For the reaction C_2H_5I + OH^- rightarrow C_ | vedclass

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

Question

(A) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$Given: $K_1 = 0.325$,$K_2 =

Vedclass · Accepted Answer

$20260$

Vedclass · Answer

(A) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{1}{T_1} - \frac{1}{T_2} ight]$Given: $K_1 = 0.325$,$K_2 = 6.735$,$T_1 = 30 + 273 = 303 \ K$,$T_2 = 60 + 273 = 333 \ K$,$R = 1.987 \ cal \ K^{-1} \ mol^{-1}$.Substituting the values:$\log \left( \frac{6.735}{0.325} ight) = \frac{E_a}{2.303 	imes 1.987} \left[ \frac{1}{303} - \frac{1}{333} ight]$$\log(20.723) = \frac{E_a}{4.575} \left[ \frac{333 - 303}{303 	imes 333} ight]$$1.3164 = \frac{E_a}{4.575} 	imes \frac{30}{100899}$$E_a = \frac{1.3164 	imes 4.575 	imes 100899}{30} \approx 20260 \ cal \ mol^{-1}$.

For the reaction $C_2H_5I + OH^- \rightarrow C_2H_5OH + I^-$,the rate constants at $30^\circ C$ and $60^\circ C$ are $0.325$ and $6.735 \ L \ mol^{-1} \ s^{-1}$ respectively. The value of activation energy $(E_a)$ is .......... calories.

Similar Questions

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:

The rate of a reaction doubles when its temperature changes from $300 \ K$ to $310 \ K$. The activation energy of such a reaction will be ........ $kJ/mol$.

The rate constant is given by the equation $k = p Z e^{-E_a/RT}$. Which factor should register a decrease for the reaction to proceed more rapidly?

At what temperature does the rate become double that at $300 \, K$ (in $, K$)? Given $\ln k = 10 - \frac{69 \, kJ}{RT}$.

In the Arrhenius equation,the pre-exponential factor represents ...

Vedclass Test Series

Exam Paper Generator

Online Exam Module