Question

A reaction takes place at $300\,{\text{K}}$ .When a catalyst is added, the rate of the reaction increases. How much temperature should be increased (in ${}^{\text{O}}\,{\text{C}}$ ) which can create the same effect as produced by the catalyst? (Experimentally it is known that the catalyst changes the activation energy by $20\,\% $ )
A. $100$
B. $50$
C. $125$
D. $75$

Answer 1

Hint: Activation energy was introduced by Swedish scientist Svante Arrhenius in 1889, and it is defined as the minimum energy required to start a chemical reaction. The activation energy is denoted by ${{\text{E}}_a}$ and is given in the units of Kilojoules per mole or Kilocalories per mole. Whereas catalysts are substances which can be added to a reaction in order to increase the rate of the reaction without getting involved in the reaction. Catalysts speed up the reaction by reducing the activation energy or by changing the reaction mechanism.

Formula used:${\text{k}}\,{\text{ = }}\,{\text{A }}{{\text{e}}^{ - \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}}}$
Where,
$
 {\text{k}}\,{\text{ = reaction rate}} \\
  {\text{A}}\,{\text{ = pre exponential factor}} \\
  {{\text{E}}_a}{\text{ = activation energy}} \\
  {\text{R}}\,{\text{ = }}\,{\text{gas constant}} \\
  {\text{T}}\,{\text{ = }}\,{\text{temperature in kelvin}} \\
 $

Complete step by step answer:
The activation energy is related to rate of the reaction by the given formula ${\text{k}}\,{\text{ = }}\,{\text{A }}{{\text{e}}^{ - \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}}}$.
Another form of this equation which relates rate constant ${\text{k}}$ at two temperatures
 ${\text{ln}}\,\dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}}\, = \, - \dfrac{{{{\text{E}}_{\text{a}}}}}{{\text{R}}}\left( {\dfrac{1}{{{{\text{T}}_2}}} - \dfrac{1}{{{{\text{T}}_1}}}} \right)\,{\text{ }} \to \left( 1 \right)$
Where,
${{\text{k}}_1}\,{\text{rate constant at }}{{\text{T}}_{1\,}}\,,\,\,{{\text{k}}_2}\,{\text{rate constant at }}{{\text{T}}_2}$
 Here it is given that the chemical reaction takes place at $300\,{\text{K}}$ , on adding the catalyst the rate of the reaction increases by decreasing the activation energy of the reaction by $20\,\% $. So we can write the equation as
${\text{ln}}\,\dfrac{{{{\text{k}}_{\text{C}}}}}{{{{\text{k}}_1}}}\, = \,\dfrac{{{{\text{E}}_a} - {{\text{E}}_{a\,c}}}}{{{\text{R}}\,{{\text{T}}_1}}}{\text{ }} \to \left( 2 \right)$
Where ${{\text{k}}_C}\,$ be the rate of the reaction after adding the catalyst at $300\,{\text{K}}$ and having activation energy ${{\text{E}}_{a\,c}}$ . Where ${{\text{k}}_1}$ be the rate of reaction without catalyst at $300\,{\text{K}}$.
So here we have to calculate the amount of temperature that should be increased for this reaction so that we can increase the rate of the reaction in a similar way as the catalyst increases the reaction rate.
Here ${{\text{k}}_2}$ be the rate of the reaction after increasing the temperature. And here we are calculating the amount which create same effect as catalyst so ${{\text{k}}_2}\, = \,{{\text{k}}_{\text{C}}}$ .
So comparing the equations $ \to \left( 1 \right)\,\,\,\& \,\, \to \left( 2 \right)$ we get
\[\dfrac{{{{\text{E}}_{a\,}} - \,{\text{E}}{\,_{a\,c}}}}{{{\text{R}}\,{\text{T}}{\,_1}}}\, = \,\,\dfrac{{{{\text{E}}_a}}}{{\text{R}}}\,\left( {\dfrac{1}{{{{\text{T}}_1}}} - \,\dfrac{1}{{{{\text{T}}_2}}}} \right)\]
By rearranging the equation
$
\dfrac{{{{\text{E}}_{{\text{a}}\,}}{\text{ - }}\,{{\text{E}}_{{\text{a}}\,{\text{c}}}}}}{{{{\text{T}}_{\text{1}}}}}\,{\text{ = }}\,\,{{\text{E}}_{\text{a}}}\left( {\dfrac{{\text{1}}}{{{{\text{T}}_{\text{1}}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{T}}_{\text{2}}}}}} \right) \\
\dfrac{{{{\text{E}}_{{\text{a}}\,}}{\text{ - }}\,{{\text{E}}_{{\text{a}}\,{\text{c}}}}}}{{{{\text{T}}_{\text{1}}}\,{{\text{E}}_{\text{a}}}}}\, = \,\dfrac{{\text{1}}}{{{{\text{T}}_{\text{1}}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{T}}_{\text{2}}}}} \\
$
Here the activation energy is given that the activation energy is decreased by $20\,\% $ after adding a catalyst. So we can take ${{\text{E}}_a}\, - \,100\,\,\,{\text{and }}{{\text{E}}_{a\,c}} - \,80$. And ${{\text{T}}_1}\, - \,300{\text{K}}$ . By adding the values we get
\[
\dfrac{{100 - 80}}{{100 \times 300}}\, = \,\dfrac{1}{{300}} - \,\dfrac{1}{{{{\text{T}}_2}}} \\
\dfrac{{20}}{{100 \times 300}}\, = \,\dfrac{1}{{300}} - \,\dfrac{1}{{{{\text{T}}_2}}} \\
\therefore \,\,\,\dfrac{1}{{{{\text{T}}_{\text{2}}}}}\, = \,\dfrac{1}{{300}}\, - \,\dfrac{{20}}{{100 \times 300}} = \dfrac{4}{{1500}} \\
{{\text{T}}_2}\, = \,\dfrac{{1500}}{4}\, = 375\,{\text{K}} \\
 \]
So here the change in temperature ${{\text{T}}_2}\,{\text{ - }}\,{{\text{T}}_1}\, = \,\,375 - 300\, = 75$
Here by increasing the temperature by $75{\,^{\text{O}}}\,{\text{C}}$ we get the same effect as produced by the catalyst.

The correct answer is option D.

Note: Catalyst is a substance which participates in a chemical reaction by forming temporary bonds with the reactants thereby forms an intermediate complex which is then decomposed to produce a catalyst. A catalyst can remain chemically and quantitatively unchanged after the reaction. By making the intermediate complex catalyst provides an alternative mechanism by reducing the activation energy between the reactants and products thereby lowering their potential energy barrier which leads to the increased reaction rate