{"id":1069,"date":"2016-01-11T20:02:43","date_gmt":"2016-01-11T20:02:43","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/end-of-chapter-material-33\/"},"modified":"2020-04-20T16:45:16","modified_gmt":"2020-04-20T16:45:16","slug":"end-of-chapter-17-material","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/end-of-chapter-17-material\/","title":{"raw":"End-of-Chapter Material","rendered":"End-of-Chapter Material"},"content":{"raw":"
\r\n

Exercises<\/h3>\r\n1. What factors affect the rate of a reaction?\r\n\r\n2. How does a decrease in temperature affect the reaction rate?\u00a0Explain the outcome by describing changes that occur at the molecular level.\r\n\r\n3. For which of the following two reactions would you expect the orientation of the molecules to be more important?\r\n

a) \u00a0AB + C \u2192\u00a0AC + B<\/p>\r\n

b) \u00a0D + E \u2192\u00a0F<\/p>\r\n4. Determine the relative rates of disappearance of reactants and formation of products for the following reactions:\r\n

a)\u00a0\u00a0N2<\/sub>(g) + 3 H2<\/sub>(g) \u2192\u00a02 NH3<\/sub>(g)<\/p>\r\n

b)\u00a0\u00a02 A + 3 B \u2192\u00a04 C<\/p>\r\n

c)\u00a0\u00a02 N2<\/sub>O5<\/sub>\u00a0\u2192 4\u00a0 NO2<\/sub> + O2<\/sub><\/p>\r\n5. Methanol reacts with hydrochloric acid to produce methyl chloride and water by the following reaction:\r\n\r\nCH3<\/sub>OH(aq) + HCl(aq) \u2192\u00a0CH3<\/sub>Cl(aq) + H2<\/sub>O(\u2113)\r\n\r\nThe following data were\u00a0obtained for this reaction at a particular temperature:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Time (min)<\/strong><\/td>\r\n[CH3<\/sub>OH] (M)<\/strong><\/td>\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

a)\u00a0\u00a0Calculate 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.<\/p>\r\n

b)\u00a0\u00a0Determine the instantaneous rate at 200 minutes.<\/p>\r\n6. For the following reaction: 3 E + 2 F \u2192\u00a02 G, the rate law is:\r\n\r\nRate = k<\/i>[F]2<\/sup>\r\n

a)\u00a0\u00a0How does the rate change if [E] is doubled?<\/p>\r\n

b)\u00a0\u00a0How does the rate change is [F] is doubled?<\/p>\r\n

c)\u00a0\u00a0What is the overall reaction order?<\/p>\r\n

d)\u00a0\u00a0What are the units for the rate constant for this reaction?<\/p>\r\n7. 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>\u00a0+ 5 Br\u2013<\/sup>\u00a0\u2192 3 Br2<\/sub>\u00a0+ 3 H2<\/sub>O<\/p>\r\nand is found to follow the rate law\r\n

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

a) \u00a0[BrO3<\/sub>\u2013<\/sup>] is doubled?<\/p>\r\n

b) \u00a0\u00a0the pH is increased by one unit?<\/p>\r\n

c) \u00a0the solution is diluted to twice its volume, with the pH kept constant by use of a buffer?\u00a0[footnote]Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0[\/footnote]<\/p>\r\n\u00a0<\/span>8. The following data was obtained using the initial rate method:\r\n\r\nA + B \u2192\u00a0C + D\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Experiment<\/strong><\/td>\r\n[A] (M)<\/strong><\/td>\r\n[B] (M)<\/strong><\/td>\r\nInitial Rate (M\/s)<\/strong><\/td>\r\n<\/tr>\r\n
\r\n

1<\/p>\r\n<\/td>\r\n

2.0 x 10-3<\/sup><\/td>\r\n2.0 x 10-3<\/sup><\/td>\r\n1.45 x 10-4<\/sup><\/td>\r\n<\/tr>\r\n
\r\n

2<\/p>\r\n<\/td>\r\n

4.0 x 10-3<\/sup><\/td>\r\n2.0 x 10-3<\/sup><\/td>\r\n2.90 x 10-4<\/sup><\/td>\r\n<\/tr>\r\n
\r\n

