Activation energy (E_a) and rate constants (k | vedclass

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

Question

(D) The Arrhenius equation is given by: $k = A e^{-E_a / RT}$.For two different temperatures $T_1$ and $T_2$ with corresponding rate constants $k_1$ a

Vedclass · Accepted Answer

Both $(b)$ and $(c)$

Vedclass · Answer

(D) The Arrhenius equation is given by: $k = A e^{-E_a / RT}$.For two different temperatures $T_1$ and $T_2$ with corresponding rate constants $k_1$ and $k_2$:$\ln k_1 = \ln A - \frac{E_a}{RT_1}$$\ln k_2 = \ln A - \frac{E_a}{RT_2}$Subtracting the first equation from the second:$\ln k_2 - \ln k_1 = \left( \ln A - \frac{E_a}{RT_2} \right) - \left( \ln A - \frac{E_a}{RT_1} \right)$$\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$Alternatively,this can be written as:$\ln \frac{k_2}{k_1} = \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)$Thus,both expressions $(b)$ and $(c)$ are correct.

Activation energy $(E_a)$ and rate constants $(k_1)$ and $(k_2)$ of a chemical reaction at two different temperatures $(T_1)$ and $(T_2)$ are related by

Similar Questions

The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

For a chemical reaction,the rate of reaction doubles when the temperature increases by $10\,^oC$. If the temperature is increased from $10\,^oC$ to $100\,^oC$,the rate of reaction will increase by a factor of .....

Which of the following would give a linear plot?
($k$ is the rate constant of an elementary reaction and $T$ is temperature in absolute scale)

According to the collision theory of reaction rates,the rate of reaction increases with temperature due to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module