<body> <p align="center"><a name="top"><font size="6"><b>Chemical Kinetics</b></font></a></p> <div align="center"><center><table border="1" width="500"> <tr> <td align="center"><a href="#kinetics">Chemical Kinetics</a></td> <td align="center"><a href="#instant">Instantaneous Rates of Reaction and the Rate Law for a Reaction</a></td> <td align="center"><a href="#constant">Rate Laws and Rate Constants</a></td> <td align="center"><a href="#ways">Different Ways of Expressing the Rate of Reaction</a></td> </tr> <tr> <td align="center"><a href="#stoich">The Rate Law Versus the Stoichiometry of a Reaction</a></td> <td align="center"><a href="#order">Order and Molecularity</a></td> <td align="center" colspan="2"><a href="#model">A Collision Theory Model of Chemical Reactions</a></td> </tr> </table> </center></div><p align="center"><a name="kinetics"><em><b>Chemical Kinetics</b></em><b> </b></a></p> <p align="left">You may be familiar with acid-base titrations that use phenolphthalein as the endpoint indicator. You might not have noticed, however, what happens when a solution that contains phenolphthalein in the presence of excess base is allowed to stand for a few minutes. Although the solution initially has a pink color, it gradually turns colorless as the phenolphthalein reacts with the OH<sup>-</sup> ion in a strongly basic solution.</p> <p align="left">The table below shows what happens to the concentration of phenolphthalein in a solution that was initially 0.005 <i>M</i> in phenolphthalein and 0.61 <i>M</i> in OH<sup>-</sup> ion. As you can see when these data are plotted in the graph below, the phenolphthalein concentration decreases by a factor of 10 over a period of about four minutes.</p> <blockquote> <blockquote> <blockquote> <blockquote> <p align="center"><img src="graphics/22_2fig.gif" alt="graph" width="336" height="229"></p> </blockquote> </blockquote> </blockquote> </blockquote> <p align="center"><i>Experimental Data for the Reaction Between Phenolphthalein and Excess Base</i></p> <div align="center"><center><table border="0"> <tr> <td>Concentration of<br> Phenolphthalein (M)</td> <td> </td> <td><i>Time (s)</i></td> </tr> <tr> <td>0.0050</td> <td> </td> <td>0.0</td> </tr> <tr> <td>0.0045 </td> <td> </td> <td>10.5</td> </tr> <tr> <td>0.0040</td> <td> </td> <td>22.3</td> </tr> <tr> <td>0.0035</td> <td> </td> <td>35.7</td> </tr> <tr> <td>0.0030</td> <td> </td> <td>51.1</td> </tr> <tr> <td>0.0025</td> <td> </td> <td>69.3</td> </tr> <tr> <td>0.0020 </td> <td> </td> <td>91.6</td> </tr> <tr> <td>0.0015</td> <td> </td> <td>120.4</td> </tr> <tr> <td>0.0010</td> <td> </td> <td>160.9</td> </tr> <tr> <td>0.00050 </td> <td> </td> <td>230.3</td> </tr> <tr> <td>0.00025 </td> <td> </td> <td>299.6</td> </tr> <tr> <td>0.00015 </td> <td> </td> <td>350.7</td> </tr> <tr> <td>0.00010 </td> <td> </td> <td>391.2</td> </tr> </table> </center></div><p align="left">Experiments such as the one that gave us the data in the above table are classified as measurements of <b>chemical kinetics</b> (from a Greek stem meaning "to move"). One of the goals of these experiments is to describe the <b>rate of reaction<img src="graphics/em.gif" alt="--" width="22" height="9"></b>the rate at which the reactants are transformed into the products of the reaction.</p> <p align="left">The term <i>rate</i> is often used to describe the change in a quantity that occurs per unit of time. The rate of inflation, for example, is the change in the average cost of a collection of standard items per year. The rate at which an object travels through space is the distance traveled per unit of time, such as miles per hour or kilometers per second. In chemical kinetics, the distance traveled is the change in the concentration of one of the components of the reaction. The rate of a reaction is therefore the change in the concentration of one of the reactants<b><img src="graphics/em.gif" alt="--" width="22" height="9"><img src="graphics/delta.gif" alt="delta" width="11" height="10"></b>(<i>X</i>)<b><img src="graphics/em.gif" alt="--" width="22" height="9"></b>that occurs during a given period of time<b><img src="graphics/em.gif" alt="--" width="22" height="9"><img src="graphics/delta.gif" alt="delta" width="11" height="10"></b><i>t</i>.