3<\/p>\r\n<\/td>\r\n

2.0 x 10-3<\/sup><\/td>\r\n4.0 x 10-3<\/sup><\/td>\r\n2.90 x 10-4<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

a) \u00a0Determine the rate law for this equation.<\/p>\r\n

b)\u00a0\u00a0Calculate the rate constant, including units.<\/p>\r\n

c)\u00a0\u00a0Determine the rate when [A] = 0.250 M and [B] = 0.100 M<\/p>\r\n9. A study of the gas-phase reduction of nitric oxide by hydrogen\r\n

2 NO + 2 H2<\/sub>\u00a0\u2192 N2<\/sub>\u00a0+ 2 H2<\/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
\r\n

Experiment<\/b><\/p>\r\n<\/td>\r\n

\r\n

P<\/em>(NO)<\/b><\/p>\r\n<\/td>\r\n

\r\n

P<\/em>(H2<\/sub>)<\/b><\/p>\r\n<\/td>\r\n

\r\n

Initial Rate\r\n(torr s\u20131<\/sup>)<\/b><\/p>\r\n<\/td>\r\n<\/tr>\r\n

\r\n

1<\/p>\r\n<\/td>\r\n

359<\/td>\r\n300<\/td>\r\n1.50<\/td>\r\n<\/tr>\r\n
\r\n

2<\/p>\r\n<\/td>\r\n

300<\/td>\r\n300<\/td>\r\n1.03<\/td>\r\n<\/tr>\r\n
\r\n

3<\/p>\r\n<\/td>\r\n

152<\/td>\r\n300<\/td>\r\n0.25<\/td>\r\n<\/tr>\r\n
\r\n

4<\/p>\r\n<\/td>\r\n

300<\/td>\r\n289<\/td>\r\n1.00<\/td>\r\n<\/tr>\r\n
\r\n

5<\/p>\r\n<\/td>\r\n

300<\/td>\r\n205<\/td>\r\n0.71<\/td>\r\n<\/tr>\r\n
\r\n

6<\/p>\r\n<\/td>\r\n

200<\/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]\r\n\r\n\u00a0<\/span>10. What should be plotted on the x-<\/em> and y<\/em>-axis to obtain a straight line for:\r\n

a)\u00a0\u00a0a\u00a0first-order reaction?<\/p>\r\n

b)\u00a0\u00a0a\u00a0zero-order reaction?<\/p>\r\n11. What does the slope represent for:\r\n

a)\u00a0\u00a0a\u00a0second-order reaction kinetic plot?<\/p>\r\n

b)\u00a0\u00a0a\u00a0first-order reaction kinetic plot?<\/p>\r\n12. Common sugar, sucrose C<\/span>12<\/sub>H<\/span>22<\/sub>O<\/span>11<\/sub>, reacts in dilute acidic solutions to produce glucose and fructose, both having the molecular formula C<\/span>6<\/sub>H<\/span>12<\/sub>0<\/span>6<\/sub>.<\/span>\r\n\r\nC12<\/sub>H22<\/sub>O11<\/sub>(aq) + H2<\/sub>O(\u2113) \u2192 2 C6<\/sub>H12<\/sub>06<\/sub>(aq)\r\n\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
\r\n

Time (hours)<\/strong><\/p>\r\n<\/td>\r\n

\r\n

[Sucrose] (mM)<\/strong><\/p>\r\n<\/td>\r\n<\/tr>\r\n

\r\n

0<\/p>\r\n<\/td>\r\n

\r\n

316<\/p>\r\n<\/td>\r\n<\/tr>\r\n

\r\n

0.65<\/p>\r\n<\/td>\r\n

\r\n

274<\/p>\r\n<\/td>\r\n<\/tr>\r\n

\r\n

1.33<\/p>\r\n<\/td>\r\n

\r\n

238<\/p>\r\n<\/td>\r\n<\/tr>\r\n

\r\n

2.33<\/p>\r\n<\/td>\r\n

\r\n

190<\/p>\r\n<\/td>\r\n<\/tr>\r\n

\r\n

3.5<\/p>\r\n<\/td>\r\n

\r\n

146<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

a)\u00a0\u00a0What is the order of the reaction?<\/p>\r\n

b)\u00a0\u00a0Determine the rate law and the rate constant.<\/p>\r\n

c)\u00a0\u00a0Calculate the expected concentration of sucrose after 100 min.<\/p>\r\n13. 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]\r\n\r\n14. The mass-241 isotope of americium, widely used as an ionizing source in smoke detectors, has a half-life of 432 years.<\/span>\r\n

