## 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}

^{-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 N_{2}O_{5} in CCl_{4} solution has been found to be first order with respect to N_{2}O_{5} with rate constant K = 6.2 x 10 ^{-4} sec^{-1}

### N_{2}O_{5} ——-> 2 NO_{2} + 1/2 O_{2}

### Calculate the rate of reaction when ,

### (i) [N_{2}O_{5}] = 2.5 mole / litre

### (ii) [N_{2}O_{5}] = 0.50 mole / litre

### (iii) What concentration of N_{2}O_{5} would give a rate of 4.2 x 10 ^{-3} mole litre^{-1}sec^{-1}.

### Solution-

*(i) K = 6.2 x 10 ^{-4} sec^{-1} , *

*[N _{2}O_{5}] = 2.5 mole / litre *

*Rate = K [N _{2}O_{5}] = 6.2 x 10 ^{-4} x 2.5 *

*Rate =1.55 x 10*^{-3} mole litre^{-1}sec^{-1} Ans

^{-3}mole litre

^{-1}sec

^{-1}Ans

*(ii) *

*K = 6.2 x 10 ^{-4} sec^{-1} , *

*[N _{2}O_{5}] = 0.50 mole / litre *

*Rate = K [N _{2}O_{5}] = 6.2 x 10 ^{-4} x 0.50*

*Rate = 3.1 x 10*^{-4} mole litre^{-1}sec^{-1} Ans

^{-4}mole litre

^{-1}sec

^{-1}Ans

*(iii) Rate = 4.2 x 10 ^{-3} mole litre^{-1}sec^{-1}*

*K = 6.2 x 10 ^{-4} sec^{-1} , *

*[N _{2}O_{5}] = 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}