</p> <p align="center"><img src="graphics/22_1eqn.gif" alt="equation" width="155" height="41"></p> <p align="center"><a href="#top"><img src="graphics/button.gif" alt="return to top" border="0" width="226" height="36"></a></p> <p align="center"><a name="instant"><em><b>Instantaneous Rates of Reaction and the Rate Law for a Reaction</b></em></a></p> <p align="left">The rate of the reaction between phenolphthalein and the OH<sup>-</sup> ion isn't constant; it changes with time. Like most reactions, the rate of this reaction gradually decreases as the reactants are consumed. This means that the rate of reaction changes while it is being measured. </p> <p align="left">To minimize the error this introduces into our measurements, it seems advisable to measure the rate of reaction over periods of time that are short compared with the time it takes for the reaction to occur. We might try, for example, to measure the infinitesimally small change in concentration<img src="graphics/em.gif" alt="--" width="22" height="9"><i>d</i>(<i>X</i>)<img src="graphics/em.gif" alt="--" width="22" height="9">that occurs over an infinitesimally short period of time<img src="graphics/em.gif" alt="--" width="22" height="9"><i>dt</i>. The ratio of these quantities is known as the <b>instantaneous rate of reaction</b>.</p> <p align="center"><img src="graphics/22_2eqn.gif" alt="equation" width="91" height="40"></p> <p align="left">The instantaneous rate of reaction at any moment in time can be calculated from a graph of the concentration of the reactant (or product) versus time. The graph below shows how the rate of reaction for the decomposition of phenolphthalein can be calculated from a graph of concentration versus time. The rate of reaction at any moment in time is equal to the slope of a tangent drawn to this curve at that moment. <a name="graph_1"></a></p> <p align="center"><img src="graphics/22_3fig.gif" alt="graph" width="363" height="223"></p> <p align="left">The instantaneous rate of reaction can be measured at any time between the moment at which the reactants are mixed and the reaction reaches equilibrium. Extrapolating these data back to the instant at which the reagents are mixed gives the <i>initial instantaneous rate of reaction</i>.</p> <p align="center"><a href="#top"><img src="graphics/button.gif" alt="return to top" border="0" width="226" height="36"></a></p> <p align="center"><a name="constant"><em><b>Rate Laws and Rate Constants</b></em><b> </b></a></p> <p align="left">An interesting result is obtained when the instantaneous rate of reaction is calculated at various points along the curve in the <a href="#graph_1">graph</a> in the previous section. The rate of reaction at every point on this curve is directly proportional to the concentration of phenolphthalein at that moment in time.</p> <p align="center"><em>Rate = k(phenolphthalein)</em></p> <p align="left">Because this equation is an experimental law that describes the rate of the reaction, it is called the <b>rate law</b> for the reaction. The proportionality constant, <i>k</i>, is known as the <b>rate constant</b>. </p> <p align="center"><a href="#top"><img src="graphics/button.gif" alt="return to top" border="0" width="226" height="36"></a></p> <p align="center"><a name="ways"><em><b>Different Ways of Expressing the Rate of Reaction </b></em></a></p> <p align="left">There is usually more than one way to measure the rate of a reaction. We can study the decomposition of hydrogen iodide, for example, by measuring the rate at which either H<sub>2</sub> or I<sub>2</sub> is formed in the following reaction or the rate at which HI is consumed.</p> <p align="center"><font size="2">2 HI(<i>g</i>) </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> H<sub>2</sub>(<i>g</i>) + I<sub>2</sub>(<i>g</i>)</font></p> <p align="left">Experimentally we find that the rate at which I<sub>2</sub> is formed is proportional to the square of the HI concentration at any moment in time.</p> <p align="center"><img src="graphics/22_4aeqn.gif" alt="equation" width="103" height="41"></p> <p align="left">What would happen if we studied the rate at which H<sub>2</sub> is formed? The balanced equation suggests that H<sub>2</sub> and I<sub>2</sub> must be formed at exactly the same rate.</p> <p align="center"><img src="graphics/22_4beqn.gif" alt="equation" width="101" height="42"></p> <p align="left">What would happen, however, if we studied the rate at which HI is consumed in this reaction? Because HI is consumed, the change in its concentration must be a negative number. By convention, the rate of a reaction is always reported as a positive number. We therefore have to change the sign before reporting the rate of reaction for a reactant that is consumed in the reaction.