a)\u00a0\u00a0What fraction of the Am241<\/sup>\u00a0in a smoke detector will have decayed after 50 years?<\/p>\r\n

b)\u00a0\u00a0How long will it take for the activity to decline to 80% of its initial value?<\/p>\r\n

c)\u00a0\u00a0What would be the \"seventh-life\" of Am241<\/sup>?[footnote]Question and solution from Chem1 Virtual Textbook, Stephen Lower\/CC-BY-SA-3.0[\/footnote]<\/p>\r\n\u00a0<\/span>15. 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]\r\n\r\n16. The temperature dependence of the rate constant, k<\/i>, 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)<\/strong><\/td>\r\nk<\/i> (M-1<\/sup> s-1<\/sup>)<\/strong><\/td>\r\n<\/tr>\r\n
500<\/td>\r\n0.025<\/td>\r\n<\/tr>\r\n
600<\/td>\r\n0.3<\/td>\r\n<\/tr>\r\n
700<\/td>\r\n1.5<\/td>\r\n<\/tr>\r\n
800<\/td>\r\n7<\/td>\r\n<\/tr>\r\n
900<\/td>\r\n28<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nCalculate the activation energy (E<\/em>a<\/sub>) and A<\/em>.\r\n\r\n\u00a0<\/i>\u00a017. State the molecularity and rate law for each of the following elementary steps:\r\n

a)\u00a0\u00a0Cl(g) + CCl3<\/sub>(g) \u2192 CCl4<\/sub>(\u2113)<\/p>\r\n

b) \u00a0Br2<\/sub> \u2192 2 Br<\/p>\r\n

c)\u00a0\u00a02 A \u2192 B<\/p>\r\n18. A reaction is believed to occur by the following two elementary steps:\r\n

a)\u00a0\u00a02 NO2<\/sub>\u00a0\u2192 NO3<\/sub>\u00a0+ NO<\/p>\r\n

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

c)\u00a0\u00a0Write the balanced overall equation of this reaction<\/p>\r\n

d)\u00a0\u00a0Identify any intermediates in the mechanism.<\/p>\r\n

e)\u00a0\u00a0If the first step is much slower than the second, what would you expect the rate law for the overall reaction to be?<\/p>\r\n19. Use the following potential energy diagram to answer questions 19.1 to 19.3:\r\n\r\n[caption id=\"attachment_2139\" align=\"alignnone\" width=\"342\"]\"Potential<\/a> Potential energy plot of a given reaction mechanism[\/caption]\r\n

a)\u00a0\u00a0How many steps are in\u00a0this mechanism?<\/p>\r\n

b)\u00a0\u00a0How many intermediates are there in this reaction?<\/p>\r\n

c)\u00a0\u00a0What is the slowest step?<\/p>\r\n20. What is a catalyst? Explain how a catalyst affects reaction rates.\r\n\r\n21. Many solid catalysts are commercially sold as fine powders or small spherical beads. Explain why this might be beneficial.<\/span>\r\n\r\n<\/div>\r\n

\r\n

Answers<\/h3>\r\n1.<\/strong>\r\n\r\ntemperature, reactant concentrations, physical state (surface area of solids), presence of a catalyst\r\n\r\n3.<\/strong>\r\n

