CHM 5423 – Atmospheric Chemistry
Problem Set 4
Due date: Tuesday, November 2 nd
1) Table 6.2, page 182 of the Chapter 6 handout list values for the rate constant for several n-alkane + OH reactions at T = 298. K. a) Plot k vs n, where n is the number of carbon atoms in the n-alkane (n = 1 for CH
4
, n = 2 for C
2
H
6
, etc). b) For the region of the plot where there appears to be a linear relationship between n and k, find an expression for k(n), that is, for the rate constant as a function of n. c) Using your expression in b find the root mean square error in your data fitt. The rms error is defined as
Error = { (1/N)
 i=1
N (k n
(exp) – k n
(calc)) 2 } 1/2 (1.1)
This gives you an idea of the quality of the fit to the experimental data. d) For those n-alkanes excluded from the fitting procedure in parts b and c of this problem, briefly explain why you might not expect them to follow the general trends observed for the other n-alkanes.
2) Consider each of the following alkanes. For each alkane, use equation A, page 183 of the Chapter 6 handout and the accompanying information to estimate the value for the alkane + OH rate constant at 298 K. Compare your result to the values given in Table 6.2. Also estimate the percentage of molecules forming each possible initial product for the alkane + OH reaction. a) propane (CH
3
CH
2
CH
3
) b) 2-methylbutane (CH
3
CH(CH
3
)CH
2
CH
3
)
3) Consider the daytime propene (CH
3
CH=CH
2
) reacts with hydroxyl radical (OH) and ozone (O
3
). a) Find the half-life of propene in the troposphere, and the fraction of propene molecules that react with hydroxyl radical and with ozone. Assume T = 298 K, p = 1.00 atm, [OH] = 1.0 x 10 6 molecule/cm 3 , and [O
3
] = 1.0 x
10 12 molecule/cm 3 . b) Why might the above results be misleading in regards to propene emitted at night?
4) Give the major daytime reactions for each of the following molecules. Begin with the molecule and end whenever a stable product is formed. For cases where there are important competing pathways for reaction give all of the important pathways involved. a) n-butane (CH
3
CH
2
CH
2
CH
3
) b) isoprene (CH
2
=C(CH
3
)-CH=CH
2
) - give initial products only c) acetaldehyde (CH
3
CHO) d) ethyl alcohol (CH
3
CH
2
OH)
5) One way in which nitrogen oxides can be transported from urban areas to remote sites is in the form of peroxyacetyl nitrates (PANs). These compounds are thermally unstable, and so will decompose over time to release
NO
2
into the troposphere, by the process
RC(O)OONO
2
+ M

RC(O)OO + NO
2
+ M (5.1) a) Using the data in Table 6.21, page 219 of the Chapter 6 handout, find the half-life for the decomposition of PnBN at T = 300 K, 280 K, 260 K, and 240 K. Note that for atmospheric conditions the reaction is unimolecular
(k has units of s -1 ). b) Briefly discuss the implications of your results in a on the long range transport of PnBN in the troposphere.

Solutions.
6) The Leighton mechanism is the following set of reactions
NO
2
+ h
 
NO + O J
1
O + O
2
+ M

O
3
+ M k
2
NO + O
3

NO
2
+ O
2 k
3
(6.1)
(6.2)
(6.3) a) Assuming the Leighton mechanism is correct, find [NO] and [NO
2
] for the following set of conditions.
[NO x
] = 10.0 ppb J
1
= 1.3 x 10 -3 s -1 .
[O
3
] = 60.0 ppb k
2
= 6.0 x 10 -34 cm 6 /molecule 2 s (low pressure limit applies)
[air] = 2.0 x 10 19 molecule/cm 3 k
3
= 1.8 x 10 -14 cm 3 /molecule s b) Assuming no other reactions involving NO, NO
2
, and O
3
are significant, what is the value for the half life for NO and NO
2
for the above set of conditions? c) Why is the above half life value misleading?

