<?xml version="1.0" encoding="UTF-8"?>
<quiz>


<!-- question: 0  -->
  <question type="category">
    <category>
        <text>$course$/Kinetics</text>

    </category>
  </question>



<!-- question: 27194  -->
  <question type="matching">
    <name><text>CompareEnergyProfile01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
 Category: Kinetics (KINETIC)
 Keywords: energy profile, reaction diagram.
 ----------------------------End Comment--------------------------->
 Consider the following reactions that take place at the same temperature, begin with the same molar concentrations, and whose energy profiles are presented below:
 <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/ActEnergy_1a.gif" border="0" />]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text>Remember, a larger activation energy means a slower reaction which must have a smaller rate constant.</text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>1</shuffleanswers>
<subquestion>
<text>Which reaction is endothermic?</text>
<answer><text>Reaction I</text>
</answer>
</subquestion>
<subquestion>
<text>Which reaction has the higher activation energy?</text>
<answer><text>Reaction II</text>
</answer>
</subquestion>
<subquestion>
<text>Assuming equal Arrhenius A factors, which reaction is faster?</text>
<answer><text>Reaction I</text>
</answer>
</subquestion>
<subquestion>
<text>Which of the reactions has a larger rate constant?</text>
<answer><text>Reaction I</text>
</answer>
</subquestion>
<subquestion>
<text></text>
<answer><text>Reaction I and Reaction II</text>
</answer>
</subquestion>
<subquestion>
<text></text>
<answer><text>neither Reaction I nor Reaction II</text>
</answer>
</subquestion>
</question>



<!-- question: 27195  -->
  <question type="matching">
    <name><text>ConcVsTimeGraph01</text>
</name>
    <questiontext format="html">
<text><![CDATA[<!----------------------------Comment---------------------------&#010; Category: Kinetics (KINETIC)&#010; Keywords: concentration, product, reactant.&#010; ----------------------------End Comment--------------------------->The following graph shows the change in concentration as a function of time for the reaction:<br />2NO<sub>2</sub>(g) + F<sub>2</sub>(g) <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif" /> 2NO<sub>2</sub>F(g) <br />What do each of the curves A, B, and C represent?<br /><img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/CHGABC1.gif" />]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text>The analysis of this question involves thinking about the relationship between reactants disappearing and products appearing. It also involves looking at the coefficients of the balanced equation to decide the relationship between the concentrations.</text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
<subquestion>
<text><![CDATA[NO<SUB>2</SUB>]]></text>
<answer><text>B</text>
</answer>
</subquestion>
<subquestion>
<text><![CDATA[F<SUB>2</SUB>]]></text>
<answer><text>C</text>
</answer>
</subquestion>
<subquestion>
<text><![CDATA[NO<SUB>2</SUB>F]]></text>
<answer><text>A</text>
</answer>
</subquestion>
</question>



<!-- question: 27198  -->
  <question type="matching">
    <name><text>Enzyme01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
 Category: Kinetics (KINETIC)
 Keywords: catalyst, enzyme.
 ----------------------------End Comment--------------------------->
 Trypsin assists in digestion by breaking proteins down into smaller peptides. However, this process is not as efficient in the presence of fluorophosphates. Identify the role of each of these components.]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text>An enzyme is a protein that is a highly efficient catalyst for one or more chemical reactions in a living system. A molecule whose reaction is catalyzed by an enzyme is referred to as a substrate. An inhibitor is a molecule that decreases the activity of an enzyme.</text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>1</shuffleanswers>
<subquestion>
<text>fluorophosphate</text>
<answer><text>inhibitor</text>
</answer>
</subquestion>
<subquestion>
<text>proteins</text>
<answer><text>substrate</text>
</answer>
</subquestion>
<subquestion>
<text>trypsin</text>
<answer><text>enzyme</text>
</answer>
</subquestion>
</question>



<!-- question: 27204  -->
  <question type="matching">
    <name><text>InterpretEnergyProfile01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
 Category: Kinetics (KINETIC)
 Keywords: energy profile, reaction diagram.
