{"id":7965,"date":"2021-06-08T21:58:20","date_gmt":"2021-06-08T21:58:20","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/end-of-chapter-17-material\/"},"modified":"2021-10-26T18:33:06","modified_gmt":"2021-10-26T18:33:06","slug":"end-of-chapter-17-material","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/end-of-chapter-17-material\/","title":{"raw":"End-of-Chapter Material","rendered":"End-of-Chapter Material"},"content":{"raw":"[latexpage]\r\n
\r\n

Additional Exercises<\/p>\r\n\r\n<\/header>\r\n

\r\n
    \r\n \t
  1. What factors affect the rate of a reaction?<\/li>\r\n \t
  2. How does a decrease in temperature affect the reaction rate? Explain the outcome by describing changes that occur at the molecular level.<\/li>\r\n \t
  3. For which of the following two reactions would you expect the orientation of the molecules to be more important?\r\n
      \r\n \t
    1. AB + C \u2192 AC + B<\/li>\r\n \t
    2. D + E \u2192 F<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
    3. Determine the relative rates of disappearance of reactants and formation of products for the following reactions:\r\n
        \r\n \t
      1. N2<\/sub>(g) + 3H2<\/sub>(g) \u2192 2NH3<\/sub>(g)<\/li>\r\n \t
      2. 2A + 3B \u2192 4C<\/li>\r\n \t
      3. 2N2<\/sub>O5<\/sub> \u2192 4 NO2<\/sub> + O2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
      4. Methanol reacts with hydrochloric acid to produce methyl chloride and water by the following reaction:\r\n

        CH3<\/sub>OH(aq) + HCl(aq) \u2192\u00a0CH3<\/sub>Cl(aq) + H2<\/sub>O(\u2113)\r\nThe following data were obtained for this reaction at a particular temperature:<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Time (min)<\/th>\r\n[CH3OH] (M)<\/th>\r\n<\/tr>\r\n
        0.0<\/td>\r\n1.95<\/td>\r\n<\/tr>\r\n
        60.0<\/td>\r\n1.50<\/td>\r\n<\/tr>\r\n
        120.0<\/td>\r\n1.24<\/td>\r\n<\/tr>\r\n
        180.0<\/td>\r\n1.04<\/td>\r\n<\/tr>\r\n
        240.0<\/td>\r\n0.85<\/td>\r\n<\/tr>\r\n
        300.0<\/td>\r\n0.70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
          \r\n \t
        1. Calculate the average reaction rate between 0 and 60 minutes and 180 and 240 minutes. Which is faster? Explain using your knowledge of factors that can affect rate.<\/li>\r\n \t
        2. Determine the instantaneous rate at 200 minutes.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
        3. For the following reaction: 3 E + 2 F \u2192 2 G, the rate law is Rate = k<\/em>[F]2<\/sup>.\r\n
            \r\n \t
          1. How does the rate change if [E] is doubled?<\/li>\r\n \t
          2. How does the rate change is [F] is doubled?<\/li>\r\n \t
          3. What is the overall reaction order?<\/li>\r\n \t
          4. What are the units for the rate constant for this reaction?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
          5. The rate of oxidation of bromide ions by bromate in an acidic aqueous solution is\r\n

            6 H+<\/sup>\u00a0+ BrO3<\/sub>\u2013<\/sup> + 5Br\u2013<\/sup> \u2192 3Br2<\/sub> + 3H2<\/sub>O<\/p>\r\nand is found to follow the rate law\r\n

            Rate\u00a0=\u00a0k<\/em>[Br\u2013<\/sup>][BrO3<\/sub>\u2013<\/sup>][H+<\/sup>]2<\/sup><\/p>\r\nWhat happens to the rate if, in separate experiments:\r\n

              \r\n \t
            1. [BrO3<\/sub>\u2013<\/sup>] is doubled?<\/li>\r\n \t
            2. the pH is increased by one unit?<\/li>\r\n \t
            3. the solution is diluted to twice its volume, with the pH kept constant by use of a buffer?[footnote]Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0[\/footnote]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
            4. The following data was obtained using the initial rate method:\r\n