1) a) The data for the n-alkanes are given below n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
The data are plotted below. k x 10 12
(cm 3 /molecule .
s)
0.00618
0.254
1.12
2.44
4.0
5.45
7.0
8.7
10.0
11.2
12.9
13.9
16.
18.
21.
23.
25
20
15
10
5
0
0 4 8 n k calc
x 10 12
(cm 3 /molecule .
s)
|
 k|
(cm 3 /molecule .
s)
1.96
3.60
5.24
6.88
8.53
10.17
11.81
13.45
15.09
16.74
18.38
20.02
21.66
0.48
0.40
0.21
0.12
0.17
0.17
0.61
0.55
1.19
0.74
0.38
0.98
1.34
12 16

b) Based on the above plot, the data appear linear for the range n = 4 – 16. If the data in this range are fit, the plot shown below is obtained.
25
20
15
10
5
0
0 4 8 n
12 16
The best fitting line is k = (1.642 n – 4.61) x 10 -12 cm 3 /molecule .
s c) The differences between the experimental and calculated values of k
|
 k| = | k – k calc
| are given in the data table in part a. Using the definition of rms error
Error = { (1/N)
 i=1
N (k n
(exp) – k n
(calc)) 2 } 1/2 the rms error is 0.68 x 10 -12 cm 3 /molecule .
s. It is interesting to note that about 40% of the rms error comes from two data points, n = 12 and n = 16, suggesting that those points might deserve to be remeasured. d) We can think of the n-alkanes included in the fit as having the formula
CH
3
CH
2
(CH
2
) p
CH
2
CH
3
, where p = 0, 1, 2, ...
The reactivity of the terminal CH
3
CH
2
- is expected to be relatively unaffected by the number of –(CH
2
)- groups separating the ends of the molecules, while each of the -CH
2
- groups in the middle of the molecules should be approximately the same reactivity. However, for smaller n-alkanes (methane, ethane, and n-propane) the small size of the molecules means that the reactivity of the hydrogens have not yet reached a constant value. This is in fact the reasoning behind equations like the modified equation on page 183 of the Chapter 6 handout.
2) a) propane (CH
3
CH
2
CH
3
) k total
= 2 (0.136) (1.23) + 1 (0.934) (1.00) (1.00) = 1.27 x 10 -12 cm 3 /molecule .
s product distribution CH
3
CH
2
CH
2
= 26 % CH
3
CHCH
3
= 74 %

b) 2-methylbutane (CH
3
CH(CH
3
)CH
2
CH
3
) k total
= 3 (0.136) (1.23) + 1 (0.934) (1.00) (1.23) + 1 (1.94) (1.00) (1.00) (1.23) = 4.04 x 10 -12 cm 3 /molecule .
s
CH
2
CH(CH
3
)CH
2
CH
3
= 8 % CH
3
CH(CH
3
)CH
2
CH
2
= 4 % product distribution
CH
3
C(CH
3
)CH
2
CH
3
= 59 % CH
3
CH(CH
3
)CHCH
3
= 28 %
3) a) There are two reactions taking place
CH
3
CH=CH
2
+ OH

“products” k = 26.3 x 10 -12 cm 3 /molecule .
s [OH] = 1.0 x 10 6 molecule/cm 3 k

= k [OH] = 26.3 x 10 -6 s -1
CH
3
CH=CH
2
+ O
3

“products” k = 10.1 x 10 -18 cm 3 /molecule .
s [O
3
] = 1.0 x 10 12 molecule/cm 3 k

= k [O
3
] = 10.1 x 10 -6 s -1
Since both processes occur simultaneously (at least during the daytime) then d[propene]/dt = - (k

+ k

) [propene] = (36.4 x 10 -6 s -1 ) [propene]
Using the expression for the half-life for a pseudo first order reaction, we get t
1/2
= ln2/(36.4 x 10 -6 s -1 ) = 1.90 x 10 4 s = 5.3 hours
% OH = (26.3/36.4) x 100 % = 72 % % O
3
= (10.1/36.4) x 100 % = 28 % b) [OH] shows a large amount of day/night variability. At night the OH concentration is very small, and only reaction with O
3
will occur. In the daytime both the OH and O
3
reaction will take place. Since the half-life is short compared to a day the time of release will therefore play a role both in the half-life and product distribution, with different results in daytime and at night.
4) The major pathways and products are given below. This will not have a lot of detail as I don’t feel like typing up all of the intermediate structures. Stable products are bolded. a) n-butane (CH
3
CH
2
CH
2
CH
3
)
Major reaction is hydrogen atom abstraction by OH
CH
3
CH
2
CH
2
CH
3
+ OH

CH
3
CH
2
CHCH
3
+ H
2
O (major)

CH
3
CH
2
CH
2
CH
2
+ H
2
O (minor)
If we follow the major initial product, we expect
CH
3
CH
2
CHCH
3
+ O
2
+ M