 ----------------------------End Comment--------------------------->
 Use the energy profile below to decide if the following statements are true or false.<BR>
 <IMG SRC="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/energyprofile1.gif">]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text>To determine whether a reaction is exothermic or endothermic, compare the energy of the products and the reactants. If the products have less energy than the reactants, some energy will transfer to the surroundings (exothermic reaction). The reaction may have more than one step. Each step is represented by a bump in the profile. The higher the bump the more energy a molecule needs in order to react and the slower the reaction at that step. To decide where products, reactants, intermediates, and transition states are located in the diagram, you need to think about how the molecules are transformed during a reaction.</text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>1</shuffleanswers>
<subquestion>
<text>This reaction is exothermic.</text>
<answer><text>True</text>
</answer>
</subquestion>
<subquestion>
<text>The reactants are represented by letter A.</text>
<answer><text>True</text>
</answer>
</subquestion>
<subquestion>
<text>The fastest step in this reaction is step D.</text>
<answer><text>False</text>
</answer>
</subquestion>
<subquestion>
<text>An intermediate is represented by letter C.</text>
<answer><text>False</text>
</answer>
</subquestion>
</question>



<!-- question: 27210  -->
  <question type="matching">
    <name><text>RxnOrder01</text>
</name>
    <questiontext format="html">
<text><![CDATA[<!----------------------------Comment---------------------------&#010; Category: Kinetics (KINETIC)&#010; Keywords: rate law, rate equation, reaction order.&#010; ----------------------------End Comment--------------------------->For the reaction below, determine the order with respect to each reactant and the overall order.<br />2NO<sub>2</sub>(g) + F<sub>2</sub>(g) <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif" /> 2NO<sub>2</sub>F(g)<br />rate = k[NO<sub>2</sub>][F<sub>2</sub>]]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text><![CDATA[The rate equation has the following general form<br />Rate = k[A]<sup>m</sup>[B]<sup>n</sup><br />... where m is the order of A, n is the order of B etc. Note, if a reactant is not included in the rate equation then its concentration doesn't affect the rate and its order is zero. The overall order is the sum of the powers (m+n+...).]]></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
<subquestion>
<text><![CDATA[NO<SUB>2</SUB>]]></text>
<answer><text>first-order</text>
</answer>
</subquestion>
<subquestion>
<text><![CDATA[F<SUB>2</SUB>]]></text>
<answer><text>first-order</text>
</answer>
</subquestion>
<subquestion>
<text>overall</text>
<answer><text>second-order</text>
</answer>
</subquestion>
<subquestion>
<text></text>
<answer><text>zero-order</text>
</answer>
</subquestion>
<subquestion>
<text></text>
<answer><text>third-order</text>
</answer>
</subquestion>
<subquestion>
<text></text>
<answer><text>fourth-order</text>
</answer>
</subquestion>
</question>



<!-- question: 27205  -->
  <question type="cloze">
    <name><text>RateConstantUnits01</text>
</name>
    <questiontext>
<text><![CDATA[The isomerization of CH<sub>3</sub>NC to CH<sub>3</sub>CN has a rate constant of 0.46 s<sup>-1</sup> at 600 K.
 <p>(1) Based on the units of the rate constant, what is the order of the reaction? {1:MULTICHOICE:zero-order#No, that is not correct.~=first-order#Correct!~second-order#No, that is not correct.~third-order#No, that is not correct.} </p>
 <p>(2) What is the concentration of CH<sub>3</sub>CN after 0.20 minutes of reaction if the initial concentration is 0.10 M? {1:NUMERICAL:=4e-4:1e-5#Correct!~%50%4e-4:5e-5#Close, but not quite.}(Do not include units!)<br /></p>]]></text>
    </questiontext>
    <generalfeedback>
<text><![CDATA[The units of the rate constant tell us the order of the reaction. In this case, <i>k</i> in units of time<sup>-1</sup> tells us it is a first-order reaction. This in turn tells us we need to use the integrated rate law: ln[A] = -kt + ln[A]<sub>o</sub>.]]></text>
    </generalfeedback>
    <shuffleanswers>0</shuffleanswers>
</question>



<!-- question: 27189  -->
  <question type="multichoice">
    <name><text>ArrheniusActEnergy01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: activation energy, Arrhenius equation.