              A + B \u2192 C + D<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
              Experiment<\/th>\r\n[A] (M)<\/th>\r\n[B] (M)<\/th>\r\nInitial Rate (M\/s)<\/th>\r\n<\/tr>\r\n
              1<\/td>\r\n2.0 \u00d7 10\u22123<\/sup><\/td>\r\n2.0 \u00d7 10\u22123<\/sup><\/td>\r\n1.45 \u00d7 10\u22124<\/sup><\/td>\r\n<\/tr>\r\n
              2<\/td>\r\n4.0 \u00d7 10\u22123<\/sup><\/td>\r\n2.0 \u00d7 10\u22123<\/sup><\/td>\r\n2.90 \u00d7 10\u22124<\/sup><\/td>\r\n<\/tr>\r\n
              3<\/td>\r\n2.0 \u00d7 10\u22123<\/sup><\/td>\r\n4.0 \u00d7 10\u22123<\/sup><\/td>\r\n2.90 \u00d7 10\u22124<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
                \r\n \t
              1. Determine the rate law for this equation.<\/li>\r\n \t
              2. Calculate the rate constant, including units.<\/li>\r\n \t
              3. Determine the rate when [A] = 0.250 M and [B] = 0.100 M.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
              4. A study of the gas-phase reduction of nitric oxide by hydrogen\r\n

                2NO + 2H2<\/sub> \u2192 N2<\/sub> + 2H2<\/sub>O<\/p>\r\nyielded the following initial-rate data (all pressures in torr):\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                Experiment<\/th>\r\nP<\/em>(NO)<\/th>\r\nP<\/em>(H2<\/sub>)<\/th>\r\nInitial Rate (torr s\u22121<\/sup>)<\/th>\r\n<\/tr>\r\n
                1<\/td>\r\n359<\/td>\r\n300<\/td>\r\n1.50<\/td>\r\n<\/tr>\r\n
                2<\/td>\r\n300<\/td>\r\n300<\/td>\r\n1.03<\/td>\r\n<\/tr>\r\n
                3<\/td>\r\n152<\/td>\r\n300<\/td>\r\n0.25<\/td>\r\n<\/tr>\r\n
                4<\/td>\r\n300<\/td>\r\n289<\/td>\r\n1.00<\/td>\r\n<\/tr>\r\n
                5<\/td>\r\n300<\/td>\r\n205<\/td>\r\n0.71<\/td>\r\n<\/tr>\r\n
                6<\/td>\r\n200<\/td>\r\n147<\/td>\r\n0.51<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFind the order of the reaction with respect to each component.[footnote]Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0[\/footnote]<\/li>\r\n \t
              5. What should be plotted on the x<\/em>- and y<\/em>-axis to obtain a straight line for:\r\n
                  \r\n \t
                1. a first-order reaction?<\/li>\r\n \t
                2. a zero-order reaction?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                3. What does the slope represent for:\r\n
                    \r\n \t
                  1. a second-order reaction kinetic plot?<\/li>\r\n \t
                  2. a first-order reaction kinetic plot?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                  3. Common sugar, sucrose C12<\/sub>H22<\/sub>O11<\/sub>, reacts in dilute acidic solutions to produce glucose and fructose, both having the molecular formula C6<\/sub>H12<\/sub>O6<\/sub>.\r\n

                    C12<\/sub>H22<\/sub>O11<\/sub>(aq) + H2<\/sub>O(\u2113) \u2192 2C6<\/sub>H12<\/sub>O6<\/sub>(aq)<\/p>\r\nDuring an experiment, the following data were obtained:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                    Time (hours)<\/th>\r\n[Sucrose] (mM)<\/th>\r\n<\/tr>\r\n
                    0.00<\/td>\r\n316<\/td>\r\n<\/tr>\r\n
                    0.65<\/td>\r\n274<\/td>\r\n<\/tr>\r\n
                    1.33<\/td>\r\n238<\/td>\r\n<\/tr>\r\n
                    2.33<\/td>\r\n190<\/td>\r\n<\/tr>\r\n
                    3.50<\/td>\r\n146<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
                      \r\n \t
                    1. What is the order of the reaction?<\/li>\r\n \t
                    2. Determine the rate law and the rate constant.<\/li>\r\n \t
                    3. Calculate the expected concentration of sucrose after 100 min.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                    4. The half-life of a first-order reaction was found to be 10 min at a certain temperature. What is its rate constant in reciprocal seconds?[footnote]Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0[\/footnote]<\/li>\r\n \t
                    5. The mass-241 isotope of americium, widely used as an ionizing source in smoke detectors, has a half-life of 432 years.\r\n
                        \r\n \t
                      1. What fraction of the Am241<\/sup>\u00a0in a smoke detector will have decayed after 50 years?<\/li>\r\n \t
                      2. How long will it take for the activity to decline to 80% of its initial value?<\/li>\r\n \t
                      3. What would be the \"seventh-life\" of Am241<\/sup>?[footnote]Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0[\/footnote]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                      4. A widely used \u201crule of thumb\u201d for the temperature dependence of a reaction rate is that a 10\u00b0C rise in the temperature approximately doubles the rate. (This is obviously not generally true, especially when a strong covalent bond must be broken.) For a reaction that shows this behaviour, what would the activation energy be?[footnote]Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0[\/footnote]<\/li>\r\n \t
                      5. The temperature dependence of the rate constant, k<\/em>, for a reaction was found to be:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                        Temperature (K)<\/th>\r\nk<\/em> (M\u22121<\/sup>s\u22121<\/sup>)<\/th>\r\n<\/tr>\r\n
                        500<\/td>\r\n0.025<\/td>\r\n<\/tr>\r\n
                        600<\/td>\r\n0.300<\/td>\r\n<\/tr>\r\n
                        700<\/td>\r\n1.500<\/td>\r\n<\/tr>\r\n
                        800<\/td>\r\n7.000<\/td>\r\n<\/tr>\r\n
                        900<\/td>\r\n28.000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nCalculate the activation energy (Ea<\/sub><\/em>) and A<\/em>.<\/li>\r\n \t
                      6. State the molecularity and rate law for each of the following elementary steps:\r\n
                          \r\n \t
                        1. Cl(g) + CCl3<\/sub>(g) \u2192 CCl4<\/sub>(\u2113)<\/li>\r\n \t
                        2. Br2<\/sub> \u2192 2Br<\/li>\r\n \t
                        3. 2A \u2192 B<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                        4. A reaction is believed to occur by the following two elementary steps:\r\n

