1st order and 2nd order reactions

(a) Rate expression of the 1st order reaction .

Let us suppose that

                                       A ---------> Product
Initial concentration  ‘a’                  '0'

Concentration after time ‘t’ (a-x)

rate, r= dx αcĀ/dt

or, dx/dt=k(a-x) where, k = rate constant dx
or , dx/(a-x)=kdt
On integration. dx )I kdt (ax) g 
or,  -ln(a-x)=kt+Ic ...(1)

Where Ic = Integration constant . Initially, when t = 0 then x = 0. Thus by putting these values in the  above equation, we can calculate the value of Ic
-ln(a-0)= k. O+Ic
-Ina=lc ....(ii)
Now, putting the value of I c in equation (i) we can get
-ln(a -.x)= kt +(-lnd)
or, -In(a-x)=kt-Ina
Or, - kt==ln(a-x)---Ina
By changing the sign on both sides-
kt =ln a-ln(a-x)

or,  kt: In a/(a-x) [ln a/b = ln a-ln b]

 Q. 53. Find out the rate expressions and the units of their constants of  the following-(a) Ist order and (1)) 2nd order reactions.
(ii) What is half life period ? Show that half iife period of
(a) 1st order reaction is independent of the initial concentration of the reactant ?

(b) 2nd order reaction is inversely proportional to the initial concentration of reactant

Ans. (i) (a) Rate expression of the 1st order reaction .

Let us suppose that

                                       A ---------> Product
Initial concentration  ‘a’                  '0'

Concentration after time ‘t’ (a-x)

rate, r= dx αcĀ/dt

or, dx/dt=k(a-x) where, k = rate constant dx
or , dx/(a-x)=kdt
On integration. dx )I kdt (ax) g 
or,  -ln(a-x)=kt+Ic ...(1)

Where Ic = Integration constant . Initially, when t = 0 then x = 0. Thus by putting these values in the  above equation, we can calculate the value of Ic
-ln(a-0)= k. O+Ic
-Ina=lc ....(ii)
Now, putting the value of I c in equation (i) we can get
-ln(a -.x)= kt +(-lnd)
or, -In(a-x)=kt-Ina
Or, - kt==ln(a-x)---Ina
By changing the sign on both sides-
kt =ln a-ln(a-x)

or,  kt: In a/(a-x) [ln a/b = ln a-ln b]

or, k =-In a/t(A- x)
By convertmg In term into log term-
k =2.303 log a/ t (a-x)

This is the Ist order rate expression. Unit for the rate constant of the Ist order reaction :

1st order rate constant, k=2 303 .log a/(a-x) t

=2. 303x log , a time Unit of the concentration /Unit of the concentration × time

=2.303x time–1 x log moles/ lltre /moles/ lltre
=2.303x time^-1

Thus, the unit of the rate constant of Ist order reaction is a reciprocal oil tune 1 e. If‘ t’ is eXpressed in second, then the unit of‘ k‘ will be set”‘, similarly, 1f the ‘t' IS expressed 1n ‘minute’, then the unit of‘ P will be ‘min^-1

(b) Rate expression of the 2nd order reaction.

2nd order reaction can be expressed by two ways :---

Case-I In which, reactants are same, e. g; . 2A '------------,----> Product
Case-II Reaction 1n which, both the reactants are different, e. g A + B .---..> Product

Case-I: Reaction in which both the reactants are same .

                                    2A -‘----.-----' ---->Product

Initial concentration: ‘a’                        O
Concentration after time ‘t’ (a-x)        'X'

dx/dt =k (A-x)^2
where, k: rate constant

it,  dx/dt=k(a-x) ^2

on integration-

.(Where, If = Integration constant) w_hen r== 0 then,x x50. Thus by putting these values in the

Initially ’ hove equation, we can calculate the value of l .._. a . a» (010) -kxo 4; 1 . '19r. ”$23246 . .‘ ~~-.(ii) .

Now, putting the value of [U in equation (i), we can get--‘ '

3k, “F“ ‘ or, . ~ “kt=-‘.

By changing the sign on both sides-

ktl ___l_ . a---(a x) ~a+x‘ 'l_ x ' (a x) a =-a(a x) =a(q--x) * a(a--x) Ic="

t.a(a-x)

Case-J11 Reaction 1n which both the reactants are different: A + MB -> Product

Initial concentration : ‘a’ ‘ .‘b ‘ 0 Concentration after time ‘t’ (a --x) (b --x) "2x ' dr (it

.rate= jams] or,' Tit/m3]

where k = rate constant.

or, E=k(a-,-x)_(b-x) or, .'cbc=dt.k(a-~x)(b-x) dx

‘°” (a-x)(b-x_)-!kdt

00integrzition-~

air ' z: I I(€I~-x)(b-~x) ”d! 0": I{(M~-L~)E{a~}m=k1dt

 .(Where, If = Integration constant) w_hen r== 0 then,x x50. Thus by putting these values in the

Initially ’ hove equation, we can calculate the value of l .._. a . a» (010) -kxo 4; 1 . '19r. ”$23246 . .‘ ~~-.(ii) .

Now, putting the value of [U in equation (i), we can get--‘ '

3k, “F“ ‘ or, . ~ “kt=-‘.

By changing the sign on both sides-

ktl ___l_ . a---(a x) ~a+x‘ 'l_ x ' (a x) a =-a(a x) =a(q--x) * a(a--x) Ic="

t.a(a-x)

Case-J11 Reaction 1n which both the reactants are different: A + MB -> Product

Initial concentration : ‘a’ ‘ .‘b ‘ 0 Concentration after time ‘t’ (a --x) (b --x) "2x ' dr (it

.rate= jams] or,' Tit/m3]

where k = rate constant.

or, E=k(a-,-x)_(b-x) or, .'cbc=dt.k(a-~x)(b-x) dx

‘°” (a-x)(b-x_)-!kdt

00integrzition-~

air ' z: I I(€I~-x)(b-~x) ”d! 0": I{(M~-L~)E{a~}m=k1dt

{ l .. l } l :{b-x-m.” ‘ (cw-x) (b-x) (b-a) (a-x)(b~x)}23:‘;$

.. (ht-R) . I l (a--x)(b-x) (Ix-h) I. (a-x)(b~x)

‘ 1 dx dr a, ----{ w

(bwa) (a-x) (b-x) or, (bio) {ln(ax)+!n(b«-x)}=kt+l m0) or, I In (b ~ x) = kr + 1‘. where [c ==.Integration constant.

(Ir-a) (av-x)

[wlm-In [”31] :1

Initially, when t =:0 then, x = 0. Thus by putting these values in the above equation, we can calculate the value of IL.

I {__ In (b-0)}=k‘0+ [c or, ___l..__.[n?_ =04.)

.(b-a) (a-O) (b---a)_ a ' ml? mom)

°” " .=(b-a) a

Now, putting the value of IL. in equation (i), we can get-.

' 1n(b"“’)=kr+‘ L 1113

(b-a) (a-x) -(b-a) a

I In“) x)__ 1 Int): ~kt 1 {ln(b~x)-ln-[Z-}=kt

“(17(2) (a2:) (Ir-a) a or _(b-a) (a--x) a l a(b-x) ' ~ I k .. m n __ mn‘ ~or, (b-a){ "17(a n}: I [ [n ,.+lu --l,; 1 01‘, k" I Ina“) 3) . ‘ _ . ~

t(b a)l nb(a--~x) . “

By converting ln term into log term...

.(Where, If = Integration constant) w_hen r== 0 then,x x50. Thus by putting these values in the

Initially ’ hove equation, we can calculate the value of l .._. a . a» (010) -kxo 4; 1 . '19r. ”$23246 . .‘ ~~-.(ii) .

Now, putting the value of [U in equation (i), we can get--‘ '

3k, “F“ ‘ or, . ~ “kt=-‘.