  ----------------------------End Comment--------------------------->
  The rate constant for the first-order dehydration of tert-butyl alcohol at 500<sup>o</sup>C is 1.20 x 10<sup> -4 </sup> s<sup> -1</sup>.  The rate constant for this process at 600<sup>o</sup>C is 6.80 x 10 <sup> -3</sup>  s<sup> -1</sup>.  Calculate the activation energy for this reaction in kJ/mol.   (R = 8.314 J/mol K)]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="0">
        <text>
-227 kJ/mol
        </text>
      <feedback>
          <text>
<![CDATA[You mixed up k<sub>1</sub> and k<sub>2</sub>.  Be sure that each rate constant is matched correctly to its temperature.]]>
          </text>
      </feedback>
    </answer>
      <answer fraction="100">
        <text>
227 kJ/mol
        </text>
      <feedback>
          <text>
Correct!
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
318 kJ/mol
        </text>
      <feedback>
          <text>
Check your arithmetic carefully.  If is easy to make an arithmetic error using this equation.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
100. kJ/mol
        </text>
      <feedback>
          <text>
You used the temperature in degrees Celsius.  You always need to use the temperature in Kelvin for calculations.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
-100 kJ/mol
        </text>
      <feedback>
          <text>
<![CDATA[You used the temperature in degrees Celsius.  You always need to use the temperature in Kelvin for calculations.  Furthermore, you mixed up k<sub>1</sub> and k<sub>2</sub>.  Be sure that each rate constant is matched correctly to its temperature.]]>
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27190  -->
  <question type="multichoice">
    <name><text>ArrheniusRateConst01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: Arrhenius equation, rate constant.
  ----------------------------End Comment--------------------------->CH<sub>3</sub>CH<sub>2</sub>NO<sub>2</sub> <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif" /> C<sub>2</sub>H<sub>2</sub> + HNO<sub>2</sub> <p>Using the graph below, calculate the rate constant for this reaction at 40.0<sup>o</sup>C. <br /><img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/const1.gif" />]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="100">
        <text>
<![CDATA[3.1 x 10<SUP>-18</SUP> sec<SUP>-1</SUP>]]>
        </text>
      <feedback>
          <text>
Correct. ln k = (slope) 1/T + intercept where the slope is -Ea/R and the intercept is lnA.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[9.56 x 10<SUP>-15</SUP> sec<SUP>-1</SUP>]]>
        </text>
      <feedback>
          <text>
Your answer does not match an obvious error, but it is incorrect. ln k = (slope) 1/T + intercept where the slope is -Ea/R and the intercept is lnA.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[0 sec<SUP>-1</SUP>]]>
        </text>
      <feedback>
          <text>
You left your temperature in Celsius. You have to convert the temperature to Kelvin to solve for the rate constant correctly. ln k = (slope) 1/T + intercept where the slope is -Ea/R and the intercept is lnA.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[-8.3 x 10<SUP>5</SUP> sec<SUP>-1</SUP>]]>
        </text>
      <feedback>
          <text>
You have not used the equation correctly. The equation for the line is ln k = (slope) 1/T + intercept where the slope is -Ea/R and the intercept is lnA, not k=(slope)T + intercept. You also need to convert the temperature to Kelvins.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[-6.5 x 10<SUP>6</SUP> sec<SUP>-1</SUP>]]>
        </text>
      <feedback>
          <text>
You have not used the equation correctly. The equation for the line is ln k = (slope) 1/T + intercept where the the slope is -Ea/R and the intercept is lnA, not k=(slope)T + intercept.
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27191  -->
  <question type="multichoice">
    <name><text>ArrheniusTemp01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: Arrhenius equation.
  ----------------------------End Comment--------------------------->
  The activation energy for the isomerization of cyclopropane to propene is 274 kJ/mol.  By what factor does the rate of reaction increase as the temperature rises from 500<sup>o</sup>C to 550<sup>o</sup>C assuming all else remains constant?    (R = 8.314 J/mol k)  
  <p> Hint:  This is asking for the ratio of rates which is proportional to the ratio of the rate constants, all else being equal.</p>]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="0">
        <text>
1.0
        </text>
      <feedback>
          <text>
<![CDATA[You have a mistake in the units.  You forgot to change 274 kJ to 274 x 10<sup>3</sup> J  <b>or</b>   8.314 J to 8.314 x 10<sup>-3</sup> 10kJ]]>
          </text>
      </feedback>
    </answer>
      <answer fraction="100">
        <text>
13
        </text>
      <feedback>
          <text>
Correct!
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
2.6
        </text>
      <feedback>
          <text>
<![CDATA[You forgot to take the antilog\: &nbsp &nbsp &nbsp k<sub>2</sub>/k<sub>1</sub> = e <sup>2.59</sup>]]>
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[4.0 x 10<sup>2</sup>]]>
        </text>
      <feedback>
          <text>
You used the temperature in degrees Celsius.  You always need to use the temperature in Kelvin for calculations.