</p> <p align="center"><img src="graphics/22_4ceqn.gif" width="132" height="41"></p> <p align="left">The negative sign does two things. Mathematically, it converts a negative change in the concentration of HI into a positive rate. Physically, it reminds us that the concentration of the reactant decreases with time.</p> <p align="left">What is the relationship between the rate of reaction obtained by monitoring the formation of H<sub>2</sub> or I<sub>2</sub> and the rate obtained by watching HI disappear? The stoichiometry of the reaction says that two HI molecules are consumed for every molecule of H<sub>2</sub> or I<sub>2</sub> produced. This means that the rate of decomposition of HI is twice as fast as the rate at which H<sub>2</sub> and I<sub>2</sub> are formed. We can translate this relationship into a mathematical equation as follows.</p> <p align="center"><img src="graphics/22_4deqn.gif" alt="equation" width="201" height="43"></p> <p align="left">As a result, the rate constant obtained from studying the rate at which H<sub>2</sub> and I<sub>2</sub> are formed in this reaction (<i>k</i>) is not the same as the rate constant obtained by monitoring the rate at which HI is consumed (<i>k'</i>)</p> <p align="center"><a href="#top"><img src="graphics/button.gif" alt="return to top" border="0" width="226" height="36"></a></p> <p align="center"><a name="stoich"><em><b>The Rate Law Versus the Stoichiometry of a Reaction</b></em></a><em><b> </b></em></p> <p align="left">In the 1930s, Sir Christopher Ingold and coworkers at the University of London studied the kinetics of substitution reactions such as the following.</p> <blockquote> <blockquote> <blockquote> <blockquote> <p align="center"><font size="2">CH<sub>3</sub>Br(<i>aq</i>) + OH<sup>-</sup>(<i>aq</i>) </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> CH<sub>3</sub>OH(<i>aq</i>) + Br<sup>-</sup>(<i>aq</i>)</font></p> </blockquote> </blockquote> </blockquote> </blockquote> <p align="left">They found that the rate of this reaction is proportional to the concentrations of both reactants.</p> <p align="center">Rate = <i>k</i>(CH<sub>3</sub>Br)(OH<sup>-</sup>)</p> <p align="left">When they ran a similar reaction on a slightly different starting material, they got similar products.</p> <p align="center"><font size="2">(CH<sub>3</sub>)<sub>3</sub>CBr(<i>aq</i>) + OH<sup>-</sup>(<i>aq</i>) </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> (CH<sub>3</sub>)<sub>3</sub>COH(<i>aq</i>) + Br<sup>-</sup>(<i>aq</i>)</font></p> <p align="left">But now the rate of reaction was proportional to the concentration of only one of the reactants.</p> <p align="center">Rate = <i>k</i>((CH<sub>3</sub>)<sub>3</sub>CBr)</p> <p align="left">These results illustrate an important point: <b>The rate law for a reaction cannot be predicted from the stoichiometry of the reaction; it must be determined experimentally</b>. Sometimes, the rate law is consistent with what we expect from the stoichiometry of the reaction. </p> <blockquote> <blockquote> <blockquote> <blockquote> <div align="center"><center><table border="0"> <tr> <td><font size="2">2 HI(<i>g</i>) </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> H<sub>2</sub>(<i>g</i>) + I<sub>2</sub>(<i>g</i>)</font></td> <td> </td> <td>Rate = <i>k</i>(HI)<sup>2</sup></td> </tr> </table> </center></div> </blockquote> </blockquote> </blockquote> </blockquote> <p align="left">Often, however, it is not.</p> <div align="center"><center><table border="0"> <tr> <td><font size="2">2 N<sub>2</sub>O<sub>5</sub>(<i>g</i>) <img src="graphics/rarrow.gif" alt="--->" width="17" height="9"> 4 NO<sub>2</sub>(<i>g</i>) + O<sub>2</sub>(<i>g</i>)</font></td> <td> </td> <td>Rate = <i>k</i>(N<sub>2</sub>O<sub>5</sub>)</td> </tr> </table> </center></div><p align="center"><a href="#top"><img src="graphics/button.gif" alt="return to top" border="0" width="226" height="36"></a></p> <p align="center"><a name="order"><em><b>Order of Molecularity </b></em></a></p> <p align="left">Some reactions occur in a single step. The reaction in which a chlorine atom is transferred from ClNO<sub>2</sub> to NO to form NO<sub>2</sub> and ClNO is a good example of a one-step reaction.</p> <blockquote> <blockquote> <blockquote> <blockquote> <p align="center"><font size="2">ClNO<sub>2</sub>(<i>g</i>) + NO(<i>g</i>) </font><font size="1"><img src="graphics/equilibr.gif" alt="<----->" width="17" height="9"></font><font size="2"> NO<sub>2</sub>(<i>g</i>) + ClNO(<i>g</i>)</font></p> </blockquote> </blockquote> </blockquote> </blockquote> <p align="left">Other reactions occur by a series of individual steps. N<sub>2</sub>O<sub>5</sub>, for example, decomposes to NO<sub>2</sub> and O<sub>2</sub> by a three-step mechanism.