\u00a0AB + C \u2192\u00a0AC + B<\/p>\r\n5.<\/strong><\/span>\r\n

a)\u00a0\u00a0average reaction rate 0-60 minutes =\u00a0- $$\\frac{{{\\rm [}{{\\rm CH}}_{{\\rm 3}}{\\rm OH]}}_{{\\rm 60}}{\\rm \\ }--{\\rm \\ }{{\\rm [}{{\\rm CH}}_{{\\rm 3}}{\\rm OH]}}_0}{60.0{\\min -\\ 0.0\\ min\\ }}\\$$ = $$\\frac{1.50\\ M-1.95\\ M}{60.0\\ min}\\$$ = 7.50 x 10$${}^{-3} M\/min\\$$<\/span><\/p>\r\n

average reaction rate 180-240 minutes = -\u00a0$$\\frac{{{\\rm [}{{\\rm CH}}_{{\\rm 3}}{\\rm OH]}}_{{\\rm 240}}{\\rm \\ }--{\\rm \\ }{{\\rm [}{{\\rm CH}}_{{\\rm 3}}{\\rm OH]}}_{{\\rm 180}}}{240.0{\\min -\\ 180.0\\ min\\ }}\\$$ = $$\\frac{0.85\\ M-1.04\\ M}{60.0\\ min}\\$$ = 3.17 x 10$${}^{-3} M\/min\\$$<\/p>\r\n

Therefore the average reaction rate between 0-60 minutes is faster. This is because, as the reaction proceeds, the reactant gets consumed to make product, lowering the concentration of reactants.<\/p>\r\n

b)<\/p>\r\n\r\n\r\n[caption id=\"attachment_2138\" align=\"alignnone\" width=\"481\"]\"instantaneous<\/a> Solution to instantaneous rate question[\/caption]\r\n

Instantaneous rate at 200 minutes.<\/p>\r\n

-$$\\frac{Rise}{Run}\\$$ = $$\\frac{1.0\\ M\\ -\\ 0.84\\ M}{230{\\min -\\ 180\\ min\\ }}$$ = $$\\frac{{\\rm 0.16\\ M}}{50\\ min}\\$$ = 3.2 x 10-3 <\/sup>M\/min<\/p>\r\n7.<\/strong>\r\n

a)\u00a0\u00a0Since the rate is first-order in bromate, doubling its concentration will double the reaction rate.<\/p>\r\n

b)\u00a0\u00a0Increasing the pH by one unit will\u00a0decrease<\/i>\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.<\/p>\r\n

c) \u00a0Dilution reduces the concentrations of both Br2<\/sub>\u00a0and BrO3<\/sub>\u2013<\/sup>\u00a0to 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.<\/p>\r\n\u00a0<\/strong>9.<\/strong>\r\n\r\nExperiments 2 and 3: Reduction of the initial partial pressure of NO by a factor of about 2 (300\/152) results in a reduction of the initial rate by a factor of about 4, so the reaction is second-order in nitric oxide.\r\n\r\nExperiments 4 and 6: Reducing the initial partial pressure of hydrogen by a factor of approximately 2 (289\/147) causes a similar reduction in the initial rate, so the reaction is first-order in hydrogen.\r\n\r\nThe rate law is:\u00a0rate\u00a0=\u00a0k<\/i>[NO]2<\/sup>[H2<\/sub>].\r\n\r\n11.<\/strong>\r\n

a)\u00a0k<\/i><\/p>\r\n

b)\u00a0- k<\/i><\/p>\r\n13.<\/strong>\r\n\r\nFrom the above equation,\u00a0k<\/i>\u00a0= \u20130.693\/(600 s) = 0.00115 s\u20131<\/sup>\u00a0<\/sup>15.<\/strong>\r\n\r\n <\/i><\/b>We will center our ten-degree interval at 300\u00a0K. Substituting into the above expression yields\r\n\r\n\"end<\/a>\r\n\r\n= (8.314)(0.693) \/ (.00339 - 0.00328)\r\n\r\n= (5.76 J mol\u20131<\/sup>\u00a0K\u20131<\/sup>) \/ (0.00011 K\u20131<\/sup>) = 52400 J mol\u20131<\/sup>\u00a0=\u00a052.4 kJ mol\u20131<\/sup>17.<\/strong>\r\n

a)\u00a0\u00a0bimolecular, rate = k <\/i>[Cl] [CCl3<\/sub>]<\/p>\r\n

b)\u00a0\u00a0unimolecular , rate = k<\/i> [Br2<\/sub>]<\/p>\r\n

c)\u00a0\u00a0bimolecular 2 A , rate = k <\/i>[A]2<\/sup><\/p>\r\n19.<\/strong>\r\n

a)\u00a0\u00a03 steps<\/p>\r\n

b)\u00a0\u00a02 intermediates<\/p>\r\n

c) \u00a0The second step is the slowest step<\/p>\r\n21.<\/strong>\r\n\r\nFinely divided powders and spherical beads have larger surface areas than big solid chunks, allowing for more sites for reactions to occur, which further\u00a0speeds up the reaction.\r\n\r\n<\/div>","rendered":"

