The reaction 2no2 → 2no + o2 obeys the rate law: rate = 1.4 x 10-2[no2]2 at 500 k . what would be the rate constant at 119 k if the activation energy is 80. kj/mol? this is a second order reaction, giving k the units of m-1s-1 this will not change with the change in temperature. do not include units in your answer. exponential numbers need to be entered like this: 2 e-1 means 2 x 10-1. the rate constant, k, at 119 k equals:

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,

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:

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

the answer is 0.42 is that is an option. this question provides no info