                          2NO2<\/sub> \u2192 NO3<\/sub> + NO<\/p>\r\n

                          NO3<\/sub>\u00a0\u2192 NO + O2<\/sub><\/p>\r\nWith this in mind:\r\n

                            \r\n \t
                          1. Write the balanced overall equation of this reaction.<\/li>\r\n \t
                          2. Identify any intermediates in the mechanism.<\/li>\r\n \t
                          3. If the first step is much slower than the second, what would you expect the rate law for the overall reaction to be?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                          4. Use the following potential energy diagram to answer questions 19a to 19c:\r\n\r\n[caption id=\"attachment_7962\" align=\"aligncenter\" width=\"342\"]\"\" Potential energy plot of a given reaction mechanism.[\/caption]\r\n
                              \r\n \t
                            1. How many steps are in this mechanism?<\/li>\r\n \t
                            2. How many intermediates are there in this reaction?<\/li>\r\n \t
                            3. What is the slowest step?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                            4. What is a catalyst? Explain how a catalyst affects reaction rates.<\/li>\r\n \t
                            5. Many solid catalysts are commercially sold as fine powders or small spherical beads. Explain why this might be beneficial.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
                              \r\n

                              Answers<\/p>\r\n\r\n<\/header>\r\n

                              \r\n
                                \r\n \t
                              1. temperature, reactant concentrations, physical state (surface area of solids), presence of a catalyst<\/li>\r\n<\/ol>\r\n
                                  \r\n \t
                                1. AB + C \u2192 AC + B<\/li>\r\n<\/ol>\r\n
                                    \r\n \t
                                  1. \r\n
                                      \r\n \t
                                    1. average reaction rate 0\u201360 minutes:\r\n

                                      [latex]-\\dfrac{\\ce{[CH3OH]_60}-\\ce{[CH3OH]_0}}{60.0\\text{ min}-0.0\\text{ min}}=\\dfrac{1.50\\text{ M}-19.5\\text{ M}}{60.0\\text{ min}}=7.50\\times 10^{-3}\\text{ M\/min}[\/latex]<\/p>\r\naverage reaction rate 180\u2013240 minutes:\r\n

                                      [latex]-\\dfrac{\\ce{[CH3OH]_240}-\\ce{[CH3OH]_180}}{240.0\\text{ min}-180.0\\text{ min}}=\\dfrac{0.85\\text{ M}-1.04\\text{ M}}{60.0\\text{ min}}=3.17\\times 10^{-3}\\text{ M\/min}[\/latex]<\/p>\r\nTherefore the average reaction rate between 0\u201360 minutes is faster. This is because, as the reaction proceeds, the reactant gets consumed to make product, lowering the concentration of reactants.<\/li>\r\n \t

