The rate of a first order reaction doubles wh | vedclass

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

Question

(B) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 \ R} [\frac{T_2 - T_1}{T_1 T_2}]$Given: $\frac{K_2}{K_1} = 2$,$T_1 = 300 \

Vedclass · Accepted Answer

$53$

Vedclass · Answer

(B) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 \ R} [\frac{T_2 - T_1}{T_1 T_2}]$Given: $\frac{K_2}{K_1} = 2$,$T_1 = 300 \ K$,$T_2 = 310 \ K$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$\log 2 = 0.3$Substituting the values:$0.3 = \frac{E_a}{2.303 	imes 8.3} [\frac{310 - 300}{300 	imes 310}]$$0.3 = \frac{E_a}{19.1149} [\frac{10}{93000}]$$0.3 = \frac{E_a}{19.1149} 	imes 0.00010753$$E_a = \frac{0.3 	imes 19.1149}{0.00010753} \approx 53341 \ J \ mol^{-1} = 53.34 \ kJ \ mol^{-1}$The closest value is $53.33 \ kJ \ mol^{-1}$.

The rate of a first order reaction doubles when the temperature changes from $300 \ K$ to $310 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is
$R=8.3 \ J \ K^{-1} \ mol^{-1}, \log 2=0.3$ (in $.33$)

Similar Questions

In the Arrhenius equation,the rate of reaction is given by $k = A{e^{ - {E_a}/RT}}$. What does $E_a$ represent?

For the reaction,$A \rightleftharpoons B$,$E_a = 50 \ kJ \ mol^{-1}$ and $\Delta H = -20 \ kJ \ mol^{-1}$. When a catalyst is added,$E_a$ decreases by $10 \ kJ \ mol^{-1}$. What is the $E_a$ for the backward reaction in the presence of the catalyst?

Give an example of a collision in the proper direction that results in the formation of a product.

In the Arrhenius equation,$k = A e^{-E_a/RT}$,the Arrhenius constant $A$ will be equal to the rate constant when

The rate of gas phase chemical reactions generally increases rapidly with a rise in temperature. This is mainly because

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The rate of a first order reaction doubles when the temperature changes from $300 \ K$ to $310 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is$R=8.3 \ J \ K^{-1} \ mol^{-1}, \log 2=0.3$ (in $.33$)