In a reaction,the rate constant K_1 is twice | vedclass

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

Question

(B) According to the Arrhenius equation,the rate constant is given by $K = A e^{-E_a/RT}$.For two reactions at the same temperature $T$,assuming the f

Vedclass · Accepted Answer

$E_1 < E_2$

Vedclass · Answer

(B) According to the Arrhenius equation,the rate constant is given by $K = A e^{-E_a/RT}$.For two reactions at the same temperature $T$,assuming the frequency factor $A$ is the same for both:$\frac{K_1}{K_2} = \frac{e^{-E_1/RT}}{e^{-E_2/RT}} = e^{(E_2 - E_1)/RT}$.Given $K_1 = 2K_2$,we have $\frac{K_1}{K_2} = 2$.Taking the natural logarithm on both sides: $\ln(2) = \frac{E_2 - E_1}{RT}$.Since $\ln(2) > 0$,it follows that $E_2 - E_1 > 0$,which means $E_2 > E_1$ or $E_1 < E_2$.

In a reaction,the rate constant $K_1$ is twice the rate constant $K_2$ of another reaction. Find the relationship between the activation energies $E_1$ and $E_2$ of the two reactions at the same temperature.

Similar Questions

The rate of a certain reaction doubles when the temperature changes from $27\,^oC$ to $37\,^oC$. The activation energy for the reaction is ........... $kJ\,mol^{-1}$.

Which is a wrong statement?

Activation energy is defined as:

Activation energy is given by the formula

Half life of a first order reaction is $900 \ \text{min}$ at $400 \ K$. Find its half life at $300 \ K$. Given: $\frac{E_a}{2.303 \ R} = 1.3056 \times 10^3 \ K$. (in $\text{min}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module