\n

Exercises<\/h3>\n

1. What factors affect the rate of a reaction?<\/p>\n

2. How does a decrease in temperature affect the reaction rate?\u00a0Explain the outcome by describing changes that occur at the molecular level.<\/p>\n

3. For which of the following two reactions would you expect the orientation of the molecules to be more important?<\/p>\n

a) \u00a0AB + C \u2192\u00a0AC + B<\/p>\n

b) \u00a0D + E \u2192\u00a0F<\/p>\n

4. Determine the relative rates of disappearance of reactants and formation of products for the following reactions:<\/p>\n

a)\u00a0\u00a0N2<\/sub>(g) + 3 H2<\/sub>(g) \u2192\u00a02 NH3<\/sub>(g)<\/p>\n

b)\u00a0\u00a02 A + 3 B \u2192\u00a04 C<\/p>\n

c)\u00a0\u00a02 N2<\/sub>O5<\/sub>\u00a0\u2192 4\u00a0 NO2<\/sub> + O2<\/sub><\/p>\n

5. Methanol reacts with hydrochloric acid to produce methyl chloride and water by the following reaction:<\/p>\n

CH3<\/sub>OH(aq) + HCl(aq) \u2192\u00a0CH3<\/sub>Cl(aq) + H2<\/sub>O(\u2113)<\/p>\n

The following data were\u00a0obtained for this reaction at a particular temperature:<\/p>\n\n\n\n\n\n\n\n\n\n
Time (min)<\/strong><\/td>\n[CH3<\/sub>OH] (M)<\/strong><\/td>\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

a)\u00a0\u00a0Calculate 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.<\/p>\n

b)\u00a0\u00a0Determine the instantaneous rate at 200 minutes.<\/p>\n

6. For the following reaction: 3 E + 2 F \u2192\u00a02 G, the rate law is:<\/p>\n

Rate = k<\/i>[F]2<\/sup><\/p>\n

a)\u00a0\u00a0How does the rate change if [E] is doubled?<\/p>\n

b)\u00a0\u00a0How does the rate change is [F] is doubled?<\/p>\n

c)\u00a0\u00a0What is the overall reaction order?<\/p>\n

d)\u00a0\u00a0What are the units for the rate constant for this reaction?<\/p>\n

7. The rate of oxidation of bromide ions by bromate in an acidic aqueous solution is<\/p>\n

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

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

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

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

a) \u00a0[BrO3<\/sub>\u2013<\/sup>] is doubled?<\/p>\n

b) \u00a0\u00a0the pH is increased by one unit?<\/p>\n

c) \u00a0the solution is diluted to twice its volume, with the pH kept constant by use of a buffer?\u00a0[1]<\/sup><\/a><\/p>\n

\u00a0<\/span>8. The following data was obtained using the initial rate method:<\/p>\n

A + B \u2192\u00a0C + D<\/p>\n\n\n\n\n\n\n
Experiment<\/strong><\/td>\n[A] (M)<\/strong><\/td>\n[B] (M)<\/strong><\/td>\nInitial Rate (M\/s)<\/strong><\/td>\n<\/tr>\n
\n

1<\/p>\n<\/td>\n

2.0 x 10-3<\/sup><\/td>\n2.0 x 10-3<\/sup><\/td>\n1.45 x 10-4<\/sup><\/td>\n<\/tr>\n
\n

2<\/p>\n<\/td>\n

4.0 x 10-3<\/sup><\/td>\n2.0 x 10-3<\/sup><\/td>\n2.90 x 10-4<\/sup><\/td>\n<\/tr>\n
\n

