Dt STEP 3 Enter the data into the formula. Because the reaction was started by mixing pure NO2 and CO, the initial concentration C0 of CO2 was 0.000 mol/L. After 50.0 seconds, the concentration Ct of CO2 was 0.0160 mol/L. Ct 2 C0 0.0160 molyL 2 0.000 molyL molyL Rate 5 5 5 3.20 3 1024 Dt 50.0 s s The answer is 3.20 3 1024 mol per liter per second, meaning that during the 50.0 second time interval, an average of 3.20 3 1024 mol CO2 was formed in each liter of reacting mixture each second. STEP 4 Check to see whether the answer makes sense. Yes. The answer has the correct units and the correct number of significant figures, and the value seems reasonable. ✔ Learning Check 8.1 The Ce41 and Fe21 ions react in solution as follows: Ce41 saqd 1 Fe21 saqd S Ce31 saqd 1 Fe31 saqd Solutions of Ce41 and Fe21 are mixed and allowed to react. After 75.0 seconds, the concentration of Ce31 is found to be 1.50 3 1025 mol/L. Calculate the average rate of the reaction. Reaction Rates and Equilibrium  239 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53384_ch08_ptg01.indd 239 22/06/21 12:46 PM  8.3 Molecular Collisions Learning Objective 3 Use the concept of molecular collisions to explain reaction characteristics. Chemical equations such as those in Example 8.1 and Learning Check 8.1 are useful in identifying reactants and products and in representing the stoichiometry of a reaction. However, they indicate nothing about how a reaction takes place—how reactants get together (or break apart) to form products. The explanation of how a reaction occurs is reaction mechanism A detailed called a reaction mechanism. Reaction mechanisms are not discussed much in this book, explanation of how a reaction but most are based on the following assumptions: actually takes place. 1. Reactant particles must collide with one another in order for a reaction to occur. 2. Particles must collide with at least a certain minimum total amount of energy if the collision is to result in a reaction. 3. In some cases, colliding reactants must be oriented in a specific way if a reaction is to occur. The validity of the first assumption is fairly obvious. Two molecules cannot react with each other if they are far apart. In order to break bonds, exchange atoms, and form new Martyn F. Chillmaid/Science Source bonds, they must come in contact. There are, however, some exceptions, such as decom- positions in which molecules break into fragments and processes in which molecules react by an internal rearrangement of their atoms. In general, reactions take place more rapidly in the gaseous or liquid state than in the solid state (see Figure 8.5). This observation verifies the first assumption because mol- ecules of gases and liquids move about freely and can undergo many more collisions than can the rigidly held molecules of solids. Reactions involving solids usually take place only Figure 8.5 The gases HCl and on the solid surface and therefore involve only a small fraction of the total molecules pres- NH3 react to form a white cloud ent in the solid. As the reaction proceeds and the products dissolve, diffuse, or fall from of NH4Cl (ammonium chloride) as the two bottles are brought close the surface, fresh solid is exposed. In this way, the reaction proceeds into the solid. The together. rusting of iron is an example of such a process. If collisions were the only factor, then, most gaseous and liquid state reactions would internal energy The energy take place almost instantaneously if every collision results in a reaction. Such high reac- associated with vibrations within tion rates are not observed, a fact that brings us to assumptions 2 and 3. molecules. One of the ways to speed up a chemical reaction is to add energy in the form of heat. activation energy Energy needed The added heat increases both the average speed (kinetic energy) and the internal energy to start some spontaneous processes. of the molecules. Internal energy is the energy associated with molecular vibrations. An Once started, the processes continue increase in internal energy increases the amplitude of molecular vibrations and, if large without further stimulus or energy enough, breaks bonds (see Figure 8.6). Internal energy is also increased by the conversion from an outside source. of some kinetic energy into internal energy during collisions. When heat is added, both the kinetic and the internal energy of molecules is increased. This increases the frequency and speed of collisions and the chances that a collision will cause a sufficient increase in Bond internal energy to break bonds and bring about a reaction (see Figure 8.7). In some reaction mixtures, the average total energy of the molecules is too low at the Low internal energy prevailing temperature for a reaction to proceed at a detectable rate; the reaction mixture is stable. However, some reactions can be started by providing activation energy. Once the reaction is started, enough energy is released to activate other molecules and keep the re- action going at a good rate. The energy needed to get the boulder over the small hump in Bond being stretched Figure 8.2(b) is a kind of activation energy. In many chemical reactions, activation energy Higher internal energy causes bonds in reactant molecules to break. When the broken bonds react to form the new bonds of the products, energy is released that can cause bonds in more reactant molecules to break and the reaction to continue. The striking of a kitchen match is a good example. Acti- vation energy is provided by rubbing the match head against a rough surface. Once started, Bond breaks the match continues to burn spontaneously. A number of gases are routinely used in hospitals. Some of