Question

[I-1] Consider the following data for the gas-phase decomposition of NO₂:  

2NO(g) ® 2NO(g) + O₂(g)

Temperature (K)	Initial [NO2] (mM)	Initial Rate of NO2 Decomposition (mM/hr)
600	1.00	1.94
600	2.00	7.92
700	2.00	187

At 650 K, how many hours will it take for the NO₂ concentration to drop from 5.00 mM to 1.00 mM

Answer 1

When the concentration of No2 is doubled, the rate of the reaction increases four fold. So, this is a second order reaction.

rate = k[NO2]^2

k = 1.94 /[1.00] ^2 = 1.94 mM^-1h^-1

Rate constant at T=700K k = 187/[2.00]^2= 46.75 mM^-1h^-1

The rate constnat of the reaction at T=650K can be calculated by using Arrhenius equation.

log k1/k2 =Ea/2.303R [T2-T1/T1T2]

log (1.94/46.75) = Ea/ 2.303 *8.314 [700-600/600*700]

Ea =-11.103 *10^4 J

calculate the rate constant at T=650

log (k2/1.94) = -11.103 *10^4/2.303 *8.314 [650-600/650*600]

k2 = 0.35

For a second order reaction the rate law is as follows.

1/[NO2]t -1/[NO2]o = kt

1/1mM-1/5mM = 0.35 mM^-1 * hr^-1 *t

t = 2.29 hrs

Time required to drop the concentration from 5mM to 1 mM is 2.29 hr