Rate of Chemical Reaction
Chemical reactions are the simplest way to convert one substance into the other substance. The feasibility of a chemical reaction is decided by the concepts of chemical thermodynamics under a defined set of conditions. The extent of a chemical reaction is determined by the principles of chemical equilibrium. But the information that remains undisclosed is the speed or rate of a chemical reaction, path or mechanism of a chemical reaction, time taken for a reaction to attain a state of equilibrium, factors controlling the rate of reaction etc. This information pertaining to the study of a chemical reaction is very pivotal, and it helps in understanding the nature (slow or fast nature) of a chemical reaction. The kinetics of a chemical reaction can be understood with the help of average rate of reaction or a differential rate law or an integrated rate law. There are various factors on which the rate of a chemical reaction depends viz. nature of reactants, temperature, catalyst, pressure and concentration of reactants and products (rate law expression).
The rate constants of a reaction at 700 K and 500 K are 0.07 s−1 and 0.02 s−1 respectively. What is the value of activation energy of the reaction?
(Given: log 7 = 0.8451, log 2 = 0.3010)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 18.230 kJ mol−1
The Arrhenius equation relates the rate constant to temperature. Taking the rate constants at two temperatures gives:
\( \displaystyle \log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1T_2}\right) \)
Here \( k_1 = 0.02\ \mathrm{s^{-1}} \) at \( T_1 = 500\ \mathrm{K} \), \( k_2 = 0.07\ \mathrm{s^{-1}} \) at \( T_2 = 700\ \mathrm{K} \), and \( R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}} \).
\( \displaystyle \frac{k_2}{k_1}=\frac{0.07}{0.02}=\frac{7}{2} \)
\( \displaystyle \log\frac{7}{2}=\log 7-\log 2=0.8451-0.3010=0.5441 \)
\( \displaystyle \frac{T_2-T_1}{T_1T_2}=\frac{200}{500\times700}=\frac{1}{1750}\ \mathrm{K^{-1}} \)
\( \displaystyle E_a=0.5441\times2.303\times8.314\times1750 \)
\( \displaystyle E_a\approx18231\ \mathrm{J\,mol^{-1}} \)
So \( E_a \) = 18230.8 J mol−1 = 18.230 kJ mol−1.
Note 1: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Temperature Dependence of the Rate of a Reaction – …
Note 2: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| 15-05 S1 11-05 S1 | – | – | – | – |