The decomposition of NO2(g) occurs by the following bimolecular elementary reaction. 2NO2(g) → 2NO(g) + O2(g) The rate constant at 273 K is 2.3 x 10-12 L mol-1 s-1, and the activation energy is 111 kJ/mol. How long will it take (in s) for the concentration of NO2(g) to decrease from an initial partial pressure of 2.80 atm to 1.90 atm at 479 K? Assume ideal gas behavior.

The time taken for the concentration of nitrogen dioxide to decrease is 54.0 seconds

Explanation:

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

where,

= equilibrium constant at 479 K = ?

= equilibrium constant at 273 K =

= Activation energy = 111 kJ/mol = 111000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

= initial temperature = 273 K

= final temperature = 479 K

Putting values in above equation, we get:

The given chemical equation follows:

As, the above equation is bimolecular elementary reaction, the order od the reaction is 2.

Integrated rate law for second order kinetics is given by:

= initial partial pressure = 2.80 atm

p = partial pressure left after time t = 1.90 atm

k = rate constant =

t = time taken = ?

Putting values in above equation, we get:

Hence, the time taken for the concentration of nitrogen dioxide to decrease is 54.0 seconds

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.