## 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}

^{-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}

^{ -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

^{ 3}= 399 min. Ans