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Order of reaction numerical part 2

Order of reaction-

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Order of reaction-

Q 1) Starting with one mole of a compound ,it is found 3/4 of the reaction is completed in one hour . Calculate the rate constant if the reaction is of (i) first order (ii) second order

Solution )

i) For first order reaction-

t  = 1 hour  , a =1 mole, x = 3/4 mole

K= 2.303 / t [log (a/a-x)]

K = 2.303 / 1 [log  1 /( 1 – 3/4)]

K = 2.303  [log 1 /( 1/4)]

K = 2.303 (log 4)

log 4 = 0.6020

K = 2.303 x 0.6020

K =  1.386 hour-1 Ans

ii) For  second order reaction –

a=1 mole , x = 3/4 mole , t = 1 hour

K =1/t [x/a(a-x)]

K =1/1 [(3/4)/1(1 – 3/4)]

K = 1/ 1 [ (3/4)/( 1/4)

K = 3.0 litre mole-1 hour-1  Ans.

Q 2) A second order reaction in which both the reactants have same concentration , is 20 % completed in 500 sec. How much time will it take for 60 % completion ?

Solution )

a=100 , x = 20,  t = 500 sec.

K =1/t [x/a(a-x)]

K = 1/ 500  [20/100 (100 – 20)

K = 1/ 500 (20/ 100 x 80)

K=20 / 500 x 100 x 80

K = 5 x 10-6 litre mole-1 sec-1

a = 100 , x = 60

t =1/K [x/a(a-x)]

t  = 1/ 5 x 10 -6  [60/100 (100 – 60)

t = 1/ 5 x 10 -6 (60/ 100 x 40)

t = 60 / 5 x 10 -6 x 100 x 40

t = 3000  sec. Ans.

Q 3) The decomposition of N2O5 in CCl4 solution has been found to be first order with respect to N2O5 with rate constant K = 6.2 x 10 -4 sec-1

N2O5 ——-> 2 NO2 + 1/2 O2

Calculate the rate of reaction when ,

(i) [N2O5] = 2.5 mole / litre

(ii) [N2O5] = 0.50  mole / litre

(iii) What concentration of N2O5 would give a rate of  4.2 x 10 -3 mole litre-1sec-1.

Solution-

(i) K = 6.2 x 10 -4 sec-1 ,  

[N2O5] = 2.5 mole / litre 

Rate  = K [N2O5] = 6.2 x 10 -4  x 2.5

Rate =1.55 x 10-3  mole litre-1sec-1 Ans

(ii) 

K = 6.2 x 10 -4 sec-1 ,  

[N2O5] = 0.50  mole / litre 

Rate  = K [N2O5] = 6.2 x 10 -4  x 0.50

Rate = 3.1 x 10-4  mole litre-1sec-1 Ans

(iii) Rate = 4.2 x 10 -3 mole litre-1sec-1

K = 6.2 x 10 -4 sec-1 ,  

[N2O5] = Rate / K

              = 4.2 x 10 -3/ 6.2 x 10 -4

Rate = 6.77 mole / litre Ans.

Q 4- find the order of reaction for the rate expression , Rate = K[A] [B] 2/3 . Also suggest the units of rate and rate constant for this expression.

Solution –

Rate = K[A] [B] 2/3

order of reaction = 1 + 2/3

order of reaction = 1.67 Ans.

Unit of rate –

Rate = dx/dt= (mole/litre)/ time

unit of rate = mole litre-1 time -1 Ans.

Unit of rate constant –

Rate = K [A] [B]2/3

K = (mole litre-1 time -1)/ (mole litre-1)(mole litre-1)2/3

unit of K = mole-2/3 litre2/3time-1 Ans.

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