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

Oxygen is available in plenty in air,yet fuels do not burn by themselves at room temperature. Explain.

At $298 \text{ K}$ temperature,the activation energy for the reaction $x_2 + y_2 \rightarrow 2xy + 20 \text{ kJ}$ is $15 \text{ kJ}$. What will be the activation energy for the reaction $2xy \rightarrow x_2 + y_2$?

According to the Arrhenius equation, which of the following statements is correct?

The rate of a reaction doubles when its temperature changes from $300 \, K$ to $310 \, K.$ Activation energy of such a reaction will be .......... $kJ \, mol^{-1}$. $(R= 8.314 \, J \, K^{-1} \, mol^{-1}$ and $\log 2=0.301)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module