3<\/p>\n<\/td>\n

2.0 x 10-3<\/sup><\/td>\n4.0 x 10-3<\/sup><\/td>\n2.90 x 10-4<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

a) \u00a0Determine the rate law for this equation.<\/p>\n

b)\u00a0\u00a0Calculate the rate constant, including units.<\/p>\n

c)\u00a0\u00a0Determine the rate when [A] = 0.250 M and [B] = 0.100 M<\/p>\n

9. A study of the gas-phase reduction of nitric oxide by hydrogen<\/p>\n

2 NO + 2 H2<\/sub>\u00a0\u2192 N2<\/sub>\u00a0+ 2 H2<\/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
\n

Experiment<\/b><\/p>\n<\/td>\n

\n

P<\/em>(NO)<\/b><\/p>\n<\/td>\n

\n

P<\/em>(H2<\/sub>)<\/b><\/p>\n<\/td>\n

\n

Initial Rate
\n(torr s\u20131<\/sup>)<\/b><\/p>\n<\/td>\n<\/tr>\n

\n

1<\/p>\n<\/td>\n

359<\/td>\n300<\/td>\n1.50<\/td>\n<\/tr>\n
\n

2<\/p>\n<\/td>\n

300<\/td>\n300<\/td>\n1.03<\/td>\n<\/tr>\n
\n

3<\/p>\n<\/td>\n

152<\/td>\n300<\/td>\n0.25<\/td>\n<\/tr>\n
\n

4<\/p>\n<\/td>\n

300<\/td>\n289<\/td>\n1.00<\/td>\n<\/tr>\n
\n

5<\/p>\n<\/td>\n

300<\/td>\n205<\/td>\n0.71<\/td>\n<\/tr>\n
\n

6<\/p>\n<\/td>\n

200<\/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><\/p>\n

\u00a0<\/span>10. What should be plotted on the x-<\/em> and y<\/em>-axis to obtain a straight line for:<\/p>\n

a)\u00a0\u00a0a\u00a0first-order reaction?<\/p>\n

b)\u00a0\u00a0a\u00a0zero-order reaction?<\/p>\n

11. What does the slope represent for:<\/p>\n

a)\u00a0\u00a0a\u00a0second-order reaction kinetic plot?<\/p>\n

b)\u00a0\u00a0a\u00a0first-order reaction kinetic plot?<\/p>\n

12. Common sugar, sucrose C<\/span>12<\/sub>H<\/span>22<\/sub>O<\/span>11<\/sub>, reacts in dilute acidic solutions to produce glucose and fructose, both having the molecular formula C<\/span>6<\/sub>H<\/span>12<\/sub>0<\/span>6<\/sub>.<\/span><\/p>\n

C12<\/sub>H22<\/sub>O11<\/sub>(aq) + H2<\/sub>O(\u2113) \u2192 2 C6<\/sub>H12<\/sub>06<\/sub>(aq)<\/p>\n

During an experiment, the following data were obtained:<\/p>\n\n\n\n\n\n\n\n\n
\n

Time (hours)<\/strong><\/p>\n<\/td>\n

\n

[Sucrose] (mM)<\/strong><\/p>\n<\/td>\n<\/tr>\n

\n

0<\/p>\n<\/td>\n

\n

316<\/p>\n<\/td>\n<\/tr>\n

\n

0.65<\/p>\n<\/td>\n

\n

274<\/p>\n<\/td>\n<\/tr>\n

\n

1.33<\/p>\n<\/td>\n

\n

238<\/p>\n<\/td>\n<\/tr>\n

\n

2.33<\/p>\n<\/td>\n

\n

190<\/p>\n<\/td>\n<\/tr>\n

\n

3.5<\/p>\n<\/td>\n

\n

146<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

a)\u00a0\u00a0What is the order of the reaction?<\/p>\n

b)\u00a0\u00a0Determine the rate law and the rate constant.<\/p>\n

c)\u00a0\u00a0Calculate the expected concentration of sucrose after 100 min.<\/p>\n

13. 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><\/p>\n

14. The mass-241 isotope of americium, widely used as an ionizing source in smoke detectors, has a half-life of 432 years.<\/span><\/p>\n

a)\u00a0\u00a0What fraction of the Am241<\/sup>\u00a0in a smoke detector will have decayed after 50 years?<\/p>\n

b)\u00a0\u00a0How long will it take for the activity to decline to 80% of its initial value?<\/p>\n

c)\u00a0\u00a0What would be the “seventh-life” of Am241<\/sup>?[4]<\/sup><\/a><\/p>\n

