Answer : The temperature will be, 392.462 K
Explanation :
According to the Arrhenius equation,
![K=A\times e^{\frac{-Ea}{RT}}](https://tex.z-dn.net/?f=K%3DA%5Ctimes%20e%5E%7B%5Cfrac%7B-Ea%7D%7BRT%7D%7D)
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at
= ![K_1](https://tex.z-dn.net/?f=K_1)
= rate constant at
= ![3K_1](https://tex.z-dn.net/?f=3K_1)
= activation energy for the reaction = 66.41 kJ/mole = 66410 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature = 293 K
= final temperature = ?
Now put all the given values in this formula, we get:
![\log (\frac{3K_1}{K_1})=\frac{66410J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{293K}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B3K_1%7D%7BK_1%7D%29%3D%5Cfrac%7B66410J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B293K%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
![T_2=392.462K](https://tex.z-dn.net/?f=T_2%3D392.462K)
Therefore, the temperature will be, 392.462 K