                                    2. \"\"\r\nInstantaneous rate at 200 minutes:\r\n

                                      [latex]-\\dfrac{\\text{Rise}}{\\text{Run}}=\\dfrac{1.0\\text{ M}-0.84\\text{ M}}{230\\text{ min}-180\\text{ min}}=\\dfrac{0.16\\text{ M}}{50\\text{ min}}=3.2\\times 10^{-3}\\text{ M\/min}[\/latex]<\/p>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n

                                        \r\n \t
                                      1. \r\n
                                          \r\n \t
                                        1. Since the rate is first-order in bromate, doubling its concentration will double the reaction rate.<\/li>\r\n \t
                                        2. Increasing the pH by one unit will\u00a0decrease<\/em>\u00a0the [H+<\/sup>] by a factor of 10. Since the reaction is second-order in [H+<\/sup>], this will decrease the rate by a factor of 100.<\/li>\r\n \t
                                        3. Dilution reduces the concentrations of both Br2<\/sub>\u00a0and BrO3<\/sub>\u2013<\/sup> to half their original values. Doing this to each concentration alone would reduce the rate by a factor of 2, so reducing both concentrations will reduce the rate by a factor of 4, to (\u00bd) \u00d7 (\u00bd) = \u00bc of its initial value.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
                                            \r\n \t
                                          1. \r\n
                                              \r\n \t
                                            1. Experiments 2 and 3<\/strong>: Reduction of the initial partial pressure of NO by a factor of about 2 [latex]\\left(\\dfrac{300}{152}\\right)[\/latex] results in a reduction of the initial rate by a factor of about 4, so the reaction is second-order in nitric oxide.<\/li>\r\n \t
                                            2. Experiments 4 and 6<\/strong>: Reducing the initial partial pressure of hydrogen by a factor of approximately 2 [latex]\\left(\\dfrac{289}{147}\\right)[\/latex] causes a similar reduction in the initial rate, so the reaction is first-order in hydrogen.<\/li>\r\n \t
                                            3. The rate law is<\/strong>:\u00a0rate\u00a0=\u00a0k<\/em>[NO]2<\/sup>[H2<\/sub>].<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
                                                \r\n \t
                                              1. \r\n
                                                  \r\n \t
                                                1. k<\/em><\/li>\r\n \t
                                                2. \u2212k<\/em><\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
                                                    \r\n \t
                                                  1. From the above equation, [latex]k=\\dfrac{-0.693}{600\\text{ s}}=0.00115\\text{ s}^{-1}[\/latex]<\/li>\r\n<\/ol>\r\n
                                                      \r\n \t
                                                    1. We will centre our ten-degree interval at 300 K. Substituting into the above expression yields:\r\n

                                                      [latex]\\begin{align*}\r\nE_a&=\\dfrac{8.314\\ln \\frac{2}{1}}{\\frac{1}{295}-\\frac{1}{305}} \\\\ \\\\\r\n&=\\dfrac{(8.314)(0.693)}{0.00339-0.00328} \\\\ \\\\\r\n&=\\dfrac{5.76\\text{ J mol}^{-1}\\text{ K}^{-1}}{0.00011\\text{ K}^{-1}} \\\\ \\\\\r\n&=52,400\\text{ J mol}^{-1}=52.4\\text{ kJ mol}^{-1}\r\n\\end{align*}[\/latex]<\/p>\r\n<\/li>\r\n<\/ol>\r\n

                                                        \r\n \t
                                                      1. \r\n
                                                          \r\n \t
                                                        1. bimolecular, rate = k<\/em>[Cl] [CCl3<\/sub>]<\/li>\r\n \t
                                                        2. unimolecular, rate = k<\/em>[Br2<\/sub>]<\/li>\r\n \t
                                                        3. bimolecular 2A, rate = k<\/em>[A]2<\/sup><\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
                                                            \r\n \t
                                                          1. \r\n
                                                              \r\n \t
                                                            1. 3 steps<\/li>\r\n \t
                                                            2. 2 intermediates<\/li>\r\n \t
                                                            3. The second step is the slowest step<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
                                                                \r\n \t
                                                              1. Finely divided powders and spherical beads have larger surface areas than big solid chunks, allowing for more sites for reactions to occur, which further speeds up the reaction.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>","rendered":"
                                                                \n
                                                                \n

