The plot of log K vs frac{1}{T} helps to calc | vedclass

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

Question

(D) According to the Arrhenius equation:$\log K = \log A - \frac{E_a}{2.303 R} \cdot \frac{1}{T}$Comparing this with the equation of a straight line $

Vedclass · Accepted Answer

The energy of activation as well as frequency factor

Vedclass · Answer

(D) According to the Arrhenius equation:$\log K = \log A - \frac{E_a}{2.303 R} \cdot \frac{1}{T}$Comparing this with the equation of a straight line $y = mx + c$,where $y = \log K$ and $x = \frac{1}{T}$:The slope of the plot is $m = -\frac{E_a}{2.303 R}$,which allows the calculation of the activation energy $(E_a)$.The y-intercept is $c = \log A$,which allows the calculation of the frequency factor $(A)$.Therefore,the plot helps to calculate both the activation energy and the frequency factor.

The plot of $\log K$ vs $\frac{1}{T}$ helps to calculate

Similar Questions

Explain: How is the value of activation energy determined based on the Arrhenius equation?

The rate of a reaction at $273 \ K$ is $R_0$. What will be the rate of the reaction at $313 \ K$? (Take the temperature coefficient as $2$)

The equilibrium constant at $27\,^{\circ}C$ and $127\,^{\circ}C$ is $30$ and $40$ respectively. What will be the ratio of activation energy $(E_{a})_{f} / (E_{a})_{b}$?

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

The Arrhenius equation can be represented as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module