Answer : The value of rate constant at temperature 119 K is 2.46 E-29
Explanation :
As we are the rate law expression as:
..........(1)
The general rate law expression will be:
............(2)
By comparing equation 1 and 2 we get:

Now we have to calculate the rate constant at temperature 119 K.
According to the Arrhenius equation,

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
= 
= rate constant at
= ?
= activation energy for the reaction = 80.0 kJ/mole = 80000 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature = 500 K
= final temperature = 119 K
Now put all the given values in this formula, we get:
![\log (\frac{K_2}{1.4\times 10^{-2}})=\frac{80000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{500}-\frac{1}{119}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B1.4%5Ctimes%2010%5E%7B-2%7D%7D%29%3D%5Cfrac%7B80000J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B500%7D-%5Cfrac%7B1%7D%7B119%7D%5D)

Therefore, the value of rate constant at temperature 119 K is 2.46 E-29