these form very flammable Internal energy high or even potentially explosive mixtures. Cyclopropane, a formerly used anesthetic, will enough to break bonds burn vigorously in the presence of the oxygen in air. However, a mixture of the two will Figure 8.6 The internal energy not react unless activation energy is provided in the form of an open flame or a spark. of molecules. This is the reason that extreme precautions are taken to avoid open flames and sparks in a 240   Chapter 8 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53384_ch08_ptg01.indd 240 22/06/21 12:47 PM  Health Connections 8.1 Energy Drinks This feature will focus on energy drinks. aware that caffeine is a diuretic that might increase the risk It is difficult to lump all energy drinks together because of dehydration. different products contain different proprietary blends of in- Recommendations: gredients, but they all contain some stimulant(s), most often 1. If you consistently lack energy, consider healthier alter- caffeine. Naturally caffeinated drinks (tea or coffee) are natives such as scheduling an additional 10–15 minutes usually not considered to be “energy drinks.” Typical popu- of sleep each night. lar soft drinks that contain 20–50 mg of caffeine per 12 oz 2. Eat a healthful diet and follow a regular exercise/recre- serving are also not considered to be energy drinks in this ation program. discussion. 3. Evaluate the amount and quality of sleep you get. Other common ingredients of energy drinks include amino Energy drinks that help you get through the day might acids (taurine and carnitine), vitamins (in nominal amounts interfere with your sleep architecture and actually de- that have little/no benefit), carbohydrates, and herbal extracts crease the quality and quantity of your sleep. That leads (ginseng, guarana, yerba maté, and ginkgo biloba in variable to a vicious cycle. amounts that are unregulated in spite of having numerous in- 4. If you intend to decrease your energy drink consump- teractions with drugs). tion for any reason, do not stop abruptly. An abrupt stop Even though the amount of caffeine in a typical energy might cause withdrawal symptoms such as headache drink is not more per ounce than brewed coffee, the mar- and fatigue. keted container size of energy drinks is much larger than a cup of coffee, and most consumers drink the entire can. This means that the total amount of caffeine consumed will have both good and bad pharmacological effects. The good effects include increased mental alertness at work or school, since caffeine stimulates the central and au- tonomic nervous systems. This combats fatigue while driv- ing, and might help prevent auto accidents. The possible bad effects are as follows: Causes nervousness, irritability, anxiety, and phobias. Causes insomnia. Might be more difficult to fall asleep Chuck Wagner/Shutterstock.Com and stay asleep and result in less deep sleep. Causes rapid heartbeat. Can induce dysrhythmias, heart attacks, and death. Causes increased blood pressure, gastrointestinal distress, and muscle tremors. Energy drinks are often consumed before exercise or ath- letic events. The additional fluid intake is healthful, but be There are both pros and cons to consuming energy drinks. hospital operating room. Even sparks from static electricity can set off such gaseous mix- tures, so special materials are used in clothing, floor coverings, and so on to prevent static electricity from building up. Oxygen gas does not burn, but it reacts with anything that is combustible. Oxygen is often used in hospitals in concentrations much higher than the 20% found in air. Thus, precautions must be taken to prevent fire or sparks in any areas where oxygen gas is used in high concentrations. Orientation effects are related to which side or end of a particle actually hits another particle during a collision. Orientation effects are unimportant in many reactions. For ex- ample, the orientation of silver ions (Ag1) and chloride ions (Cl2) toward each other during a collision has no effect on the rate of forming AgCl: Ag1saqd 1 Cl2saqd S AgClssd(8.2) Reaction Rates and Equilibrium  241 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53384_ch08_ptg01.indd 241 22/06/21 12:47 PM  © Cengage Learning/Charles D. Winters © Cengage Learning/Charles D. Winters Figure 8.7 The reaction that makes a Cyalume light stick glow takes place more rapidly in hot water (left) than in ice water (right). The light sticks outside the container are at room temperature. Which pair of sticks would glow for a longer time? Explain your reasoning. The reason is that Ag1 and Cl2 are both essentially spherical charged particles. How- A C ever, collision orientation may be important in reactions that involve nonspherical molecules (assumption 3). Consider the following hypothetical reaction: B D AiB 1 CiD S AiC 1 BiD(8.3) a Favorable orientation: A is near C, It is clear that the collision orientation of AiB and CiD shown in Figure 8.8(a) is and B is near D during collision. more favorable to reaction than the orientations shown in Figures 8.8(b) and 8.8(c). A D B C 8.4 Energy Diagrams Learning Objective 4 Represent and interpret the energy relationships for reactions b Unfavorable orientation: A is not by using energy diagrams. near C, and B is not near D during collision. Energy relationships for reactions can be represented by energy diagrams like the one A B D C in Figure 8.9. Notice the similarity to the earlier example of rolling boulders. The energy diagrams for most reactions look generally alike, but there are some dif- orientation: B is near D, ferences. Typical diagrams for exothermic (exergonic) and endothermic (endergonic) c Unfavorable but A and C are far removed during reactions are given in Figure 8.10. It is clear that the products of exothermic reactions collision. have lower energy than the reactants and that the products of endothermic reactions Figure 8.8 Molecular orientations have higher energy than the reactants. When exothermic reactions occur, the energy during collisions. difference between reactants and products is released. This energy often appears as Average energy of reactants Activation energy Increasing energy Energy difference between reactants Average energy and products of products Reaction progress Figure 8.9 A typical energy ­diagram for chemical reactions. 