The rate constant is doubled when temperature | vedclass

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

Question

(B) The Arrhenius equation is given by: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$Given: $\frac{K_2}{K_1} =

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(B) The Arrhenius equation is given by: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$Given: $\frac{K_2}{K_1} = 2$,$T_1 = 300 \, K$,$T_2 = 310 \, K$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.Substituting the values: $\log 2 = \frac{E_a}{2.303 	imes 8.314} \left[ \frac{310 - 300}{300 	imes 310} ight]$$0.3010 = \frac{E_a}{19.147} 	imes \frac{10}{93000}$$E_a = \frac{0.3010 	imes 19.147 	imes 93000}{10} \approx 53598 \, J \, mol^{-1} \approx 53.6 \, kJ \, mol^{-1}$.Rounding to the nearest integer,the activation energy is $54 \, kJ \, mol^{-1}$.

The rate constant is doubled when temperature increases from $27\,^oC$ to $37\,^oC.$ Activation energy in $kJ$ is

Similar Questions

The following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

Which of the following plots is in accordance with the Arrhenius equation?

The rate constant of a reaction at $500 \ K$ and $700 \ K$ are $0.02 \ s^{-1}$ and $0.2 \ s^{-1}$ respectively. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R=8.3 \ J \ K^{-1} \ mol^{-1})$

In collision theory of chemical reaction,$Z_{AB}$ represents

Fill in the blanks:
$1.$ Common collision means ......... phenomenon in species.
$2.$ In collision theory,activation energy and proper orientation of the molecules together determine the ......... for an effective collision.

Vedclass Test Series

Exam Paper Generator

Online Exam Module