NUMERICALS ON 2ND ORDER REACTION

TOPIC: - NUMERICALS ON 2ND ORDER REACTION











NUMERICALS


Example 1: Butadiene dimerizes to form C8H12. This reaction is 2nd order in butadiene. If the rate constant for the reaction is 0.84 L/mol min, how long will it take for a 0.500 M sample of butadiene to dimerize until the butadiene concentration is 0.200 M?
Solution 1: Use the integrated rate law above
1/[C4H6] - 1/[C4H6]0 = kt
The intial concentration is 0.500 M, and the final concentration is 0.200 M. K is 0.84 L/mol min, so
1/0.200 mol/L - 1/0.500 mol/L = 0.84 L/mol min * t

t = 3.6 minutes



Example 2: A reaction 2A --> P has a second order rate law with k = 3.5E-4 L/mol-s. Calculate the time required for the concentration of A to change from 0.260 mol/L to 0.011 mol/L.

Solution 2: K1 =0.206 mol/l
K2 =0.011mol/l

1/ 0.011 - 1 / 0.260 = 3.5 x 10^-4 t

90.9 - 3.85 = 3.5 x 10^-4 t

t =248714 s = 69 h 5 min 14 s



Example 3: The rate constant for the SECOND ORDER reaction ; 2NO2 --> 2NO + O2 .Rate = 0.54 ? M.s at 300 Celsius. How long in seconds would it take for the concentration of NO2 to decrease from 0.62 M to 0.28 M?
Solution 3: time elapsed until the concentration has dropped from [NO2] to [NO2] is given by
t = ( 1/[NO2] - 1/[NO2] ) / k

Hence the time to decrease from 0.62M to 0.28M is
t = ( 1/.28M - 1/0.62M ) / 0.54M⁻¹s⁻¹
= 3.63S.
Example 4: Decomposition of a gas is a 2nd order reaction. It takes 40 min. for 40% of the gas to decompose. When its initial concentration is 0.04mol/l. Cal. Specific rate?
Solution 4: let the initial conc. Is a
 x=o.4a
 K=1/t * x/a(a-x)
 K=1/40* 0.4/0.6*0.04
 K=0.416mol/l/sec.



Example 5: a 2nd order reaction in which a=b is 20% comp. in 40 sec. How long will it take to comp. 60% ?
Solution 5: we know k=1/t* x/a(a-x)

 x=20* a/100
 x=1/160a…….( I )
now for x=60%
 k=1/t* 0.6a/a(a-0.6a)
 k=1/t* 3/2a…….( II )
from ( I ) ( II )
1/160a=1/t*3/2a
 t=240 sec.


Example 6: A second order reaction where a=b is 25% in 300sec. how long will it take for the reaction for 50 % completion?
Solution 6: a=100, T=300sec, x=25
a-x=75
now K=1/t* x/a(a-x)
k=1/300* 25/100(75)
 k=1/90,000
 t=1/k* x/a(a-x)
 t=1/1/90,000 * 50/5000
 t=900 sec.


Example 7: A second order reaction in which 25% of reaction comp. in 60 sec. How long will it take to comp. 75%.

Solution 7: a=100, t=600sec, a-x= 75
K=1/t * x/a(a-x)
K=1/60* 25/100(75)
K=1/18000
Now t=1/k* x/a(a-x)
T=18000* 75/2500
T=540 sec.


Example 8: 60% of the 2nd order reaction was completed in 60 min. when was it half comp.?
Solution 8: a=100, x=60, a-x=40
K=1/t*x/a(a-x)
K=1/60* 60/100(40)
K=0.00024
Now t=1/k* x/a(a-x)
T1/2= 1/ka
T1/2=0.024min.


REFERENCE
This topic has been referred from the following links :
1. ^ IUPAC Gold Book definition of rate law. See also: According to IUPAC Compendium of Chemical Terminology.

2. ^ Kenneth A. Connors Chemical Kinetics, the study of reaction rates in solution, 1991, VCH Publishers. This book contains all the rate equations in this article and their derivation.

3. ^ For a worked out example see: Determination of the Rotational Barrier for Kinetically Stable Conformational Isomers via NMR



4. 2D TLC An Introductory Organic Chemistry Experiment Gregory T. Rushton, William G.

5. Burns, Judi M. Lavin, Yong S. Chong, Perry Pellechia, and Ken D. Shimizu J. Chem. Educ. 2007, 84, 1499. Abstract

6. ^ José A. Manso et al."A Kinetic Approach to the Alkylating Potential of Carcinogenic Lactones" Chem. Res. Toxicol. 2005, 18, (7) 1161-1166

1. ^ Ruthenium(VI)-Catalyzed Oxidation of Alcohols by Hexacyanoferrate(III): An Example of Mixed Order Mucientes, Antonio E,; de la Peña, María A. J. Chem. Educ. 2006 83 1643. Abstract
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