Chem 301 Problem Day #11
Fall 2008
1. The gas-phase reaction 2 NO3 → 2 NO2 + O2 is a second order reaction with respect to NO3.
R = k[NO3]2
a. At a temperature of 293 K and an initial concentration of NO3 equal to 0.0500 mol/L, the
concentration after 60.0 min is equal to 0.0358 mol/L. Find the value of the rate constant.
1/[NO3] – 1/[NO3]o = kt 27.93 – 20 = k (60)
1/(0.0358) – 1/(0.0500) = k(60) 7.93 = k (60)
k = 0.1322 M-1 min-1
= 0.00220 M-1 sec-1
b. Find the concentration of NO2 after 145 min if the initial concentration of NO3 is 0.100
mole/L.
1/[NO3] – 1/(0.100) = (0.1322) (145) [NO2] = [NO3]o – [NO3]
1/[NO3] – 10 = 19.169 [NO2] = 0.100 – 0.03428
1/[NO3] = 29.169 [NO2] = 0.0657
[NO3] = 0.03428
2. The destruction of the ozone layer of the atmosphere might involve the reaction: NO + O3 → NO2
+ O2. This reaction is first order in each reactant and the rate constant is equal to 1.3 x106 L
mol-1 sec-1 at 298 K. If the initial concentration of NO is 1.00 x10-6 mol/L and the initial
concentration of O3 is 5.00 x10-7 mol/L. Find the concentration of ozone after 3.50 sec.
R = k[NO][O3] (2nd order, Type II)
1/([O3]o-[NO]o) ln ([O3][NO]o/([NO][O3]o) = kt
1/(5.0x10-7-1x10-6) ln ([O3]/[NO] * 1x10-6/5x10-7) = 1.3x106 (3.50)
ln (2 [O3]/[NO] ) = -2.275
[O3]/[NO] = 0.051398
[O3] = 0.051398 ( [NO]o – [O3]o + [O3]) = 0.051398(1x10-6 – 5x10-7+[O3])
[O3] = 2.7 x 10-8
[NO] = [O3]/0.051398 = 5.3 x 10-7
3. Nitric oxide (NO) and chlorine (Cl2) efficiently react in the gas phase to form NOCl. This reaction
was studied at 298 K and the following data was found:
[NO] [Cl2] Initial rate (M/sec)
0.020 0.020 7.1x10-5
0.040 0.020 2.8x10-4
0.020 0.040 1.4x10-4
Find the order of the reaction with respect to each reactant and the rate constant.
ln(R1/R2) = α ln ([NO]1/[NO]2) ln (R1/R3)= β ln([Cl2]1/[Cl2]3)
ln(7.1x10-5/2.8x10-4) = α ln (0.020/0.040) ln (7.1x10-5/1.4x10-4)= β ln(0.020/0.040)
-1.372 = α (-.0693) -0.679 = β (-0.693)
α = 2 β = 1
7.1x10-5=k(0.20)2(0.02)
k = 8.88 M-2 sec-1 R=k[NO]2[Cl2]