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27192  -->
  <question type="multichoice">
    <name><text>CatalystTrueFalse01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: catalyst, enzyme.
  ----------------------------End Comment--------------------------->
  Check the box for each statement about catalysts that is <b>true</b>.]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>false</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="50">
        <text>
An enzyme is a catalyst.
        </text>
      <feedback>
          <text>
Correct.
          </text>
      </feedback>
    </answer>
      <answer fraction="50">
        <text>
The active site of an enzyme is specific and only binds certain substrates.
        </text>
      <feedback>
          <text>
Correct.
          </text>
      </feedback>
    </answer>
      <answer fraction="-33.333">
        <text>
Catalysts increase the rate of a reaction by altering the mechanism thereby increasing the activation energy.
        </text>
      <feedback>
          <text>
Catalysts increase the rate of a reaction by altering the mechanism thereby increasing the activation energy.
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27193  -->
  <question type="multichoice">
    <name><text>CatalyticConverters01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[The following table illustrates the changes in emissions requirements for automobiles: <table cellspacing="5" cellpadding="5" border="1"><tbody><tr><th>Pollutant</th><th>1970 Emissions</th><th>1990 Emissions</th><th>CA ultra-low emissions</th></tr><tr><td>C<sub>x</sub>H<sub>y</sub></td><td>1.5 g/mile</td><td>0.25 g/mile</td><td>0.04 g/mile</td></tr><tr><td>CO</td><td>15 g/mile</td><td>3.4 g/mile</td><td>1.7 g/mile</td></tr><tr><td>NO<sub>x</sub></td><td>3.1 g/mile</td><td>0.4 g/mile</td><td>0.2 g/mile</td></tr></tbody></table>Catalytic converters are used in automobiles to convert pollutants in engine emissions to benign products. Catalytic converters in today's automobiles are made of a porous ceramic brick that has been coated with catalysts, additives, stabilizers, and support materials. Corning has developed bricks that have very thin-walled channels with 600 channels per square inch. Which of the following research goals did they target?]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="100">
        <text>
Catalytic converters need to have a high surface area.
        </text>
      <feedback>
          <text>
Correct. By having more channels, they have increased the surface area of the brick.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
Catalytic converters need to withstand rapid temperature fluctuations.
        </text>
      <feedback>
          <text>
By having more channels, they have increased the surface area of the brick.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
The catalyst has to last at least 100,000 miles.
        </text>
      <feedback>
          <text>
By having more channels, they have increased the surface area of the brick.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
The catalyst needs to be stable at high temperatures.
        </text>
      <feedback>
          <text>
By having more channels, they have increased the surface area of the brick.
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27196  -->
  <question type="multichoice">
    <name><text>ElementaryRxnOrNot01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: bimolecular process, elementary process, mechanism, unimolecular process.
  ----------------------------End Comment--------------------------->
  Which of the following reactions could be elementary and unimolecular reactions?]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>false</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="50">
        <text>
<![CDATA[<IMG SRC="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/unimolecular1.gif">]]>
        </text>
      <feedback>
          <text>
Correct.  A unimolecular elementary reaction is an elementary reaction in which one reactant molecule produces the product molecule or molecules.
          </text>
      </feedback>
    </answer>
      <answer fraction="50">
        <text>
<![CDATA[<IMG SRC="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/unimolecular2.gif">]]>
        </text>
      <feedback>
          <text>
Correct.  A unimolecular elementary reaction is an elementary reaction in which one reactant molecule produces the product molecule or molecules.
          </text>
      </feedback>
    </answer>
      <answer fraction="-20">
        <text>
<![CDATA[<IMG SRC="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/bimolecular1.gif">]]>
        </text>
      <feedback>
          <text>
This a bimolecular elementary reaction. A bimolecular elementary reaction is an elementary reaction in which two reactant molecules produce the product molecule or molecules.
          </text>
      </feedback>
    </answer>
      <answer fraction="-20">
        <text>
<![CDATA[<IMG SRC="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/notelementary1.gif">]]>
        </text>
      <feedback>
          <text>
This is not an elementary reaction. Reactions involving more than two reactant molecules are not elementary.
