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 temperature coefficient of a reaction is $2$. When the temperature is increased from $30^{\circ}C$ to $90^{\circ}C$, the rate of reaction is increased by (in $\times$)

Collision theory is used to explain how chemical species undergo a reaction. Using this theory and the kinetic molecular model,which of the following does $NOT$ influence the rate of a chemical reaction?

The rate of the chemical reaction doubles for an increase of $10 \, K$ in absolute temperature from $298 \, K$. Calculate $E_{a}$.

Which of the following statements is in accordance with the Arrhenius equation?

The activation energy of a reaction at a given temperature is found to be $2.303 \ RT \ J \ mol^{-1}$. The ratio of rate constant to the Arrhenius factor is

Vedclass Test Series

Exam Paper Generator

Online Exam Module