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Order of reaction numerical part 1

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

t = 0.399 x 10 3= 399 min. Ans

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