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27197  -->
  <question type="multichoice">
    <name><text>ElementRxnRateLaws01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: bimolecular process, elementary process, mechanism, rate equation, rate law, unimolecular process.
  ----------------------------End Comment--------------------------->Assume that the equation below represents an elementary reaction. Predict the rate expression (rate law) for this reaction. <br /><p>CO(g) + Cl<sub>2</sub>(g) <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif" /> COCl<sub>2</sub>(g) </p>]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="0">
        <text>
Rate = k[CO]
        </text>
      <feedback>
          <text>
You have included just one of the reactants. This is a bimolecular elementary reaction, so both reactants should be included in the rate law.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[Rate = k[Cl<SUB>2</SUB>]]]>
        </text>
      <feedback>
          <text>
You have included just one of the reactants. This is a bimolecular elementary reaction, so both reactants should be included in the rate law.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[Rate = k[CO][Cl<SUB>2 </SUB>][CoCl<SUB>2</SUB>]]]>
        </text>
      <feedback>
          <text>
You have included both products and reactants. To write a rate law for an elementary reaction, you only include the reactants.
          </text>
      </feedback>
    </answer>
      <answer fraction="100">
        <text>
<![CDATA[Rate = k[CO][Cl<SUB>2</SUB>]]]>
        </text>
      <feedback>
          <text>
<![CDATA[Correct. This is a bimolecular elementary reaction of the form A+B<IMG SRC="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif"> products. The rate law for this type of reaction has the general formula rate= k[A][B].]]>
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27199  -->
  <question type="multichoice">
    <name><text>ExpDataRateConst01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: rate constant.
  ----------------------------End Comment--------------------------->Use the experimental data below to determine the rate constant for the following reaction. <br /><img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/initialrate1.gif" /> <br /><img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/ktable1.gif" /> <br />]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="100">
        <text>
<![CDATA[18 L<SUP>2</SUP>/mol<SUP>2</SUP>&#183;s]]>
        </text>
      <feedback>
          <text>
Correct!
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[0.06 L<SUP>2</SUP>/mol<SUP>2</SUP>&#183;s]]>
        </text>
      <feedback>
          <text>
You have the correct units, but you inverted the relationship.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[3.6 L<SUP>2</SUP>/mol<SUP>2</SUP>&#183;s]]>
        </text>
      <feedback>
          <text>
You have the correct units but you forgot to square the concentration of B.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[18 L/mol&#183;s]]>
        </text>
      <feedback>
          <text>
Be careful of your units!
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[0.06 L/mol&#183;s]]>
        </text>
      <feedback>
          <text>
You have the wrong units, and you inverted the relationship.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[3.6 L/mol&#183;s]]>
        </text>
      <feedback>
          <text>
You have the wrong units, and you forgot to square the concentration of B.
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27200  -->
  <question type="multichoice">
    <name><text>FactorsAffectingRate01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: rates of reaction, reaction rates.
  ----------------------------End Comment--------------------------->Which of the following changes would <u>NOT</u> affect the rate of the reaction of cisplatin with water?<br />Pt(NH<sub>3</sub>)<sub>2</sub>Cl<sub>2</sub>(aq) + H<sub>2</sub>O(l) <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif" /> [Pt(NH<sub>3</sub>)<sub>2</sub>(H<sub>2</sub>O)Cl]<sup>+</sup> + Cl<sup>-</sup>]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="100">
        <text>
Using a 500 mL Erlenmeyer flask for the reaction instead of a 250 mL Erlenmeyer flask.
        </text>
      <feedback>
          <text>
Correct.  The size of the container has no affect on the rate of reaction.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
Using 0.20 M cisplatin in the reaction instead of 0.10 M cisplatin.
        </text>
      <feedback>
          <text>
The rate of reaction is very much dependent on the concentration of the reactants.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[Performing the reaction at 100&deg;C instead of at 25&deg;C.]]>
        </text>
      <feedback>
          <text>
The rate of reaction is very much dependent on the temperature of the system.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
Adding a catalyst to the reaction system.
        </text>
      <feedback>
          <text>
The use of a catalyst will change the rate of reaction.
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27201  -->
  <question type="multichoice">
    <name><text>InitialRateMethod01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: initial rate, rate equation, rate law.