                                                                Additional Exercises<\/p>\n<\/header>\n

                                                                \n
                                                                  \n
                                                                1. What factors affect the rate of a reaction?<\/li>\n
                                                                2. How does a decrease in temperature affect the reaction rate? Explain the outcome by describing changes that occur at the molecular level.<\/li>\n
                                                                3. For which of the following two reactions would you expect the orientation of the molecules to be more important?\n
                                                                    \n
                                                                  1. AB + C \u2192 AC + B<\/li>\n
                                                                  2. D + E \u2192 F<\/li>\n<\/ol>\n<\/li>\n
                                                                  3. Determine the relative rates of disappearance of reactants and formation of products for the following reactions:\n
                                                                      \n
                                                                    1. N2<\/sub>(g) + 3H2<\/sub>(g) \u2192 2NH3<\/sub>(g)<\/li>\n
                                                                    2. 2A + 3B \u2192 4C<\/li>\n
                                                                    3. 2N2<\/sub>O5<\/sub> \u2192 4 NO2<\/sub> + O2<\/sub><\/li>\n<\/ol>\n<\/li>\n
                                                                    4. Methanol reacts with hydrochloric acid to produce methyl chloride and water by the following reaction:\n

                                                                      CH3<\/sub>OH(aq) + HCl(aq) \u2192\u00a0CH3<\/sub>Cl(aq) + H2<\/sub>O(\u2113)
                                                                      \nThe following data were obtained for this reaction at a particular temperature:<\/p>\n\n\n\n\n\n\n\n\n\n
                                                                      Time (min)<\/th>\n[CH3OH] (M)<\/th>\n<\/tr>\n
                                                                      0.0<\/td>\n1.95<\/td>\n<\/tr>\n
                                                                      60.0<\/td>\n1.50<\/td>\n<\/tr>\n
                                                                      120.0<\/td>\n1.24<\/td>\n<\/tr>\n
                                                                      180.0<\/td>\n1.04<\/td>\n<\/tr>\n
                                                                      240.0<\/td>\n0.85<\/td>\n<\/tr>\n
                                                                      300.0<\/td>\n0.70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                                        \n
                                                                      1. Calculate the average reaction rate between 0 and 60 minutes and 180 and 240 minutes. Which is faster? Explain using your knowledge of factors that can affect rate.<\/li>\n
                                                                      2. Determine the instantaneous rate at 200 minutes.<\/li>\n<\/ol>\n<\/li>\n
                                                                      3. For the following reaction: 3 E + 2 F \u2192 2 G, the rate law is Rate = k<\/em>[F]2<\/sup>.\n
                                                                          \n
                                                                        1. How does the rate change if [E] is doubled?<\/li>\n
                                                                        2. How does the rate change is [F] is doubled?<\/li>\n
                                                                        3. What is the overall reaction order?<\/li>\n
                                                                        4. What are the units for the rate constant for this reaction?<\/li>\n<\/ol>\n<\/li>\n
                                                                        5. The rate of oxidation of bromide ions by bromate in an acidic aqueous solution is\n

                                                                          6 H+<\/sup>\u00a0+ BrO3<\/sub>\u2013<\/sup> + 5Br\u2013<\/sup> \u2192 3Br2<\/sub> + 3H2<\/sub>O<\/p>\n

                                                                          and is found to follow the rate law<\/p>\n

                                                                          Rate\u00a0=\u00a0k<\/em>[Br\u2013<\/sup>][BrO3<\/sub>\u2013<\/sup>][H+<\/sup>]2<\/sup><\/p>\n

                                                                          What happens to the rate if, in separate experiments:<\/p>\n

                                                                            \n
                                                                          1. [BrO3<\/sub>\u2013<\/sup>] is doubled?<\/li>\n
                                                                          2. the pH is increased by one unit?<\/li>\n
                                                                          3. the solution is diluted to twice its volume, with the pH kept constant by use of a buffer?[1]<\/sup><\/a><\/li>\n<\/ol>\n<\/li>\n
                                                                          4. The following data was obtained using the initial rate method:\n

                                                                            A + B \u2192 C + D<\/p>\n\n\n\n\n\n\n
                                                                            Experiment<\/th>\n[A] (M)<\/th>\n[B] (M)<\/th>\nInitial Rate (M\/s)<\/th>\n<\/tr>\n
                                                                            1<\/td>\n2.0 \u00d7 10\u22123<\/sup><\/td>\n2.0 \u00d7 10\u22123<\/sup><\/td>\n1.45 \u00d7 10\u22124<\/sup><\/td>\n<\/tr>\n
                                                                            2<\/td>\n4.0 \u00d7 10\u22123<\/sup><\/td>\n2.0 \u00d7 10\u22123<\/sup><\/td>\n2.90 \u00d7 10\u22124<\/sup><\/td>\n<\/tr>\n
                                                                            3<\/td>\n2.0 \u00d7 10\u22123<\/sup><\/td>\n4.0 \u00d7 10\u22123<\/sup><\/td>\n2.90 \u00d7 10\u22124<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                                              \n
                                                                            1. Determine the rate law for this equation.<\/li>\n
                                                                            2. Calculate the rate constant, including units.<\/li>\n
                                                                            3. Determine the rate when [A] = 0.250 M and [B] = 0.100 M.<\/li>\n<\/ol>\n<\/li>\n
                                                                            4. A study of the gas-phase reduction of nitric oxide by hydrogen\n

