Question

The rate constant of a reaction is \[1.5{\text{ }} \times {\text{ 1}}{{\text{0}}^7}{\text{ }}{{\text{s}}^{ - 1}}\] at \[{50^ \circ }C\] and \[4.5{\text{ }} \times {\text{ 1}}{{\text{0}}^7}{\text{ }}{{\text{s}}^{ - 1}}\] at \[{100^ \circ }C\]. Calculate the Arrhenius parameter A and \[{E_a}\].

Answer 1

Hint: We have the value of rate constant for both reactions. Both the reactions are of first order. We have to find Arrhenius parameter, A and activation energy \[{E_a}\] for the reaction. We will use the relation of temperature with activation energy and then by using activation energy we will find the Arrhenius parameter.
Formula Used:
\[(i){\text{ lo}}{{\text{g}}_{10}}\dfrac{{{k_2}}}{{{k_1}}}{\text{ = }}\dfrac{{{E_a}}}{{2.303R}}\left( {\dfrac{1}{{{T_1}}}{\text{ - }}\dfrac{1}{{{T_2}}}} \right)\]
\[(ii){\text{ lo}}{{\text{g}}_{10}}{\text{k = lo}}{{\text{g}}_{10}}{\text{A - }}\dfrac{{{E_a}}}{{2.303RT}}\]

Complete answer:
Since rate of the reaction is given at different temperature we can find the value of activation energy by using the given relation,
\[{\text{ lo}}{{\text{g}}_{10}}\dfrac{{{k_2}}}{{{k_1}}}{\text{ = }}\dfrac{{{E_a}}}{{2.303R}}\left( {\dfrac{1}{{{T_1}}}{\text{ - }}\dfrac{1}{{{T_2}}}} \right)\]
Let us consider rate of reaction at temperature \[{T_{1{\text{ }}}}\] be \[1.5{\text{ }} \times {\text{ 1}}{{\text{0}}^7}{\text{ }}{{\text{s}}^{ - 1}}\] and the rate of reaction at temperature \[{T_2}\] be \[4.5{\text{ }} \times {\text{ 1}}{{\text{0}}^7}{\text{ }}{{\text{s}}^{ - 1}}\] . Hence,
\[{T_{1{\text{ }}}} = {\text{ 50 + 273 = 323 K}}\]
\[{T_{{\text{2 }}}} = {\text{ 100 + 273 = 373 K}}\]
\[{\text{ }}{{\text{k}}_1}{\text{ = }}1.5{\text{ }} \times {\text{ 1}}{{\text{0}}^7}{\text{ }}{{\text{s}}^{ - 1}}\]
\[{k_2}{\text{ = }}4.5{\text{ }} \times {\text{ 1}}{{\text{0}}^7}{\text{ }}{{\text{s}}^{ - 1}}\]
On substituting the above values we get the result as,
\[{\text{ lo}}{{\text{g}}_{10}}\dfrac{{{k_2}}}{{{k_1}}}{\text{ = }}\dfrac{{{E_a}}}{{2.303R}}\left( {\dfrac{1}{{{T_1}}}{\text{ - }}\dfrac{1}{{{T_2}}}} \right)\]
\[{\text{ lo}}{{\text{g}}_{10}}\dfrac{{4.5{\text{ }} \times {\text{ 1}}{{\text{0}}^7}}}{{1.5{\text{ }} \times {\text{ 1}}{{\text{0}}^7}}}{\text{ = }}\dfrac{{{E_a}}}{{2.303R}}\left( {\dfrac{1}{{323}}{\text{ - }}\dfrac{1}{{373}}} \right)\]
\[{\text{ lo}}{{\text{g}}_{10}}{\text{3 = }}\dfrac{{{E_a}}}{{2.303{\text{ }} \times {\text{ 8}}{\text{.314}}}}\left( {\dfrac{{50}}{{323{\text{ }} \times {\text{ 373}}}}} \right)\]
On solving the equation we get the result as,
\[{\text{ }}{{\text{E}}_a}{\text{ = 2}}{\text{.2 }} \times {\text{ 1}}{{\text{0}}^4}{\text{ J mo}}{{\text{l}}^{ - 1}}\]
Thus we get the value of activation energy for the reaction. Now we will calculate the value of Arrhenius constant by using the value of activation energy as,
\[{\text{ lo}}{{\text{g}}_{10}}{\text{k = lo}}{{\text{g}}_{10}}{\text{A - }}\dfrac{{{E_a}}}{{2.303RT}}\]
Here the value of k is \[1.5{\text{ }} \times {\text{ 1}}{{\text{0}}^7}{\text{ }}{{\text{s}}^{ - 1}}\] and value of T is \[ = {\text{ 50 + 273 = 323 K}}\]. On substituting the values we get the result as,
\[{\text{ lo}}{{\text{g}}_{10}}{\text{1}}{\text{.5 }} \times {\text{ 1}}{{\text{0}}^7}{\text{ = lo}}{{\text{g}}_{10}}{\text{A - }}\dfrac{{2.2{\text{ }} \times {\text{ }}{{10}^4}}}{{2.303{\text{ }} \times {\text{ R = 8}}{\text{.314 }} \times {\text{ 323}}}}\]
\[{\text{ lo}}{{\text{g}}_{10}}{\text{1}}{\text{.5 }} \times {\text{ 1}}{{\text{0}}^7}{\text{ = lo}}{{\text{g}}_{10}}{\text{A - 3}}{\text{.55}}\]
On solving the above equation we get,
\[{\text{A = 5}}{\text{.42 }} \times {\text{ 1}}{{\text{0}}^{10}}{\text{ }}{{\text{s}}^{ - 1}}\]
Thus we get the value of Arrhenius constant and activation energy of the reaction by using the relation between them.

Note:
We use the value of \[{\text{R = 8}}{\text{.314}}\] in the above solution. We must convert the temperature into Kelvin scale before putting in the formula we used. Since activation energy is a form of energy, therefore its unit is Joule. For finding the values of logarithmic functions, we can use a log table. Also some basic values of logarithmic function must be remembered.