  ----------------------------End Comment--------------------------->Use the experimental data below to determine the rate equation for the following reaction. <br /><img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/initialrate2.gif" /> <br /><img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/iratetable1.gif" /> <br />]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="0">
        <text>
Rate = k[A]
        </text>
      <feedback>
          <text>
A is first order and B is zero order.  This is not the case here.  The rate law must be determined by analyzing the effect of changing the concentration of reactants on the rate of reaction.  What happened to the rate when the concentration of A was doubled (while the concentration of B was held constant)?  What happened to the rate when the concentration of B was doubled (while the concentration of A was held constant)?
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
Rate = k[B]
        </text>
      <feedback>
          <text>
A is zero order and B is first order.  This is not the case here.  The rate law must be determined by analyzing the effect of changing the concentration of reactants on the rate of reaction.  What happened to the rate when the concentration of A was doubled (while the concentration of B was held constant)?  What happened to the rate when the concentration of B was doubled (while the concentration of A was held constant)?
          </text>
      </feedback>
    </answer>
      <answer fraction="100">
        <text>
Rate = k[A][B]
        </text>
      <feedback>
          <text>
Correct!  A is first order and B is first order. The rate law must be determined by analyzing the effect of changing the concentration of reactants on the rate of reaction.  What happened to the rate when the concentration of A was doubled (while the concentration of B was held constant)?  What happened to the rate when the concentration of B was doubled (while the concentration of A was held constant)?
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[Rate = k[A]<SUP>2</SUP>[B]]]>
        </text>
      <feedback>
          <text>
A is second order and B is first order.  This is not the case here.  The rate law must be determined by analyzing the effect of changing the concentration of reactants on the rate of reaction.  What happened to the rate when the concentration of A was doubled (while the concentration of B was held constant)?  What happened to the rate when the concentration of B was doubled (while the concentration of A was held constant)?
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[Rate = k[A][B]<SUP>2</SUP>]]>
        </text>
      <feedback>
          <text>
A is first order and B is second order.  This is not the case here.  The rate law must be determined by analyzing the effect of changing the concentration of reactants on the rate of reaction.  What happened to the rate when the concentration of A was doubled (while the concentration of B was held constant)?  What happened to the rate when the concentration of B was doubled (while the concentration of A was held constant)?
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[Rate = k[A]<SUP>2</SUP>[B]<SUP>2</SUP>]]>
        </text>
      <feedback>
          <text>
A is second order and B is second order.  This is not the case here.  The rate law must be determined by analyzing the effect of changing the concentration of reactants on the rate of reaction.  What happened to the rate when the concentration of A was doubled (while the concentration of B was held constant)?  What happened to the rate when the concentration of B was doubled (while the concentration of A was held constant)?
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[Rate = k[A]<SUP>2</SUP>]]>
        </text>
      <feedback>
          <text>
A is second order and B is zero order.  This is not the case here.  The rate law must be determined by analyzing the effect of changing the concentration of reactants on the rate of reaction.  What happened to the rate when the concentration of A was doubled (while the concentration of B was held constant)?  What happened to the rate when the concentration of B was doubled (while the concentration of A was held constant)?
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[Rate= k[B]<SUP>2</SUP>]]>
        </text>
      <feedback>
          <text>
A is zero order and B is second order.  This is not the case here.  The rate law must be determined by analyzing the effect of changing the concentration of reactants on the rate of reaction.  What happened to the rate when the concentration of A was doubled (while the concentration of B was held constant)?  What happened to the rate when the concentration of B was doubled (while the concentration of A was held constant)?
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[Rate = k[A]<SUP>2</SUP>[B]<SUP>3</SUP>]]>
        </text>
      <feedback>
          <text>
A is second order and B is third order.  This is not the case here. You cannot use the coefficients of the equation to determine the rate law. The rate law must be determined by analyzing the effect of changing the concentration of reactants on the rate of reaction.  What happened to the rate when the concentration of A was doubled (while the concentration of B was held constant)?  What happened to the rate when the concentration of B was doubled (while the concentration of A was held constant)?
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27202  -->
  <question type="multichoice">
    <name><text>IntegratedRateLaw01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: integrated rate equation.
