Question

According to Arrhenius equation, the rate constant (k) is related to temperature (T) as:
A. $\ln \left( \frac{{{k}_{2}}}{{{k}_{1}}} \right)=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]$
B. $\ln \left( \frac{{{k}_{2}}}{{{k}_{1}}} \right)=\frac{-{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]$
C. $\ln \left( \frac{{{k}_{2}}}{{{k}_{1}}} \right)=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}+\frac{1}{{{T}_{2}}} \right]$
D. $\ln \left( \frac{{{k}_{2}}}{{{k}_{1}}} \right)=\frac{-{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}+\frac{1}{{{T}_{2}}} \right]$

Answer 1

Hint: Arrhenius equation is basically a formula which describes the temperature dependence of the reaction rates. The formula of Arrhenius equation is \[k=A{{e}^{-Ea/RT}}\]

Complete step by step answer:
- We can see that Arrhenius equation is:
\[k=A{{e}^{-Ea/RT}}\]
- Where, k is the rate constant
  ${{E}_{a}}$ is the activation energy
   T is the temperature
   R is the universal gas constant
   A is the pre-exponential factor which represents the frequency of collisions that takes place in between reactants at a particular concentration.
- Here, we are considering two temperatures, so as we know that the value of rate constant depends on temperature. As we vary the temperature, the value of rate constant will also vary.
-So, we can write the equation for two different temperatures as:
For temperature ${{T}_{1}}$, \[{{k}_{1}}=A{{e}^{-Ea/R{{T}_{1}}}}\]
For temperature ${{T}_{2}}$, \[{{k}_{2}}=A{{e}^{-Ea/R{{T}_{2}}}}\]
By taking log on both equations we get:
$\ln \left( \frac{{{k}_{2}}}{{{k}_{1}}} \right)=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}+\frac{1}{{{T}_{2}}} \right]$\[\log {{k}_{1}}=\log A-\frac{{{E}_{a}}}{R{{T}_{1}}}\]

\[\log {{k}_{2}}=\log A-\frac{{{E}_{a}}}{R{{T}_{2}}}\]
Now, by subtracting both the equations we get:
\[\log \frac{{{k}_{1}}}{{{k}_{2}}}=\frac{-{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\]
- We can write this equation as:
\[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\]
- Hence, we can conclude that the correct option is (), that is according to Arrhenius equation, the rate constant (k) is related to temperature (T) as: $\ln \left( \frac{{{k}_{2}}}{{{k}_{1}}} \right)=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]$

Note: It is found that the unit of rate constant in Arrhenius equation is: ${{\sec }^{-1}}$. As the value of activation energy increases, the rate constant k decreases. And as the temperature increases the value of rate constant increases. The rates of uncatalysed reactions are more affected by temperature than those of the rates of the catalysed reactions.