Question

Peroxyacetyl Nitrate (PAN) is formed by addition of NO2 to the corresponding peroxyacetyl radical. PAN is considered to by a sink for NO2 in the afternoon during a smog incident

                CH3C(=O)O2 + NO2                               CH3C(=O)O2NO

k' = 1.86 x 1016 exp(-13500/T) (k' is the rate const. for the decomposition of PAN)

a. Calculate the half-life of PAN as it releases NO2 at 5 and 35°C.

b. Based on these calculated half-lives, what would favor PAN as a sink for NO2?

c. If conditions did favor release of NO2 from PAN in the afternoon, why might NO2 concentrations still decrease? Are there other NO2 sinks?

Answer 1

Since NO2 is present in excess so this reaction become first order reaction.

Rate constant k' is given as:

k' = 1.86 × 10^16 exp(-13500/T)

T½= 0.693 / k'

Taking log on both sides,

Ln T½ = ln 0.693 - ln k'

Ln k'= ln 1.86 + 16 ln 10 -13500/T

Ln k' = 2.303 log 1.86 + 2.303 × 16 log 10 - 13500/T

Log 1.86=0.2695

Log 10= 1

Ln k' = 2.303 × 0.2695 + 2.303×16 - 13500/T

Ln k' = 0.6206 + 36.84 - 13500/T

1. When T=5°C =278 K, ln k' = (0.6206(278) + 36.84(278)-13500)/278=-11.100
2. When T= 35°C=308K, ln k' =( 0.6206(308)+36.84(308)-13500)/308= -6.370

In first case, ln T½=2.303 log 0.693+11.100

Ln T½= -0.3667+11.100= 10.7332

T½= antilog 10.7332

= 5.4× 10^10 sec

In second case, ln T½ = -0.3667+6.370=6.0030

T½ = antilog 6.0030

=1.006×10^6sec

B). First one will favour PAN as sink for NO2 because it has longer half life as compared to second one

C). The concentration of NO2 still decreases because of movement of winds and also NO2 lead to production of ozone.