242   Chapter 8 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53384_ch08_ptg01.indd 242 22/06/21 12:47 PM  Exothermic reaction Endothermic reaction Figure 8.10 Energy diagrams for exothermic and endothermic Average energy Average reactions. of reactants energy of products Increasing energy Increasing energy Average energy of Average reactants energy of Energy difference Energy difference products between reactants between reactants and products and products Reaction progress Reaction progress heat, so heat is given up to the surroundings. When endothermic reactions occur, the energy (heat) added to the products is absorbed from the surroundings. Also, different reactions generally have different activation energies, a fact easily represented by energy diagrams. White phosphorus, a nonmetallic element, spontaneously bursts into flame if left in a rather warm room (34°C): 4Pssd 1 5O2sgd S P4O10ssd(8.4) Sulfur, another nonmetallic element, will also burn, but it does not ignite until heated to about 232°C: Sssd 1 O2sgd S SO2sgd(8.5) The different activation energies for these two exothermic reactions are represented in Figure 8.11. Low activation energy, High activation energy, Figure 8.11 Differences in high heat of reaction low heat of reaction ­activation energies. Low activation High energy activation 4P + 5O2 energy S + O2 Increasing energy Increasing energy 71 kcal SO2 720 kcal P4O10 Reaction progress Reaction progress It should now be clear why we stressed the notion that some substances that are “stable” at normal living conditions undergo spontaneous changes at other conditions. It is simply that a higher temperature would provide the necessary activation energy. In a room hotter than 232°C, for example, both white phosphorus and sulfur would be “unstable” in the presence of oxygen. 8.5 Factors That Influence Reaction Rates Learning Objective 5 Explain how factors such as reactant concentrations, temperature, and catalysts influence reaction rates. Reaction rates are influenced by a number of different factors. Four factors that affect the rates of all reactions are: 1. The nature of the reactants. 2. The concentration of the reactants. Reaction Rates and Equilibrium  243 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53384_ch08_ptg01.indd 243 22/06/21 12:47 PM  3. The temperature of the reactants. 4. The presence of catalysts. The formation of insoluble AgCl by mixing together solutions containing Ag1 and Cl2 ions is represented by Equation 8.2, given earlier. The white solid AgCl forms the instant the two solutions are mixed. This behavior is typical of reactions involving ionic reac- tants. The high reaction rate results from the attraction of the charged reactants to each other. In contrast, reactions that require covalent bonds to be broken or formed often pro- ceed slowly. The production of water gas, a mixture of hydrogen (H2) and carbon monox- ide (CO), is a reaction involving covalent bonds: Heat H2O(g) 1 C(s) CO(g) 1 H2(g) (8.6) In addition, other structural characteristics of reactants, such as bond polarity or molecular size, may also be important factors in reaction rates. The influence of reactant concentration on reaction rates can be illustrated by using the concept of molecular collisions. Suppose a reaction takes place between the hypothetical molecules A and B, which are mixed in a 1:1 ratio. The proposed reaction is effective collision A collision that A 1 B S products(8.7) causes a reaction to occur between the colliding molecules. Collisions with the capability to cause a reaction to occur are called effective collisions. Only collisions between A and B molecules can be classified as effective, since collisions between two A molecules or between two B molecules cannot possibly yield products. Imagine the reaction is begun with two A molecules and two B molecules as shown in Figure 8.12. The example is simplified by looking only at the collisions of a single A mol- ecule. Initially, two of every three collisions of the A molecule will be effective (part (a)). On doubling the number (concentration) of B molecules, the number of effective collisions also doubles, and four of every five collisions is effective (part (b)). e e A B A B e e e B ne ne e B B B A e = possibly effective collisions A ne = noneffective collisions a Low concentration of B b Higher concentration of B Figure 8.12 The effect of concentration on reaction rates. When larger numbers of molecules are used, the results approach more closely those actually observed experimentally for simple chemical reactions. Imagine a mixture con- taining 1000 A molecules and 1000 B molecules. On the average, 1000 out of every 1999 collisions of a single A molecule will involve a B molecule. (Each A molecule has almost a 50–50 chance of bumping into a B molecule.) This gives essentially a 1:1 ratio of effective to noneffective collisions. When the number of B molecules is doubled to 2000, the ratio becomes 2:1, because 2000 of every 2999 collisions will be effective. Therefore, the reac-