The rate of a reaction decreased by 3.555 tim | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Using the Arrhenius equation: $\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left[\frac{1}{T_1} - \frac{1}{T_2}\right]$Given: $T_1 = 313 \, K$

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(D) Using the Arrhenius equation: $\ln \left(\frac{k_2}{k_1}ight) = \frac{E_a}{R} \left[\frac{1}{T_1} - \frac{1}{T_2}ight]$Given: $T_1 = 313 \, K$ $(40^{\circ} C)$,$T_2 = 303 \, K$ $(30^{\circ} C)$.The rate decreases by $3.555$ times,so $\frac{k_1}{k_2} = 3.555$,which means $\frac{k_2}{k_1} = \frac{1}{3.555}$.$\ln \left(\frac{1}{3.555}ight) = \frac{E_a}{8.314} \left[\frac{1}{313} - \frac{1}{303}ight]$$-1.268 = \frac{E_a}{8.314} \left[\frac{303 - 313}{313 	imes 303}ight]$$-1.268 = \frac{E_a}{8.314} \left[\frac{-10}{94839}ight]$$E_a = \frac{1.268 	imes 8.314 	imes 94839}{10} \approx 99980 \, J \, mol^{-1} = 99.98 \, kJ \, mol^{-1} \approx 100 \, kJ \, mol^{-1}$.

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} C$ to $30^{\circ} C$. The activation energy of the reaction is.........$kJ \, mol^{-1}$.
[Take; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$,$\ln(3.555) = 1.268$]

Similar Questions

Which of the following statements is not true according to the collision theory of reaction rates?

Calculate the activation energy of a reaction,whose rate constant doubles on raising the temperature from $300 \ K$ to $600 \ K$.

The minimum energy necessary to permit a reaction is

$A$ graph plotted between $\log \,K$ vs $\frac{1}{T}$ for calculating activation energy is shown by:

The rate constant of a reaction at $25\,^oC$ is $1 \times 10^{-3}\,s^{-1}$. If the rate of the reaction doubles when the temperature is increased to $35\,^oC$,the activation energy of the reaction is .......... $kJ\, mol^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} C$ to $30^{\circ} C$. The activation energy of the reaction is.........$kJ \, mol^{-1}$.[Take; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$,$\ln(3.555) = 1.268$]