                                                                              2NO + 2H2<\/sub> \u2192 N2<\/sub> + 2H2<\/sub>O<\/p>\n

                                                                              yielded the following initial-rate data (all pressures in torr):<\/p>\n\n\n\n\n\n\n\n\n\n
                                                                              Experiment<\/th>\nP<\/em>(NO)<\/th>\nP<\/em>(H2<\/sub>)<\/th>\nInitial Rate (torr s\u22121<\/sup>)<\/th>\n<\/tr>\n
                                                                              1<\/td>\n359<\/td>\n300<\/td>\n1.50<\/td>\n<\/tr>\n
                                                                              2<\/td>\n300<\/td>\n300<\/td>\n1.03<\/td>\n<\/tr>\n
                                                                              3<\/td>\n152<\/td>\n300<\/td>\n0.25<\/td>\n<\/tr>\n
                                                                              4<\/td>\n300<\/td>\n289<\/td>\n1.00<\/td>\n<\/tr>\n
                                                                              5<\/td>\n300<\/td>\n205<\/td>\n0.71<\/td>\n<\/tr>\n
                                                                              6<\/td>\n200<\/td>\n147<\/td>\n0.51<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                                              Find the order of the reaction with respect to each component.[2]<\/sup><\/a><\/li>\n

                                                                            5. What should be plotted on the x<\/em>– and y<\/em>-axis to obtain a straight line for:\n
                                                                                \n
                                                                              1. a first-order reaction?<\/li>\n
                                                                              2. a zero-order reaction?<\/li>\n<\/ol>\n<\/li>\n
                                                                              3. What does the slope represent for:\n
                                                                                  \n
                                                                                1. a second-order reaction kinetic plot?<\/li>\n
                                                                                2. a first-order reaction kinetic plot?<\/li>\n<\/ol>\n<\/li>\n
                                                                                3. Common sugar, sucrose C12<\/sub>H22<\/sub>O11<\/sub>, reacts in dilute acidic solutions to produce glucose and fructose, both having the molecular formula C6<\/sub>H12<\/sub>O6<\/sub>.\n

                                                                                  C12<\/sub>H22<\/sub>O11<\/sub>(aq) + H2<\/sub>O(\u2113) \u2192 2C6<\/sub>H12<\/sub>O6<\/sub>(aq)<\/p>\n

                                                                                  During an experiment, the following data were obtained:<\/p>\n\n\n\n\n\n\n\n\n
                                                                                  Time (hours)<\/th>\n[Sucrose] (mM)<\/th>\n<\/tr>\n
                                                                                  0.00<\/td>\n316<\/td>\n<\/tr>\n
                                                                                  0.65<\/td>\n274<\/td>\n<\/tr>\n
                                                                                  1.33<\/td>\n238<\/td>\n<\/tr>\n
                                                                                  2.33<\/td>\n190<\/td>\n<\/tr>\n
                                                                                  3.50<\/td>\n146<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                                                    \n
                                                                                  1. What is the order of the reaction?<\/li>\n
                                                                                  2. Determine the rate law and the rate constant.<\/li>\n
                                                                                  3. Calculate the expected concentration of sucrose after 100 min.<\/li>\n<\/ol>\n<\/li>\n
                                                                                  4. The half-life of a first-order reaction was found to be 10 min at a certain temperature. What is its rate constant in reciprocal seconds?[3]<\/sup><\/a><\/li>\n
                                                                                  5. The mass-241 isotope of americium, widely used as an ionizing source in smoke detectors, has a half-life of 432 years.\n
                                                                                      \n
                                                                                    1. What fraction of the Am241<\/sup>\u00a0in a smoke detector will have decayed after 50 years?<\/li>\n
                                                                                    2. How long will it take for the activity to decline to 80% of its initial value?<\/li>\n
                                                                                    3. What would be the “seventh-life” of Am241<\/sup>?[4]<\/sup><\/a><\/li>\n<\/ol>\n<\/li>\n
                                                                                    4. A widely used \u201crule of thumb\u201d for the temperature dependence of a reaction rate is that a 10\u00b0C rise in the temperature approximately doubles the rate. (This is obviously not generally true, especially when a strong covalent bond must be broken.) For a reaction that shows this behaviour, what would the activation energy be?[5]<\/sup><\/a><\/li>\n
                                                                                    5. The temperature dependence of the rate constant, k<\/em>, for a reaction was found to be:
                                                                                      \n\n\n\n\n\n\n\n\n
                                                                                      Temperature (K)<\/th>\nk<\/em> (M\u22121<\/sup>s\u22121<\/sup>)<\/th>\n<\/tr>\n
                                                                                      500<\/td>\n0.025<\/td>\n<\/tr>\n
                                                                                      600<\/td>\n0.300<\/td>\n<\/tr>\n
                                                                                      700<\/td>\n1.500<\/td>\n<\/tr>\n
                                                                                      800<\/td>\n7.000<\/td>\n<\/tr>\n
                                                                                      900<\/td>\n28.000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                                                      Calculate the activation energy (Ea<\/sub><\/em>) and A<\/em>.<\/li>\n

                                                                                    6. State the molecularity and rate law for each of the following elementary steps:\n
                                                                                        \n
                                                                                      1. Cl(g) + CCl3<\/sub>(g) \u2192 CCl4<\/sub>(\u2113)<\/li>\n
                                                                                      2. Br2<\/sub> \u2192 2Br<\/li>\n
                                                                                      3. 2A \u2192 B<\/li>\n<\/ol>\n<\/li>\n
                                                                                      4. A reaction is believed to occur by the following two elementary steps:\n

                                                                                        2NO2<\/sub> \u2192 NO3<\/sub> + NO<\/p>\n

                                                                                        NO3<\/sub>\u00a0\u2192 NO + O2<\/sub><\/p>\n

                                                                                        With this in mind:<\/p>\n

                                                                                          \n
                                                                                        1. Write the balanced overall equation of this reaction.<\/li>\n
                                                                                        2. Identify any intermediates in the mechanism.<\/li>\n
                                                                                        3. If the first step is much slower than the second, what would you expect the rate law for the overall reaction to be?<\/li>\n<\/ol>\n<\/li>\n
                                                                                        4. Use the following potential energy diagram to answer questions 19a to 19c:
                                                                                          \n
                                                                                          \"\"
                                                                                          Potential energy plot of a given reaction mechanism.<\/figcaption><\/figure>\n
                                                                                            \n
                                                                                          1. How many steps are in this mechanism?<\/li>\n
                                                                                          2. How many intermediates are there in this reaction?<\/li>\n
                                                                                          3. What is the slowest step?<\/li>\n<\/ol>\n<\/li>\n
                                                                                          4. What is a catalyst? Explain how a catalyst affects reaction rates.<\/li>\n
                                                                                          5. Many solid catalysts are commercially sold as fine powders or small spherical beads. Explain why this might be beneficial.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
                                                                                            \n
                                                                                            \n

                                                                                            Answers<\/p>\n<\/header>\n

                                                                                            \n
                                                                                              \n
                                                                                            1. temperature, reactant concentrations, physical state (surface area of solids), presence of a catalyst<\/li>\n<\/ol>\n
                                                                                                \n
                                                                                              1. AB + C \u2192 AC + B<\/li>\n<\/ol>\n
                                                                                                  \n
                                                                                                1. \n
                                                                                                    \n
                                                                                                  1. average reaction rate 0\u201360 minutes:\n

                                                                                                    \"-\dfrac{\ce{[CH3OH]_60}-\ce{[CH3OH]_0}}{60.0\text{ min}-0.0\text{ min}}=\dfrac{1.50\text{ M}-19.5\text{ M}}{60.0\text{ min}}=7.50\times 10^{-3}\text{ M/min}\"<\/p>\n

                                                                                                    average reaction rate 180\u2013240 minutes:<\/p>\n

                                                                                                    \"-\dfrac{\ce{[CH3OH]_240}-\ce{[CH3OH]_180}}{240.0\text{ min}-180.0\text{ min}}=\dfrac{0.85\text{ M}-1.04\text{ M}}{60.0\text{ min}}=3.17\times 10^{-3}\text{ M/min}\"<\/p>\n

                                                                                                    Therefore the average reaction rate between 0\u201360 minutes is faster. This is because, as the reaction proceeds, the reactant gets consumed to make product, lowering the concentration of reactants.<\/li>\n

                                                                                                  2. \"\"
                                                                                                    \nInstantaneous rate at 200 minutes:<\/p>\n

