Order of reaction
source: Socratic
Order of reaction numerical –
Q 1) The first order reaction is 15% complete in 20 minutes. How long it will take to be 60 % complete ?
Solution )
t = 20 min. , a = 100 , x = 15
K= 2.303 / t [log (a/a-x)]
K= 2.303 / 20 [log 100/(100 – 15)]
K= 2.303 / 20 [log( 100/85)]
K= 0.115 [log 100-log 85)]
K = 0.115 (2 – 1.9294)
K = 0.115 x 0.o806 = 0.009269
K = 9.27 x 10 -3
t= 2.303 / K [log (a/a-x)]
a = 100 , x = 60
t= 2.303 / 9.27 x 10-3 [log 100/(100 – 60)]
t= 248 [log( 100/40)]
t= 248 [log 100-log 40)]
t = 248 (2 – 1.6020)
t = 248 x 0.398 = 98.7 minute Ans
Q 2) In a first order reaction it takes 100 second for half the reaction to complete.How long will it take for the 99% change. What will this time be if the reaction is of second order and involves only one reactant?
Solution )
For first order reaction-
t 1/2 =100 sec.
K = 0.693 / t 1/2 = 0.693/100
K = 6.93 x 10 -3 sec -1
t= 2.303 / K [log (a/a-x)]
a = 100 , x = 99
t = 2.303 / 6.93 x 10 -3 [log ( 100/100-99)
t = 2.303 / 6.93 x 10 -3 [log 100]
log 100 = 2
t = 2.303 x 2 / 6.93 x 10 -3
t = 0.665 x 10 3= 665 sec. Ans
For second order reaction –
t 1/2 = 1 / Ka
t 1/2 = 100 , a = 100
K = 1 /(a. t 1/2)= 1/100 x 100
K = 10 -4
a=100 , x = 99
t =1/K [x/a(a-x)]
t = 1/ 10 -4 [99/100 (100 – 99)
t = 10 4 x 99 / 100
t = 9900 sec. Ans.
Q 3) A first order reaction is 20 % complete in 10 minutes.Calculate the time taken for the reaction to 75 % completion ?
Solution )
t = 10 min. , a = 100 , x = 20
K= 2.303 / t [log (a/a-x)]
K= 2.303 / 10 [log 100/(100 – 20)]
K= 2.303 / 10 [log( 100/80)]
K= 0.2303 [log (5 – log 4)]
K = 0.2303 (0.6990 – 0.6020)
K = 0.2303 x 0.o97
K = 0.0223 min-1
t= 2.303 / K [log (a/a-x)]
a = 100 , x = 75
t= 2.303 / 0.0223 [log 100/(100 – 75)]
t= 103.27 [log( 100/25)]
t= 103.27 [log 4]
t = 103.27 x 0.6020
t = 62.16 = 62.2 minute Ans.
Q 4) A first order reaction has a rate constant of 15 x 10 -3 sec -1. How long will 5.0 gm of this reactant to take to reduce 3.0 gm?
Solution )
K = 15 x 10-3 sec-1
a = 5.0 gm , (a – x) = 3 gm
t= 2.303 / K [log (a/a-x)]
t= 2.303 / 15 x 10-3 [log ( 5/3 )]
t= 1535 [log 5 -log 3 ]
t = 1535 ( 0.6990 – 0.4771)
t = 1535 x 0.2219
t = 34.06 second Ans.
Q 5)Thermal decomposition of a compound is of first order .If 50 % of the compound is decomposed in 120 minutes. How long will it take to complete 90 % of the reaction?
Solution)
t 1/2 =120 min.
K = 0.693 / t 1/2 = 0.693/120
K = 5.77 x 10 -3 min -1
t= 2.303 / K [log (a/a-x)]
a = 100 , x = 90
t = 2.303 / 5.77 x 10 -3 [log ( 100/100-90)
t = 2.303 / 5.77 x 10 -3 [log 10]
log 10 = 1
t = 2.303 x 1 / 5.77 x 10 -3