  ----------------------------End Comment--------------------------->The following kinetics information may be useful in answering this problem.<p><a title="KineticsTable" onclick="return openpopup('http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/KineticsTable.gif', 'new_window', ','width=480,height=360,menubar=0,toolbar=0,scrollbars=1,status=1,location=0,resizable=1,resizable=1',0))" href="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/KineticsTable.gif" target="new_window"><img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/KineticsTag.gif" align="absMiddle" border="0" /></a> </p><p>Use the plots below to determine the order of the reaction with respect to A.<br /><img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/ratelawzero1.gif" /><br /><img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/ratelawzero2.gif" /><br /><img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/ratelawzero3.gif" /><br /></p>]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="100">
        <text>
zero-order
        </text>
      <feedback>
          <text>
Correct. A zero-order reaction will result in a straight line when [A] is plotted vs time.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
first-order
        </text>
      <feedback>
          <text>
A first-order reaction will result in a straight line when ln[A] is plotted vs time. This is not the case here.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
second-order
        </text>
      <feedback>
          <text>
A second-order reaction will result in a straight line when 1/[A] is plotted vs time.  This is not the case here.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
There is not enough information given to determine the order.
        </text>
      <feedback>
          <text>
The order of a reaction can be determined by collecting concentration/time data and then plotting [A], ln[A], and 1/[A] vs time.  If one of the plots results in a straight line, then the order of reaction can be determined.
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27203  -->
  <question type="multichoice">
    <name><text>IntermediateInMech01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: intermediate, mechanism.
  ----------------------------End Comment--------------------------->Suppose that for the reaction of nitrogen dioxide and carbon monoxide<br />NO<sub>2</sub> + CO <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif" /> NO + CO<sub>2 </sub><br /><p>the following mechanism has been proposed at high temperatures:<br />(1) NO<sub>2</sub> + CO <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif" /> O-N-O-C-O (slow) <br />(2) O-N-O-C-O <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif" /> NO + CO<sub>2</sub> (fast) <br />Check the box for each species that is an intermediate in this mechanism. </p>]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>false</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="0">
        <text>
<![CDATA[NO<SUB>2</SUB>]]>
        </text>
      <feedback>
          <text>
This is a reactant in the overall equation. Therefore, it cannot be an intermediate.
          </text>
      </feedback>
    </answer>
      <answer fraction="100">
        <text>
O-N-O-C-O
        </text>
      <feedback>
          <text>
Correct! This is an intermediate. An intermediate is produced in one step of a mechanism and used up in a later step. It does not appear in the overall equation.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
NO
        </text>
      <feedback>
          <text>
This is a product in the overall equation. Therefore, it cannot be an intermediate.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
<![CDATA[CO<SUB>2</SUB>]]>
        </text>
      <feedback>
          <text>
This is a product in the overall equation. Therefore, it cannot be an intermediate.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
CO
        </text>
      <feedback>
          <text>
This is a reactant in the overall equation. Therefore, it cannot be an intermediate.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
None of the above is an intermediate.
        </text>
      <feedback>
          <text>
There is an intermediate in this mechanism. An intermediate is produced in one step of a mechanism and used up in a later step. It does not appear in the overall equation.
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27208  -->
  <question type="multichoice">
    <name><text>RateLawChangeConc01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: concentration, reaction rates.
  ----------------------------End Comment--------------------------->The rate law for a reaction is rate=k[A][B]. If the concentration of A doubles, the rate of the reaction will do which of the following?]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="0">
        <text>
not change
        </text>
      <feedback>
          <text>
This would indicate the order of A is zero order.  This is not the case here.
          </text>
      </feedback>
    </answer>
      <answer fraction="100">
        <text>
increase by a factor of 2
        </text>
      <feedback>
          <text>
Correct!  The order of A is first order, which means the concentration of A is proportional to the rate of the reaction.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
increase by a factor of 4
        </text>
      <feedback>
          <text>
This would indicate the order of A is second order.  This is not the case here.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
increase by a factor of 8
        </text>
      <feedback>
          <text>
This would indicate the order of A is third order.  This is not the case here.
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27209  -->
  <question type="multichoice">
    <name><text>RxnMechanisms01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: elementary process, mechanism.
  ----------------------------End Comment--------------------------->Experiments show that the reaction of nitrogen dioxide and carbon monoxide<br />NO<sub>2</sub> + CO <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif" /> NO + CO<sub>2 </sub><br /><p>has the following rate expression at high temperatures:<br /></p><p>Rate=k[NO<sub>2</sub>][CO] <br /></p><p>Is the following mechanism compatible with the experimental information?<br />(1) NO<sub>2</sub> + CO <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif" /> O-N-O-C-O (slow) <br />(2) O-N-O-C-O <img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/arrow.gif" /> NO + CO<sub>2</sub> (fast) <br />Choose the best answer. </p>]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="50">
        <text>
Yes, the mechanism is compatible, because the mechanism sums to the overall reaction.
