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    <name>
      <text>6.4.3.1 Engineering Example 1</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3>Arrhenius' law</h3>
<h4>Introduction</h4>
<p> Chemical reactions are very sensitive to temperature; normally,
the rate of reaction increases as temperature increases. For
example, the corrosion of iron and the spoiling of food are more
rapid at higher temperatures. Chemically, the probability of
collision between two molecules increases with temperature, and an
increased collision rate results in higher kinetic energy, thus
increasing the proportion of molecules that have the
<b> activation energy</b> for the reaction, i.e. the minimum energy
required for a reaction to occur. Based upon his observations, the
Swedish chemist, Svante Arrhenius, proposed that the rate of a
chemical reaction increases exponentially with temperature. This
relationship, now known as Arrhenius' law, is written as</p>
\[k = k_0 \, \exp \left( \frac{-E_a}{R\,T} \right) \qquad (1)\]
<p>where \(k\) is the reaction rate constant, \(k_0\) is the frequency
factor, \(E_a\)is the activation energy, \(R\) is the universal gas
constant and \(T\) is the absolute temperature. Thus, the reaction
rate constant, \(k\), depends on the quantities \(k_0\) and \(E_a\),
which characterise a given reaction, and are generally assumed to
be temperature independent.</p>

<h4> Problem in words</h4>

<p>In a laboratory, ethyl acetate is reacted with sodium
hydroxide to investigate the reaction kinetics. Calculate the
frequency factor and activation energy of the reaction from
Arrhenius' Law, using the experimental measurements of temperature
and reaction rate constant in the table:</p>
<center>
<table frame="void" rules="all" border="1">
<thead>
<tr class="header">
<th style="text-align: center;"><span class="math inline">\(T\)</span></th>
<th style="text-align: center;">310</th>
<th style="text-align: center;">350</th>
</tr>
</thead>
<tbody>
<tr class="odd">
<td style="text-align: center;"><span class="math inline">\(k\)</span></td>
<td style="text-align: center;">7.757192</td>
<td style="text-align: center;">110.9601</td>
</tr>
</tbody>
</table>
</center>


<h4> Mathematical statement of problem</h4>

<p>Given that \(k = 7.757192 s^{-1}\) at \(T = 310 K\) and \(k =
110.9601 s^{-1}\) at \(T = 350 K\), use Equation (1) to produce two
linear equations in \(E_a\) and \(k_0\). Solve these to find \(E_a\) and
\(k_0\). (Assume that the gas constant \(R = 8.314 J K^{-1}
\text{mol}^{-1}.\))</p>

<h4> Mathematical analysis</h4>

<p>Taking the natural logarithm of both sides of (1)</p>

\[\ln k = \ln \left\{k_0 \, \exp\left(\frac{-E_a}{R\,T}\right)\right\}= \ln k_0 - \frac{E_a}{R\,T}.\]
Now inserting the experimental data gives the two linear equations in \(E_a\) and \(k_0\)

\[\ln k_1 = \ln k_0 - {\frac{E_a}{R\,T_1}} \qquad (2)\]

\[\ln k_2 = \ln k_0 - \frac{E_a}{R\,T_2} \qquad (3)\]

where \(k_1 = 7.757192\), \(T_1 = 310\) and \(k_2 = 110.9601\), \(T_2 =
350\).

<p>Firstly, to find \(E_a\), subtract Equation (2) from Equation (3)
\[\ln k_2-\ln k_1= \dfrac{E_a}{R\,T_1}-\dfrac{E_a}{R\,T_2}= \dfrac{E_a}{R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right)\]
so that
\[E_a = \frac{R\,(\ln k_2-\ln k_1)}{\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}\]
and substituting the values gives
\[E_a= 60 000 \mbox{ J mol}^{-1}  = 60 \mbox{ kJ mol}^{-1}\]
Secondly, to find \(k_0\), from (2)
\[\ln k_0 = \ln k_1 + \frac{E_a}{R\,T_1}\]
which implies
\[k_0 = \exp \left( \ln k_1 + \frac{E_a}{R\,T_1}\right)= k_1 \, \exp \left(\frac{E_a}{R\,T_1}\right)\]
and substituting the values gives
\[k_0= 1.0 \times 10^{11} \mbox{ s}^{-1}.\]</p>]]></text>
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