</p> <div align="center"><center><table border="0" cellpadding="5"> <tr> <td><b>Step 1</b><strong>:</strong> </td> <td> </td> <td><font size="2">N<sub>2</sub>O<sub>5</sub></font><font size="1"> <img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> NO<sub>2</sub> + NO<sub>3</sub></font></td> </tr> <tr> <td><b>Step 2:</b></td> <td> </td> <td><font size="2">NO<sub>2</sub> + NO<sub>3</sub> </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> NO<sub>2</sub> + NO + O<sub>2</sub></font></td> </tr> <tr> <td><b>Step 3:</b></td> <td> </td> <td><font size="2">NO + NO<sub>3</sub> </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> 2 NO<sub>2</sub> </font></td> </tr> </table> </center></div><p align="left">The steps in a reaction are classified in terms of <b>molecularity</b>, which describes the number of molecules consumed. When a single molecule is consumed, the step is called <b>unimolecular</b>. When two molecules are consumed, it is <b>bimolecular</b>. </p> <p align="left">Reactions can also be classified in terms of their <b>order</b>. The decomposition of N<sub>2</sub>O<sub>5</sub> is a <b>first-order reaction</b> because the rate of reaction depends on the concentration of N<sub>2</sub>O<sub>5</sub> raised to the first power.</p> <p align="center">Rate = <i>k</i>(N<sub>2</sub>O<sub>5</sub>)</p> <p align="left">The decomposition of HI is a <b>second-order reaction</b> because the rate of reaction depends on the concentration of HI raised to the second power.</p> <p align="center">Rate = <i>k</i>(HI)<sup>2</sup></p> <p align="left">When the rate of a reaction depends on more than one reagent, we classify the reaction in terms of the order of each reagent.The difference between the molecularity and the order of a reaction is important. </p> <p align="left">The molecularity of a reaction, or a step within a reaction, describes what happens on the molecular level. The order of a reaction describes what happens on the macroscopic scale. We determine the order of a reaction by watching the products of a reaction appear or the reactants disappear. The molecularity of the reaction is something we deduce to explain these experimental results.</p> <p align="center"><a href="#top"><img src="graphics/button.gif" alt="return to top" border="0" width="226" height="36"></a></p> <p align="center"><a name="model"><em><b>Collision Theory Model of Chemical Reactions </b></em></a></p> <p align="left">The <b>collision theory model </b>of chemical reactions can be used to explain the observed rate laws for both one-step and multi-step reactions. This model assumes that the rate of any step in a reaction depends on the frequency of collisions between the particles involved in that step.</p> <p align="left">The figure below provides a basis for understanding the implications of the collision theory model for simple, one-step reactions, such as the following.</p> <p align="center"><font size="2">ClNO<sub>2</sub>(<i>g</i>) + NO(<i>g</i>) </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> NO<sub>2</sub>(<i>g</i>) + ClNO(<i>g</i>)</font></p> <p align="center"><img src="graphics/22_4fig.gif" alt="graphic" width="151" height="297"></p> <p align="left">The kinetic molecular theory assumes that the number of collisions per second in a gas depends on the number of particles per liter. The rate at which NO<sub>2</sub> and ClNO are formed in this reaction should therefore be directly proportional to the concentrations of both ClNO<sub>2</sub> and NO.</p> <p align="center">Rate = <i>k</i>(ClNO<sub>2</sub>)(NO)</p> <p align="left">The collision theory model suggests that the rate of any step in a reaction is proportional to the concentrations of the reagents consumed in that step. The rate law for a one-step reaction should therefore agree with the stoichiometry of the reaction.</p> <p align="left">The following reaction, for example, occurs in a single step.</p> <p align="center"><font size="2">CH<sub>3</sub>Br(<i>aq</i>) + OH<sup>-</sup>(<i>aq</i>) </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> CH<sub>3</sub>OH(<i>aq</i>) + Br<sup>-</sup>(<i>aq</i>)</font></p> <p align="left">When these molecules collide in the proper orientation, a pair of nonbonding electrons on the OH<sup>-</sup> ion can be donated to the carbon atom at the center of the CH<sub>3</sub>Br molecule, as shown in the figure below.