By changing the sign on both sides-

ktl ___l_ . a---(a x) ~a+x‘ 'l_ x ' (a x) a =-a(a x) =a(q--x) * a(a--x) Ic="

t.a(a-x)

Case-J11 Reaction 1n which both the reactants are different: A + MB -> Product

Initial concentration : ‘a’ ‘ .‘b ‘ 0 Concentration after time ‘t’ (a --x) (b --x) "2x ' dr (it

.rate= jams] or,' Tit/m3]

where k = rate constant.

or, E=k(a-,-x)_(b-x) or, .'cbc=dt.k(a-~x)(b-x) dx

‘°” (a-x)(b-x_)-!kdt

00integrzition-~

air ' z: I I(€I~-x)(b-~x) ”d! 0": I{(M~-L~)E{a~}m=k1dt

{ l .. l } l :{b-x-m.” ‘ (cw-x) (b-x) (b-a) (a-x)(b~x)}23:‘;$

.. (ht-R) . I l (a--x)(b-x) (Ix-h) I. (a-x)(b~x)

‘ 1 dx dr a, ----{ w

(bwa) (a-x) (b-x) or, (bio) {ln(ax)+!n(b«-x)}=kt+l m0) or, I In (b ~ x) = kr + 1‘. where [c ==.Integration constant.

(Ir-a) (av-x)

[wlm-In [”31] :1

Initially, when t =:0 then, x = 0. Thus by putting these values in the above equation, we can calculate the value of IL.

I {__ In (b-0)}=k‘0+ [c or, ___l..__.[n?_ =04.)

.(b-a) (a-O) (b---a)_ a ' ml? mom)

°” " .=(b-a) a

Now, putting the value of IL. in equation (i), we can get-.

' 1n(b"“’)=kr+‘ L 1113

(b-a) (a-x) -(b-a) a

I In“) x)__ 1 Int): ~kt 1 {ln(b~x)-ln-[Z-}=kt

“(17(2) (a2:) (Ir-a) a or _(b-a) (a--x) a l a(b-x) ' ~ I k .. m n __ mn‘ ~or, (b-a){ "17(a n}: I [ [n ,.+lu --l,; 1 01‘, k" I Ina“) 3) . ‘ _ . ~

t(b a)l nb(a--~x) . “

By converting ln term into log term...

L“

Unit for the rate constant of the 2nd order reaction :-

(a

2nd order rate constant, K ’1‘ x t arm-30 § l moles/litre) l l

'3 ““1"“33‘7“”. M time (moles / litre) (moles / litre) time moles I [m e

* time" nmles”'1itre" " 'moles" litre" time" 1 Thus, if time is expressed in seconds then the unit of ‘1: ’ will be moles" . litre"l sec" and similarly if the time is expressed into minute, then the unit of ‘1” will be moles" litre“ minute“. Ans. (ii) Half life Period :

-it is denoted by ‘ (”3 ’

--The time in which the concentration of the reactant is reduced to half of its initial concentration is called half life period. ‘

. . 0.693 ’ --. The Half life period ( t”) of the lst order reaction IS k and 2nd

l

older reaction is Emspectively. Where k = rate constant, and ‘a’ = initial

concentration of the reactant. ' (a) The half life period of the lst order reaction is independent of the

initial concentration ot‘the reactant :

' 2.303 a

Since, for the lst order reaction, rate constant, k = .log . I (a x) ‘_ 2.303 loo 0 '_ . k c"(a-av) ' ' , (I)

Let us assume that when (=1’ '3 then the initial concentration of the

reactant ‘a’ becomes a ,12 i.e. x=u/2 Putting these values into the above agitation-

.(Where, If = Integration constant) w_hen r== 0 then,x x50. Thus by putting these values in the

Initially ’ hove equation, we can calculate the value of l .._. a . a» (010) -kxo 4; 1 . '19r. ”$23246 . .‘ ~~-.(ii) .

Now, putting the value of [U in equation (i), we can get--‘ '

3k, “F“ ‘ or, . ~ “kt=-‘.

By changing the sign on both sides-

ktl ___l_ . a---(a x) ~a+x‘ 'l_ x ' (a x) a =-a(a x) =a(q--x) * a(a--x) Ic="

t.a(a-x)

Case-J11 Reaction 1n which both the reactants are different: A + MB -> Product

Initial concentration : ‘a’ ‘ .‘b ‘ 0 Concentration after time ‘t’ (a --x) (b --x) "2x ' dr (it

.rate= jams] or,' Tit/m3]

where k = rate constant.

or, E=k(a-,-x)_(b-x) or, .'cbc=dt.k(a-~x)(b-x) dx

‘°” (a-x)(b-x_)-!kdt

00integrzition-~

air ' z: I I(€I~-x)(b-~x) ”d! 0": I{(M~-L~)E{a~}m=k1dt

{ l .. l } l :{b-x-m.” ‘ (cw-x) (b-x) (b-a) (a-x)(b~x)}23:‘;$

.. (ht-R) . I l (a--x)(b-x) (Ix-h) I. (a-x)(b~x)

‘ 1 dx dr a, ----{ w

(bwa) (a-x) (b-x) or, (bio) {ln(ax)+!n(b«-x)}=kt+l m0) or, I In (b ~ x) = kr + 1‘. where [c ==.Integration constant.

(Ir-a) (av-x)

[wlm-In [”31] :1

Initially, when t =:0 then, x = 0. Thus by putting these values in the above equation, we can calculate the value of IL.

I {__ In (b-0)}=k‘0+ [c or, ___l..__.[n?_ =04.)

.(b-a) (a-O) (b---a)_ a ' ml? mom)

°” " .=(b-a) a

Now, putting the value of IL. in equation (i), we can get-.

' 1n(b"“’)=kr+‘ L 1113

(b-a) (a-x) -(b-a) a

I In“) x)__ 1 Int): ~kt 1 {ln(b~x)-ln-[Z-}=kt

“(17(2) (a2:) (Ir-a) a or _(b-a) (a--x) a l a(b-x) ' ~ I k .. m n __ mn‘ ~or, (b-a){ "17(a n}: I [ [n ,.+lu --l,; 1 01‘, k" I Ina“) 3) . ‘ _ . ~

t(b a)l nb(a--~x) . “

By converting ln term into log term...

L“

Unit for the rate constant of the 2nd order reaction :-

(a

2nd order rate constant, K ’1‘ x t arm-30 § l moles/litre) l l

'3 ““1"“33‘7“”. M time (moles / litre) (moles / litre) time moles I [m e

* time" nmles”'1itre" " 'moles" litre" time" 1 Thus, if time is expressed in seconds then the unit of ‘1: ’ will be moles" . litre"l sec" and similarly if the time is expressed into minute, then the unit of ‘1” will be moles" litre“ minute“. Ans. (ii) Half life Period :

-it is denoted by ‘ (”3 ’

--The time in which the concentration of the reactant is reduced to half of its initial concentration is called half life period. ‘

. . 0.693 ’ --. The Half life period ( t”) of the lst order reaction IS k and 2nd

l

older reaction is Emspectively. Where k = rate constant, and ‘a’ = initial

concentration of the reactant. ' (a) The half life period of the lst order reaction is independent of the

initial concentration ot‘the reactant :

' 2.303 a

Since, for the lst order reaction, rate constant, k = .log . I (a x) ‘_ 2.303 loo 0 '_ . k c"(a-av) ' ' , (I)

Let us assume that when (=1’ '3 then the initial concentration of the

reactant ‘a’ becomes a ,12 i.e. x=u/2 Putting these values into the above agitation-

2.303 log a(b-x)

“1"“). g b(a-x) Unit for the rate constant of the 2nd order reaction t-

-‘

‘t t t K -‘ x . 2nd order ra e cons an , t . “(a_x) l ' (moles/ litre) __ 1 ___..__.l.._._..

----.

= time-(moles/ litre)(moles/ litre) time "10133 / 1"" e

= time"1 moles”l Iitre"‘ = 'moles’I litre"I time” 1 Thus, if time is expressed in seconds then the unit of ‘k ’ will be moles~ , litre‘l sec‘l and similarly if the time is expressed into minute, then theiunit of ‘k’ Will be moles“1 litre~1 minute”l . Ans. (ii) Half life Period : -It is denoted by ‘ [11" 3 ’

-The time in which the concentration of the reactant is reduced to half of its initial concentration is called half life period. ‘

. 0.693 ’ --The Half life period (11"?) of the lst order reaction is k and 2nd

; . 1 " . . . . order reaction 18 Erespectwely. Where k = rate constant'and ‘a’ = initial

concentration .of the reactant.

(a) The half life period of the [st order reaction is independent of the initial concentration of the reactant :---' ' ‘

Since, for the Ist order reaction, rate constant, 1:: 23% log a . ' t (a ~ X) 2303 ‘ a I: log k (a -.t) ‘ (l)

1st order and 2nd order reactions
1st order and 2nd order reactions
1st order and 2nd order reactions
1st order and 2nd order reactions