The number of given statement/s which is/are | vedclass

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

Question

(B) According to the Arrhenius equation,$k = A \cdot e^{-E_a / RT}$.Taking the natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Takin

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) According to the Arrhenius equation,$k = A \cdot e^{-E_a / RT}$.Taking the natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Taking the base-$10$ logarithm: $\log k = \log A - \frac{E_a}{2.303 R} \cdot \frac{1}{T}$.Evaluating the statements:$(A)$ Correct: $A$ higher $E_a$ leads to a steeper slope $(-\frac{E_a}{2.303 R})$ in the $\log k$ vs $\frac{1}{T}$ plot,meaning the rate constant $k$ changes more rapidly with temperature.$(B)$ Correct: If $E_a = 0$,then $k = A \cdot e^0 = A$,which is constant and independent of temperature.$(C)$ Incorrect: This is the opposite of statement $(A)$.$(D)$ Incorrect: If there is no correlation between temperature and rate constant,it implies $E_a = 0$,not a negative activation energy.Thus,there are $2$ correct statements ($A$ and $B$).

Similar Questions

What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$,where $k$ is the rate constant of a reaction at temperature $T$?

The number of molecules with energy greater than the threshold energy for a reaction increases five-fold by a rise of temperature from $27^{\circ} C$ to $42^{\circ} C$. Its energy of activation in $J \ mol^{-1}$ is.............. (Take $\ln 5 = 1.6094$; $R = 8.314 \ J \ mol^{-1} K^{-1}$)

Calculate the activation energy of a reaction,whose rate constant doubles on raising the temperature from $300 \ K$ to $600 \ K$.

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 ........ $kCal$.

For a reaction,the temperature increases from $10\,^\circ\text{C}$ to $50\,^\circ\text{C}$. The rate of the reaction will increase by a factor of ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module