Medium
The rate of the chemical reaction doubles for an increase of $10 \, K$ in absolute temperature from $298 \, K$. Calculate $E_{a}$.
(N/A) Given: $T_{1} = 298 \, K$,$T_{2} = 308 \, K$,$k_{2} = 2k_{1}$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
Using the Arrhenius equation: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \, R} \left[ \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right]$.
Substituting the values: $\log 2 = \frac{E_{a}}{2.303 \times 8.314} \left[ \frac{10}{298 \times 308} \right]$.
Solving for $E_{a}$: $E_{a} = \frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10}$.
$E_{a} \approx 52897.78 \, J \, mol^{-1} = 52.9 \, kJ \, mol^{-1}$.
Using the Arrhenius equation: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \, R} \left[ \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right]$.
Substituting the values: $\log 2 = \frac{E_{a}}{2.303 \times 8.314} \left[ \frac{10}{298 \times 308} \right]$.
Solving for $E_{a}$: $E_{a} = \frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10}$.
$E_{a} \approx 52897.78 \, J \, mol^{-1} = 52.9 \, kJ \, mol^{-1}$.
Explore More
Similar Questions
If $E_f$ and $E_r$ are the activation energies of forward and reverse reactions and the reaction is known to be exothermic,then
Easy
View SolutionSubtract $(i)$ $\ln \, k_1 = - \frac{E_a}{R T_1} + \ln A$ and $(ii)$ $\ln \, k_2 = - \frac{E_a}{R T_2} + \ln A$ and write the resulting equation.
Easy
View SolutionFrom the given data for the reaction $H_2 + I_2 \rightarrow 2HI$,calculate the activation energy $(E_a)$:
$T_1 = 769 \ K, \ 1/T_1 = 1.3 \times 10^{-3} \ K^{-1}, \ \log_{10} K_1 = 2.9$
$T_2 = 667 \ K, \ 1/T_2 = 1.5 \times 10^{-3} \ K^{-1}, \ \log_{10} K_2 = 1.1$
$T_1 = 769 \ K, \ 1/T_1 = 1.3 \times 10^{-3} \ K^{-1}, \ \log_{10} K_1 = 2.9$
$T_2 = 667 \ K, \ 1/T_2 = 1.5 \times 10^{-3} \ K^{-1}, \ \log_{10} K_2 = 1.1$
Difficult
View SolutionThe rate constant of a reaction increases by five times on increase in temperature from $27^{\circ} C$ to $52^{\circ} C$. The value of activation energy in $kJ \, mol^{-1}$ is $....$ . (Rounded-off to the nearest integer) $[R = 8.314 \, J \, K^{-1} \, mol^{-1}]$
For an endothermic reaction,the energy of activation is $E_a$ and the enthalpy of reaction is $\Delta H$ (both in $kJ/mol$). The minimum value of $E_a$ will be:
MediumAIPMT 2010
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo