Two reactions R_1 and R_2 have identical pre- | vedclass

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

Question

(D) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.For reactions $R_1$ and $R_2$ with identical pre-exponential factors $A$:$k_1 = A e^{-E_

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.For reactions $R_1$ and $R_2$ with identical pre-exponential factors $A$:$k_1 = A e^{-E_{a1} / RT}$$k_2 = A e^{-E_{a2} / RT}$Dividing $k_2$ by $k_1$:$\frac{k_2}{k_1} = \frac{A e^{-E_{a2} / RT}}{A e^{-E_{a1} / RT}} = e^{(E_{a1} - E_{a2}) / RT}$Taking the natural logarithm on both sides:$\ln(k_2/k_1) = \frac{E_{a1} - E_{a2}}{RT}$Given $E_{a1} - E_{a2} = 10 \, kJ \, mol^{-1} = 10,000 \, J \, mol^{-1}$,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$,and $T = 300 \, K$:$\ln(k_2/k_1) = \frac{10,000}{8.314 	imes 300} \approx \frac{10,000}{2494.2} \approx 4.009 \approx 4$.

Similar Questions

The decomposition of a hydrocarbon follows the equation $k = (4.5 \times 10^{11} \ s^{-1}) e^{-28000 \ K / T}$. Calculate the activation energy $E_a$.

The minimum energy required for molecules to enter into the reaction is called

$A$ catalyst

In a reaction,the rate of reaction doubles with a $10^\circ C$ rise in temperature. If the temperature is increased from $10^\circ C$ to $100^\circ C$,by how many times will the rate of reaction increase?

The influence of temperature on the rate of reaction can be found out by

Vedclass Test Series

Exam Paper Generator

Online Exam Module