</p> <p align="center"><img src="graphics/22_5fig.gif" alt="reaction" width="372" height="59"></p> <p align="left">When this happens, a carbon-oxygen bond forms at the same time that the carbon-bromine bond is broken. The net result of this reaction is the substitution of an OH<sup>-</sup> ion for a Br <sup>-</sup> ion. Because the reaction occurs in a single step, which involves collisions between the two reactants, the rate of this reaction is proportional to the concentration of both reactants.</p> <p align="center">Rate = <i>k</i>(CH<sub>3</sub>Br)(OH<sup>-</sup>)</p> <p align="left">Not all reactions occur in a single step. The following reaction occurs in three steps, as shown in the figure below.</p> <p align="center"><font size="2">(CH<sub>3</sub>)<sub>3</sub>CBr(<i>aq</i>) + OH<sup>-</sup>(<i>aq</i>) </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> (CH<sub>3</sub>)<sub>3</sub>COH(<i>aq</i>) + Br<sup>-</sup>(<i>aq</i>)</font></p> <p align="center"><img src="graphics/22_6fig.gif" alt="reaction" width="183" height="362"></p> <p align="left">In the first step, the (CH<sub>3</sub>)<sub>3</sub>CBr molecule dissociates into a pair of ions.</p> <div align="center"><center><table border="0" cellpadding="0"> <tr> <td>First step</td> <td> </td> <td><img src="graphics/22_7aeqn.GIF" alt="reaction" width="249" height="74"></td> </tr> </table> </center></div><p align="left">The positively charged (CH<sub>3</sub>)<sub>3</sub>C<sup>+</sup> ion then reacts with water in a second step.</p> <div align="center"><center><table border="0" cellpadding="0"> <tr> <td>Second step</td> <td> </td> <td><img src="graphics/22_7beqn.gif" alt="reaction" width="265" height="74"></td> </tr> </table> </center></div><p align="left">The product of this reaction then loses a proton to either the OH<sup>-</sup> ion or water in the final step.</p> <div align="center"><center><table border="0" cellpadding="0"> <tr> <td>Third step</td> <td> </td> <td><img src="graphics/22_7ceqn.gif" alt="reaction" width="343" height="74"></td> </tr> </table> </center></div><p align="left">The second and third steps in this reaction are very much faster than first.</p> <div align="center"><center><table border="0" cellpadding="5"> <tr> <td><font size="2">(CH<sub>3</sub>)<sub>3</sub>CBr </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"> </font><font size="2">(CH<sub>3</sub>)<sub>3</sub>C<sup>+</sup> + Br<sup>-</sup></font></td> <td> </td> <td>Slow step</td> </tr> <tr> <td><font size="2">(CH<sub>3</sub>)<sub>3</sub>C<sup>+</sup> + H<sub>2</sub>O </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> (CH<sub>3</sub>)<sub>3</sub>COH<sub>2</sub><sup>+</sup></font></td> <td> </td> <td>Fast step</td> </tr> <tr> <td><font size="2">(CH<sub>3</sub>)<sub>3</sub>COH<sub>2</sub><sup>+</sup> + OH<sup>-</sup> </font><font size="1"><img src="graphics/equilibr.gif" alt="<---->" width="17" height="9"></font><font size="2"> (CH<sub>3</sub>)<sub>3</sub>COH + H<sub>3</sub>O</font></td> <td> </td> <td>Fast step</td> </tr> </table> </center></div><p align="left">The overall rate of reaction is therefore more or less equal to the rate of the first step. The first step is therefore called the <b>rate-limiting step</b> in this reaction because it literally limits the rate at which the products of the reaction can be formed. Because only one reagent is involved in the rate-limiting step, the overall rate of reaction is proportional to the concentration of only this reagent.</p> <p align="center">Rate = <i>k</i>((CH<sub>3</sub>)<sub>3</sub>CBr)</p> <p align="left">The rate law for this reaction therefore differs from what we would predict from the stoichiometry of the reaction. Although the reaction consumes both (CH<sub>3</sub>)<sub>3</sub>CBr and OH<sup>-</sup>, the rate of the reaction is only proportional to the concentration of (CH<sub>3</sub>)<sub>3</sub>CBr.</p> <p align="left">The rate laws for chemical reactions can be explained by the following general rules.</p> <blockquote> <ul> <li><p align="left">The rate of any step in a reaction is directly proportional to the concentrations of the reagents consumed in that step.</p> </li> <li><p align="left">The overall rate law for a reaction is determined by the sequence of steps, or the <b>mechanism</b>, by which the reactants are converted into the products of the reaction.</p> </li> <li><p align="left">The overall rate law for a reaction is dominated by the rate law for the slowest step in the reaction.</p> </li> </ul> <blockquote> <p align="center"><a href="#top"><img src="graphics/button.gif" alt="return to top" border="0" width="226" height="36"></a></p> </blockquote> </blockquote> </body>