The differential form of the Arrhenius equati | vedclass

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

Question

(B) The Arrhenius equation is given by $K = A e^{-\frac{E_a}{RT}}$.Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.

Vedclass · Accepted Answer

$\frac{d \ln K}{dT} = \frac{\Delta E^*}{RT^2}$

Vedclass · Answer

(B) The Arrhenius equation is given by $K = A e^{-\frac{E_a}{RT}}$.Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.Differentiating both sides with respect to temperature $T$,we get $\frac{d \ln K}{dT} = 0 - \frac{E_a}{R} 	imes (-\frac{1}{T^2})$.Therefore,$\frac{d \ln K}{dT} = \frac{E_a}{RT^2}$.Comparing this with the given options,where $\Delta E^*$ represents the activation energy $E_a$,the correct expression is $\frac{d \ln K}{dT} = \frac{\Delta E^*}{RT^2}$.

The differential form of the Arrhenius equation is:

Similar Questions

Which of the following statements is true for a catalyst?

The temperature dependence of the rate constant $k$ is expressed as $k = A e^{-E_a / RT}$. When a plot between $\log k$ and $1/T$ is plotted,we get the graph as shown. What is the value of the slope in the graph?

When temperature is increased from $27\,{}^{\circ}C$ to $127\,{}^{\circ}C$,the rate of reaction becomes doubled. The activation energy $E_a$ will be ........ $kCal$.

The rate constant for a reaction can be increased by $\underline{a}$ the stability of the reactant or by $\underline{b}$ the stability of the transition state. Select the correct choice for $a$ and $b$.

For a first-order reaction $A \to P$,the rate constant equation is given by $\log K = -2000 \, (1/T) + 6.0$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module