For a first-order reaction A rightarrow P,the | vedclass

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

Question

(D) The Arrhenius equation is given by $\log \, k = \log \, A - \frac{E_a}{2.303 \, RT}$.Comparing this with the given equation $\log \, k = - \frac{2

Vedclass · Accepted Answer

$1.0 	imes 10^6 \, s^{-1}$ and $38.3 \, kJ \, mol^{-1}$

Vedclass · Answer

(D) The Arrhenius equation is given by $\log \, k = \log \, A - \frac{E_a}{2.303 \, RT}$.Comparing this with the given equation $\log \, k = - \frac{2000}{T} + 0.6$:$\log \, A = 0.6 \implies A = 10^{0.6} \approx 3.98 \, s^{-1}$ (Note: The provided option $10^6$ suggests the equation was $\log \, k = - \frac{2000}{T} + 6$).Assuming the equation is $\log \, k = - \frac{2000}{T} + 6$:$\log \, A = 6 \implies A = 10^6 \, s^{-1}$.$\frac{E_a}{2.303 \, R} = 2000$.$E_a = 2000 	imes 2.303 	imes 8.314 \, J \, mol^{-1} \approx 38297 \, J \, mol^{-1} = 38.3 \, kJ \, mol^{-1}$.

For a first-order reaction $A \rightarrow P$,the rate constant $k$ depends on temperature $(T)$ according to the equation $\log \, k = - \frac{2000}{T} + 0.6$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively:

Similar Questions

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 $\log 2=0.301)$

The Arrhenius equation expressing the effect of temperature on the rate constant of a reaction is

The value of the rate constant for the reaction $A + B \to \text{products}$ depends on:

$A$ first order reaction is $50\%$ completed in $20\, \text{minutes}$ at $27\, ^oC$ and in $10\, \text{minutes}$ at $47\, ^oC$. The energy of activation of the reaction is ....... $Kcal$.

The velocity constant of a reaction at $290 \ K$ was found to be $3.2 \times 10^{-3}$. At $310 \ K$,it will be about:

Vedclass Test Series

Exam Paper Generator

Online Exam Module