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

The temperature coefficient of the saponification reaction of an ester by $NaOH$ is $1.75$. The activation energy of the reaction is .......... $kcal \ mol^{-1}$. (in $.21$)

Consider the given energy profile diagram for a reaction and choose the correct option:

The relation between the rate constant and temperature according to the Arrhenius equation is:

The following energy profile diagram is given for the reaction $A + B \to C + D$. The enthalpy change and activation energy for the reverse reaction $C + D \to A + B$ are respectively:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module