CH
3
CH
2
CH(O
2
)CH
3
+ M
CH
3
CH
2
CH(O
2
)CH
3
+ NO

CH
3
CH
2
CH(O)CH
3
+ NO
2

At this point there are two possible pathways – fragmentation or reaction with O
2
.
CH
3
CH
2
CH(O)CH
3

CH
3
CHO + CH
3
CH
2
CH
3
CH
2
  
CH3CHO where the reactions are the usual ones for alkyl radicals (+ O
2
, +
NO, + O
2
)
CH
3
CH
2
CH(O)CH
3
+ O
2

CH
3
CH
2
CH(O)CH
3
+ HO
2 b) isoprene (CH
2
=C(CH
3
)-CH=CH
2
) - give initial products only
There are two sets of initial products
OH reaction (addition)
CH
2
OH-C(CH
3
)-CH=CH
2
or CH
2
=C(CH
3
)-CH-CH
2
OH
O
3
reaction (addition)
Formation of ozonide across either C=C, followed by its decomposition, to give
HCHO + CH
HCHO + CH
3
2
-C(O
2
)-CH=CH
=C(CH c) acetaldehyde (CH
3
3
)-CH(O
CHO)
2
2
)
Major reaction is hydrogen abstraction by OH
CH
3
CHO + OH

CH
3
C=O + H
2
O
CH
3
C=O + O
2
+ M

CH
3
C(O
2
)=O + M
CH
3
C(O
2
)=O + NO

CH
3
COO + NO
2
CH
3
COO

CH
3
+ CO
2
The CH
3
(methyl radical) will react as discussed in class to form formaldehyde (HCHO). d) ethyl alcohol (CH
3
CH
2
OH)
Major reaction is hydrogen abstraction by OH. There are three possible products, with the major product as indicated below
CH
3
CH
2
OH + OH

CH
3
CHOH + H
2
O
CH
3
CHOH + O
2
+ M

CH
3
CH(O
2
)OH + M
CH
3
CH(O
2
)OH + NO

CH
3
CH(O)OH + NO
2
CH
3
CH(O)OH + O
2

CH
3
COOH + HO
2

5) a) For PnBN, k = (3.2 x 10 18 s -1 ) exp(-15150/T). Data for various temperatures are tabulated below
T(K) k(s -1 ) t
1/2
300.
280.
3.74 x 10 -4
1.02 x 10 -5
31 min
19 hours
260.
240.
1.58 x 10 -7
1.23 x 10 -9
51 days
17.9 years b) If PnBN can migrate into the mid-troposphere, where temperatures are cooler, it can be transported long distances before decomposing.
6) a) Based on the Leighton mechanism
[NO]/[NO
2
] = J
1
/k
3
[O
3
]
But [air] = pN
A
/RT = (1.00 atm) (6.022 x 10 23 molecule/mol) = 2.46 x 10 22 molecule/L = 2.46 x 10 19 molecule/cm 3
(0.08206 L .
atm/mol .
K) (298. K)
So [O 3 ] = (60.0 x 10 -9 ) (2.46 x 10 19 molecule/cm 3 ) = 1.48 x 10 12 molecule/cm 3
And so [NO] = (1.3 x 10 -3 s -1 ) = 0.049
[NO
2
] (1.8 x 10 -14 cm 3 /molecule .
s) (1.48 x 10 12 molecule/cm 3 )
[NO x
] = 10.0 ppb. Since [NO]/[NO
2
] << 1, then [NO]/[NO
2
]

[NO]/[NO x
]
[NO
2
] = (10.0 – 0.5) = 9.5 ppb and so [NO] = (0.049) (10.0 ppb) = 0.5 ppb b) For the half lives
[NO] t
1/2
= ln(2)/k
3
[O
3
] = ln(2)/[(1.8 x 10 -14 cm 3 /molecule .
s)(1.48 x 10 12 molecule/cm
3
) = 26. s
[NO
2
] t
1/2
= ln(2)/J
1
= ln(2)/1.3 x 10 -3 s -1 ) = 8.9 min c) These half-lives are misleading ince all tht is happening is that NO is being transformed into NO
2
and
NO
2
into NO, with the total NO x
concentration remaining the same. We should really focus on processes removing
NO x
from the troposphere. The most important of these is likely
NO
2
+ OH + M

HNO
3
+ M