Question

The activation energy of a reaction can be determined from the slope of which of the following groups?
A. $\ln K~ vs.\dfrac{1}{T}$
B. $\frac{T}{\ln K}~ vs.\dfrac{1}{T}$
C. $\ln K~ vs.T$
D. $\dfrac{\ln K}{T}~ vs.T$

Answer 1

Hint: Activation energy is the minimum amount of energy required by a molecule to change its chemical state or to put in simpler words, it is the energy that is required to change reactants into products during a chemical reaction. It is measured in Joules per mole or Kilojoules per mole.

Complete Solution :
Activation energy is represented with the help of Arrhenius equation:
$k = A{{e}^{\dfrac{-Ea}{RT}}}$ , where
k = Rate constant
A = pre-exponential factor
Ea = Activation energy
R= Gas constant
T= Temperature
So, if we do the logarithm of both the sides, the equation will be:
$\ln k = \ln \left( A{{e}^{\dfrac{-Ea}{RT}}} \right)$
$\ln k = \ln A+\ln {{e}^{\dfrac{-Ea}{RT}}}$
$\ln k = \ln A-\dfrac{Ea}{RT}$ $\left[ \log _{e}^{e} ={{\ln }_{e}}=1 \right]$
$\ln k=\left( \dfrac{-Ea}{R} \right)\dfrac{1}{T}+\ln A$ $\left[ \ln _{a}^{n}=n\ln a \right]$
 Y = mx + c
We have compared the last equation with the straight-line equation which is y=mx + c and we find that the slope which is m and activation energy here can be found out with the graph which is plotted between the $\ln K~ vs.\dfrac{1}{T}$.
So, the correct answer is “Option A”.

Note: A very small thing which a student must remember is that the Temperature used in the Arrhenius equation is in Kelvin. Although, the activation energy doesn’t depend upon temperature, pressure, volume or coefficients of reactants but regarding this question it becomes very crucial as sometimes, you can get confused because of the fact that temperature is used here and therefore, activation energy depends upon the temperature. Also, the slope of the graph is negative which indicates that line will be plotted downside.