Looking for classes? Ksquare Career Institute, Bengaluru →
Chemical Kinetics
Arrhenius Equation and Activation Energy
2 questions from 2 papers, each with its answer.
- questions
- 2
- papers
- 2
- years asked, 2017–2026
- 2 of 10
- of Chemical Kinetics
- 2.5%
Appears in
Concepts tested
- Calculating activation energy from rate constants at two temperatures
All Questions
2026 1 question
-
Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, T (K). $\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$ The energy of activation in kcal $mol^{-1}$ for the reaction is: [Given: k in $s^{-1}$, $R = 1.987$ cal $mol^{-1}\,K^{-1}$]$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$ Comparing with $\ln k = \ln A - \frac{E_a}{RT}$: $\frac{E_a}{R} = 1.25 \times 10^4$ $E_a = 1.25 \times 10^4 \times \frac{1.987}{1000}\ kcal\,mol^{-1}$ $E_a = 24.84\ kcal\,mol^{-1}$
2024 1 question
-
Activation energy of any chemical reaction can be calculated if one knows the value ofThe activation energy of a reaction can be determined by evaluating the rate constants at two different temperatures, using the Arrhenius equation: $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1T_2}\right)$