        </text>
      <feedback>
          <text>
The mechanism reactions do sum to the overall reaction, but this is not the only test for the compatibility of a mechanism.
          </text>
      </feedback>
    </answer>
      <answer fraction="100">
        <text>
Yes, the mechanism is compatible, because the mechanism sums to the overall reaction and the mechanism predicts a rate law that agrees with the experimentally observed rate law.
        </text>
      <feedback>
          <text>
<![CDATA[Correct! In order for a mechanism to be compatible, the reactions of the mechanism must sum to give the overall reaction equation. In addition, the mechanism must predict a rate law that agrees with the experimentally observed rate law. In this case the rate law (determined by the slowest step in the mechanism) is rate = k[NO<SUB>2</SUB>][CO].]]>
          </text>
      </feedback>
    </answer>
      <answer fraction="50">
        <text>
Yes, the mechanism is compatible, because the mechanism predicts a rate law that agrees with the experimentally observed rate law.
        </text>
      <feedback>
          <text>
The mechanism does predict the experimental rate law, but this is not the only test for the compatibility of a mechanism.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
No, the mechanism is not compatible, because the mechanism does not sum to the overall reaction.
        </text>
      <feedback>
          <text>
<![CDATA[The mechanism <B>does</B> sum to the overall reaction. If you add the equations of the mechanism together, canceling each substance that is the same (including being in the same state) on both sides of the equation, they <B>do</B> sum to the overall equation given from the experiment.]]>
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
No, the mechanism is not compatible, because the mechanism does not predict a rate law that agrees with the experimentally observed rate law.
        </text>
      <feedback>
          <text>
<![CDATA[The rate law predicted by a mechanism is determined from the rate-determining step (the slowest step) of the mechanism. In this case, the slowest step is a bimolecular elementary reaction. As a result, the rate law is rate = k[NO<SUB>2</SUB>][CO]. This <B>agrees</B> with the experimentally observed rate law.]]>
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
The compatibility of the mechanism cannot be determined from the information given.
        </text>
      <feedback>
          <text>
There is enough information to determine the compatibility of the mechanism. Compatibility is determined by comparing the mechanism to the experimental data. First, add the equations of the mechanism together and compare the overall equation to the experimental equation. Second, use the rate-determining step (the slowest step) of the mechanism to predict a rate law. Compare this rate law to the experimentally observed rate law. If both agree with the experimental data, then the mechanism is compatible.
          </text>
      </feedback>
    </answer>
</question>



<!-- question: 27211  -->
  <question type="multichoice">
    <name><text>RxnProfileEaDeltaE01</text>
</name>
    <questiontext format="moodle_auto_format">
<text><![CDATA[<!----------------------------Comment---------------------------
  Category: Kinetics (KINETIC)
  Keywords: energy profile, reaction diagram.
  ----------------------------End Comment--------------------------->Based on the reaction profile, what is the energy change for the reaction? <br /><img src="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/catrxn3.gif" />]]></text>
    </questiontext>
    <image></image>
    <generalfeedback>
<text></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <shuffleanswers>0</shuffleanswers>
    <single>true</single>
    <shuffleanswers>false</shuffleanswers>
    <correctfeedback>      <text></text>
</correctfeedback>
    <partiallycorrectfeedback>      <text></text>
</partiallycorrectfeedback>
    <incorrectfeedback>      <text></text>
</incorrectfeedback>
    <answernumbering>abc</answernumbering>
      <answer fraction="100">
        <text>
-17 kJ/mole
        </text>
      <feedback>
          <text>
<![CDATA[Correct! The difference between lines E (products) and D (reactants) gives the energy change of the system (<IMG SRC="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/delta.gif">E or <IMG SRC="http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Kinetics/Images/delta.gif">H).]]>
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
50 kJ/mol
        </text>
      <feedback>
          <text>
This is the activation energy for the uncatalyzed reaction.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
23 kJ/mol
        </text>
      <feedback>
          <text>
This is the energy of the products.
          </text>
      </feedback>
    </answer>
      <answer fraction="0">
        <text>
32 kJ/mol
        </text>
      <feedback>
          <text>
This is the activation energy of the catalyzed reaction.
          </text>
      </feedback>
    </answer>
</question>


</quiz>