                                                                                                    \"-\dfrac{\text{Rise}}{\text{Run}}=\dfrac{1.0\text{ M}-0.84\text{ M}}{230\text{ min}-180\text{ min}}=\dfrac{0.16\text{ M}}{50\text{ min}}=3.2\times 10^{-3}\text{ M/min}\"<\/p>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                                      \n
                                                                                                    1. \n
                                                                                                        \n
                                                                                                      1. Since the rate is first-order in bromate, doubling its concentration will double the reaction rate.<\/li>\n
                                                                                                      2. Increasing the pH by one unit will\u00a0decrease<\/em>\u00a0the [H+<\/sup>] by a factor of 10. Since the reaction is second-order in [H+<\/sup>], this will decrease the rate by a factor of 100.<\/li>\n
                                                                                                      3. Dilution reduces the concentrations of both Br2<\/sub>\u00a0and BrO3<\/sub>\u2013<\/sup> to half their original values. Doing this to each concentration alone would reduce the rate by a factor of 2, so reducing both concentrations will reduce the rate by a factor of 4, to (\u00bd) \u00d7 (\u00bd) = \u00bc of its initial value.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
                                                                                                          \n
                                                                                                        1. \n
                                                                                                            \n
                                                                                                          1. Experiments 2 and 3<\/strong>: Reduction of the initial partial pressure of NO by a factor of about 2 \"\left(\dfrac{300}{152}\right)\" results in a reduction of the initial rate by a factor of about 4, so the reaction is second-order in nitric oxide.<\/li>\n
                                                                                                          2. Experiments 4 and 6<\/strong>: Reducing the initial partial pressure of hydrogen by a factor of approximately 2 \"\left(\dfrac{289}{147}\right)\" causes a similar reduction in the initial rate, so the reaction is first-order in hydrogen.<\/li>\n
                                                                                                          3. The rate law is<\/strong>:\u00a0rate\u00a0=\u00a0k<\/em>[NO]2<\/sup>[H2<\/sub>].<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
                                                                                                              \n
                                                                                                            1. \n
                                                                                                                \n
                                                                                                              1. k<\/em><\/li>\n
                                                                                                              2. \u2212k<\/em><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
                                                                                                                  \n
                                                                                                                1. From the above equation, \"k=\dfrac{-0.693}{600\text{ s}}=0.00115\text{ s}^{-1}\"<\/li>\n<\/ol>\n
                                                                                                                    \n
                                                                                                                  1. We will centre our ten-degree interval at 300 K. Substituting into the above expression yields:\n

                                                                                                                    \n

                                                                                                                      <\/span>   <\/span>\"\begin{align*}<\/p>\n<\/li>\n<\/ol>\n

                                                                                                                      \n
                                                                                                                    1. \n
                                                                                                                        \n
                                                                                                                      1. bimolecular, rate = k<\/em>[Cl] [CCl3<\/sub>]<\/li>\n
                                                                                                                      2. unimolecular, rate = k<\/em>[Br2<\/sub>]<\/li>\n
                                                                                                                      3. bimolecular 2A, rate = k<\/em>[A]2<\/sup><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
                                                                                                                          \n
                                                                                                                        1. \n
                                                                                                                            \n
                                                                                                                          1. 3 steps<\/li>\n
                                                                                                                          2. 2 intermediates<\/li>\n
                                                                                                                          3. The second step is the slowest step<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
                                                                                                                              \n
                                                                                                                            1. Finely divided powders and spherical beads have larger surface areas than big solid chunks, allowing for more sites for reactions to occur, which further speeds up the reaction.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
                                                                                                                              1. Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0 ↵<\/a><\/li>
                                                                                                                              2. Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0 ↵<\/a><\/li>
                                                                                                                              3. Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0 ↵<\/a><\/li>
                                                                                                                              4. Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0 ↵<\/a><\/li>
                                                                                                                              5. Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0 ↵<\/a><\/li><\/ol><\/div>","protected":false},"author":90,"menu_order":8,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["jessie-a-key"],"pb_section_license":"cc-by"},"chapter-type":[],"contributor":[49],"license":[54],"class_list":["post-7965","chapter","type-chapter","status-publish","hentry","contributor-jessie-a-key","license-cc-by"],"part":7933,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7965","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":6,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7965\/revisions"}],"predecessor-version":[{"id":9024,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7965\/revisions\/9024"}],"part":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/7933"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7965\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/media?parent=7965"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=7965"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/contributor?post=7965"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/license?post=7965"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}