Question

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 1

Answer : The value of rate constant at temperature 119 K is 2.46 E-29

Explanation :

As we are the rate law expression as:

$Rate=1.4\times 10^{-2}[NO_2]^2$  ..........(1)

The general rate law expression will be:

$Rate=k[NO_2]^2$       ............(2)

By comparing equation 1 and 2 we get:

$k=1.4\times 10^{-2}$

Now we have to calculate the rate constant at temperature 119 K.

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $T_1$  = $1.4\times 10^{-2}$

$K_2$ = rate constant at $T_2$ = ?

$Ea$ = activation energy for the reaction = 80.0 kJ/mole = 80000 J/mole

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 500 K

$T_2$ = final temperature = 119 K

Now put all the given values in this formula, we get:

$\log (\frac{K_2}{1.4\times 10^{-2}})=\frac{80000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{500}-\frac{1}{119}]$

$K_2=2.46\times 10^{-29}=2.46E-29$

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