\u00a0<\/span>15. 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><\/p>\n

16. The temperature dependence of the rate constant, k<\/i>, for a reaction was found to be:<\/p>\n\n\n\n\n\n\n\n\n
Temperature (K)<\/strong><\/td>\nk<\/i> (M-1<\/sup> s-1<\/sup>)<\/strong><\/td>\n<\/tr>\n
500<\/td>\n0.025<\/td>\n<\/tr>\n
600<\/td>\n0.3<\/td>\n<\/tr>\n
700<\/td>\n1.5<\/td>\n<\/tr>\n
800<\/td>\n7<\/td>\n<\/tr>\n
900<\/td>\n28<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Calculate the activation energy (E<\/em>a<\/sub>) and A<\/em>.<\/p>\n

\u00a0<\/i>\u00a017. State the molecularity and rate law for each of the following elementary steps:<\/p>\n

a)\u00a0\u00a0Cl(g) + CCl3<\/sub>(g) \u2192 CCl4<\/sub>(\u2113)<\/p>\n

b) \u00a0Br2<\/sub> \u2192 2 Br<\/p>\n

c)\u00a0\u00a02 A \u2192 B<\/p>\n

18. A reaction is believed to occur by the following two elementary steps:<\/p>\n

a)\u00a0\u00a02 NO2<\/sub>\u00a0\u2192 NO3<\/sub>\u00a0+ NO<\/p>\n

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

c)\u00a0\u00a0Write the balanced overall equation of this reaction<\/p>\n

d)\u00a0\u00a0Identify any intermediates in the mechanism.<\/p>\n

e)\u00a0\u00a0If the first step is much slower than the second, what would you expect the rate law for the overall reaction to be?<\/p>\n

19. Use the following potential energy diagram to answer questions 19.1 to 19.3:<\/p>\n

\"Potential<\/a>
Potential energy plot of a given reaction mechanism<\/figcaption><\/figure>\n

a)\u00a0\u00a0How many steps are in\u00a0this mechanism?<\/p>\n

b)\u00a0\u00a0How many intermediates are there in this reaction?<\/p>\n

c)\u00a0\u00a0What is the slowest step?<\/p>\n

20. What is a catalyst? Explain how a catalyst affects reaction rates.<\/p>\n

21. Many solid catalysts are commercially sold as fine powders or small spherical beads. Explain why this might be beneficial.<\/span><\/p>\n<\/div>\n

\n

Answers<\/h3>\n

1.<\/strong><\/p>\n

temperature, reactant concentrations, physical state (surface area of solids), presence of a catalyst<\/p>\n

3.<\/strong><\/p>\n

\u00a0AB + C \u2192\u00a0AC + B<\/p>\n

5.<\/strong><\/span><\/p>\n

a)\u00a0\u00a0average reaction rate 0-60 minutes =\u00a0– $$\\frac{{{\\rm [}{{\\rm CH}}_{{\\rm 3}}{\\rm OH]}}_{{\\rm 60}}{\\rm \\ }–{\\rm \\ }{{\\rm [}{{\\rm CH}}_{{\\rm 3}}{\\rm OH]}}_0}{60.0{\\min -\\ 0.0\\ min\\ }}\\$$ = $$\\frac{1.50\\ M-1.95\\ M}{60.0\\ min}\\$$ = 7.50 x 10$${}^{-3} M\/min\\$$<\/span><\/p>\n

average reaction rate 180-240 minutes = –\u00a0$$\\frac{{{\\rm [}{{\\rm CH}}_{{\\rm 3}}{\\rm OH]}}_{{\\rm 240}}{\\rm \\ }–{\\rm \\ }{{\\rm [}{{\\rm CH}}_{{\\rm 3}}{\\rm OH]}}_{{\\rm 180}}}{240.0{\\min -\\ 180.0\\ min\\ }}\\$$ = $$\\frac{0.85\\ M-1.04\\ M}{60.0\\ min}\\$$ = 3.17 x 10$${}^{-3} M\/min\\$$<\/p>\n

