Question

On increasing temperature from \[200K\] to \[220K\] , rate constant of reaction A increases by $3$ times and rate constant of reaction B increases by $9$ times then correct relationship between activation energy of A and B is:
(A) ${E_A}\,\, = \,\,3{E_B}$
(B) $3{E_A}\,\, = \,\,{E_B}$
(C) ${E_B}\,\, = \,\,2{E_A}$
(D) ${E_A}\,\, = \,\,2{E_B}$

Answer 1

Hint: To solve this question, we must first understand the concept of Chemical Kinetics and must know the basics of Rate Constant of a Reaction. Then we need to use the correct formulae including both activation energy and rate constant to evaluate the relation and then only we can conclude the correct answer.

Complete step-by-step solution:Activation Energy: Activation energy is defined as the minimum amount of extra energy required by a reacting molecule to get converted into product. It can also be described as the minimum amount of energy needed to activate or energize molecules or atoms so that they can undergo a chemical reaction or transformation.
The rate constant is defined as the proportionality constant which explains the relationship between the molar concentration of the reactants and the rate of a chemical reaction.
The rate constant is denoted by $k$ and is also known as reaction rate constant or reaction rate coefficient. It is dependent on the temperature.
Step 1: In this step, we will enlist all the given parameters:
${T_1}\,\, = \,\,200K$
${T_2}\,\, = \,\,220K$
Let, $k\, = $ rate constant and $E\,\, = $ Activation Energy
Step 2: We know the formula: \[\ln \frac{{{k_2}}}{{{k_1}}} = \,\,\frac{E}{{2.303 \times R}}\,\left[ {\frac{1}{{{T_1}}}\,\, - \,\,\frac{1}{{{T_2}}}} \right]\]
For reaction A:
\[\ln \frac{{3k}}{k} = \,\,\frac{{{E_A}}}{{2.303 \times R}}\,\left[ {\frac{1}{{200}}\,\, - \,\,\frac{1}{{220}}} \right]\]
\[ \Rightarrow \ln 3\,\, = \,\,\frac{{{E_A}}}{{2.303 \times R}}\,\left[ {\frac{1}{{200}}\,\, - \,\,\frac{1}{{220}}} \right]\]
Let, \[\,\frac{1}{{2.303 \times R}}\,\left[ {\frac{1}{{200}}\,\, - \,\,\frac{1}{{220}}} \right]\,\, = \,\,x\]
$\therefore \,\,\ln 3\,\, = \,\,x \times {E_A}\,\,\,\,\,\,\,\,\,\,\, \to eq.1$
For Reaction B:
\[\ln \frac{{9k}}{k} = \,\,\frac{{{E_a}}}{{2.303 \times R}}\,\left[ {\frac{1}{{200}}\,\, - \,\,\frac{1}{{220}}} \right]\]
\[ \Rightarrow \ln 9\,\, = \,\,x\]
$\begin{gathered}
   \Rightarrow \ln {3^2}\,\, = \,\,x \times {E_B} \\
  \therefore \,\,\,2\ln 3\,\, = \,\,x \times {E_B}\,\,\,\,\,\,\,\,\,\,\, \to eq.2 \\
\end{gathered} $
 Step 3: On dividing equation 1 by Equation 2 we get:
$\begin{gathered}
  \frac{{\ln 3}}{{2\ln 3}} = \frac{{x \times {E_A}}}{{x \times {E_B}}}\,\, \\
   \Rightarrow \frac{1}{2}\,\, = \,\,\frac{{{E_A}}}{{{E_B}}} \\
  \therefore \,\,{E_B} = \,\,2{E_A} \\
\end{gathered} $
So, we got our required relation between the Activation Energies of Reaction A and B i.e. ${E_B} = \,\,2{E_A}$

Hence, we can conclude that the correct answer is Option C.

Note:A catalyst is a chemical substance that either increases or decreases the rate of a chemical reaction. In the case of activation energy, a catalyst lowers it. However, the energies of the original reactants remain the same. A catalyst only alters the activation energy.