Therefore the average reaction rate between 0-60 minutes is faster. This is because, as the reaction proceeds, the reactant gets consumed to make product, lowering the concentration of reactants.<\/p>\n

b)<\/p>\n

\"instantaneous<\/a>
Solution to instantaneous rate question<\/figcaption><\/figure>\n

Instantaneous rate at 200 minutes.<\/p>\n

-$$\\frac{Rise}{Run}\\$$ = $$\\frac{1.0\\ M\\ -\\ 0.84\\ M}{230{\\min -\\ 180\\ min\\ }}$$ = $$\\frac{{\\rm 0.16\\ M}}{50\\ min}\\$$ = 3.2 x 10-3 <\/sup>M\/min<\/p>\n

7.<\/strong><\/p>\n

a)\u00a0\u00a0Since the rate is first-order in bromate, doubling its concentration will double the reaction rate.<\/p>\n

b)\u00a0\u00a0Increasing the pH by one unit will\u00a0decrease<\/i>\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.<\/p>\n

c) \u00a0Dilution reduces the concentrations of both Br2<\/sub>\u00a0and BrO3<\/sub>\u2013<\/sup>\u00a0to 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.<\/p>\n

\u00a0<\/strong>9.<\/strong><\/p>\n

Experiments 2 and 3: Reduction of the initial partial pressure of NO by a factor of about 2 (300\/152) results in a reduction of the initial rate by a factor of about 4, so the reaction is second-order in nitric oxide.<\/p>\n

Experiments 4 and 6: Reducing the initial partial pressure of hydrogen by a factor of approximately 2 (289\/147) causes a similar reduction in the initial rate, so the reaction is first-order in hydrogen.<\/p>\n

The rate law is:\u00a0rate\u00a0=\u00a0k<\/i>[NO]2<\/sup>[H2<\/sub>].<\/p>\n

11.<\/strong><\/p>\n

a)\u00a0k<\/i><\/p>\n

b)\u00a0– k<\/i><\/p>\n

13.<\/strong><\/p>\n

From the above equation,\u00a0k<\/i>\u00a0= \u20130.693\/(600 s) = 0.00115 s\u20131<\/sup>\u00a0<\/sup>15.<\/strong><\/p>\n

<\/i><\/b>We will center our ten-degree interval at 300\u00a0K. Substituting into the above expression yields<\/p>\n

\"end<\/a><\/p>\n

= (8.314)(0.693) \/ (.00339 – 0.00328)<\/p>\n

= (5.76 J mol\u20131<\/sup>\u00a0K\u20131<\/sup>) \/ (0.00011 K\u20131<\/sup>) = 52400 J mol\u20131<\/sup>\u00a0=\u00a052.4 kJ mol\u20131<\/sup>17.<\/strong><\/p>\n

a)\u00a0\u00a0bimolecular, rate = k <\/i>[Cl] [CCl3<\/sub>]<\/p>\n

b)\u00a0\u00a0unimolecular , rate = k<\/i> [Br2<\/sub>]<\/p>\n

c)\u00a0\u00a0bimolecular 2 A , rate = k <\/i>[A]2<\/sup><\/p>\n

19.<\/strong><\/p>\n

a)\u00a0\u00a03 steps<\/p>\n

b)\u00a0\u00a02 intermediates<\/p>\n

c) \u00a0The second step is the slowest step<\/p>\n

21.<\/strong><\/p>\n

Finely divided powders and spherical beads have larger surface areas than big solid chunks, allowing for more sites for reactions to occur, which further\u00a0speeds up the reaction.<\/p>\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":124,"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":[61],"license":[52],"class_list":["post-1069","chapter","type-chapter","status-publish","hentry","contributor-jessie-a-key","license-cc-by"],"part":1028,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/1069","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/users\/124"}],"version-history":[{"count":2,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/1069\/revisions"}],"predecessor-version":[{"id":1454,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/1069\/revisions\/1454"}],"part":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/parts\/1028"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/1069\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/media?parent=1069"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapter-type?post=1069"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/contributor?post=1069"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/license?post=1069"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}