Show that order is 1.

V₀ = 4.61



Ans:-



Since volue of K comes out to be nearly constant. So, it is Ist order.



Psendo I^st order rxn:- such reaction which are not purly of I^st order but under setain condition becomes reactions of I^st order.

eg.

ethyacetate (excess)



Hydrolysis of ethylacetate:-



in this rxn; acetic acid can be tirated against standard Naoh sol.but being acid catalysed, acid present orginally as catalysed, acid present orginally as catalyst also rect with naon sol.



vol. of Naoh sol used at any instant of time t (V_t) Amt. of acid present as catalyst t amt. of CH₃COOH product .......................... (ii)

by (i) & (ii)

Amt. of CH₃COOH Produced at any instant of time } (v_t-v₀ ............................(iii)

But Amt. of CH₃cooH Produced at any instant of time } amt of ch₃cooh that has reacted(x)

i.e x (V_t - V₀) .................................. (iv)

vol of Naon sol. used after infinite time Amt of acid present as catalyst + max. amt of CH₃COOH produced. ................................... (v)

By (i) and (v)

Max. amt of CH₃COOH Produced

But Max. amt. of CH₃COOH produced intial conc. of CH₃COOC₂H₅(a) ................................. (vi)

Hence a

(a-x) - (V_t - V₀)

(a-x) ...................................(vii)

Substituting in I^st order eq.





Question:-



Show that rxn is Ist order.

Ans:- V₀ = 20.24ml V = 43.95







since value of k is nearly constant. so it is ist order



Inversion of cane sugar:-

C₁₂H₂₂O₁₁+H₂O C₆H₁₂O₆ + C₆H₁₂O₆



+ 66.5₀(Dexro) + 52.5₀(dexro) - 92₀(dexro)

Reading of polarimeter at zero time = r₀

Reading of polarimeter at any time = V_t

Reading of Polarimeter at infinite time =

It is evident that reading at zero time will be +ve & would with passage of time & ultimately become -ve.

Angle of rotation at any instant of time i.e (r₀ - r_t) amt of sucrose hydrolysed.(x)

i.e x (r₀ - r_t) ...............................(1)

angle of rotation at infinite time i.e (r₀ - ) Intial

a-x) (r₀ - ) - (r₀ - r_t)

(a-x) (r _t - )





Illustration:- inversion of singal studied.

time (min)      0     7.18   18   

reading(deg) + 24.09     + 21.41    + 17.74    -10.74

= + 24.09

= - 10.74



Ans:- r₀ - = + 24.09 - (- 10.74) = 34.83



K is nearly constant so, rxn is I^st order.



Collusion theory of Rxn. rates:-



Collusion frequency(Z):- No. of collusions that takes place per second per unit volume of rxn. mix.



Barrier to rxn:-



(1)Energy factor:-

Threshold energy:- min k.E. which colliding molecules must have in order that collision b/w then may be effective. threshold but due to frequency oricutation collisions effective.

NO₂ + No₂ N₂O₄





Rotation b/w rate of rxn. & collision frequency:-

Rate = = z X f ..................... (v) F Fraction of molecule whose collisions are effective.

Ea Activation energy

Since rate is directly related to rate constant K

K =

Taking orientation factor in to account

P Static factor.(orientation)



fraction of molecule having effective collesin.:-







reactant molecules so that becomes equal to thresholad value.

Activation energy = threshold energy -avg. K.E. of reactant molecules

Activation energy = threshhold energy - avg ke of reactant molecule A - A + B - B diagram2AB

Arrehevius eqn



taking log on both sides

lnk=ln^A

Letvalue of rate constant at temp t1&t2 are k1 &k2 respectively

lnk1=lnA

lnk2=lnA

(iv)-(iii)

lnk3-lnk1















on solves

E_a=44.13kjmol^-1



Easy type:-



Q.1 for rxn 2NO₂+F₂2N0₂F

exp. rate law is r=kNO₂ Propose mechanism Of RXN

ANS:- RATE = K[NO₂][F₂] Implies that slow step

of rxn involves only one molecule of each reacting species so, mechanism.

NO₂+F₂NO₂F + F

NO₂+F₂NO₂ F

Over all rxn is 2NO₂+F₂2NO₂

Q 2:- Given that temp coeff for sponification of ch₃cooc₂h₅

by NAOH is 1.75. calculate activation energy?

Dumb question:-



Q : what is temp coeff?

Ans:- it is rate of constant at 35c to 25c

given,

2.303 log₁₀

2.303 log₁₀

Ea = 10.207kcal/mol Q.3:- A hydrogenation rxn. is carries out at 500k. if some

rxn. is carried out in presence of catalyst at same rat, temp. required is 400k.

calculate activation energy if catalyst lowers activation energy berrier by

20kj/mol Ans.:- let Ea & Ea' be energy of activation is presence of catalyst

for hydrogenation rxn,; then k= A_e^{-Ea / RT}

inpresence of catalyst, k1=A_e^{-Ea/ (rx500)}

in absence of catalyst, k2= A_e^{-Ea / Rx400)}

given two rates are same i.e r₁ =r ₂ k₁ =k₂





but, E<sub>a'=(Ea-20)kj/mol





Q4:-for A+B C+D,=20kj/mol.

Eaf for forwardrxn is 85 kj/mol . calculate a activation energy of reverse rxn?

Ans:- Energy of activation for farward rxn= energy of activation for backward



rxn +

Eb=85-20=65kj/mol







derive arelation in b/w k1,k2& k3



Ans:- for agiven change





2k1=k2=4k3 Q6:rate law has form , rate=k[A][B]^3/e can rxn be an elementry process?

Ans: Dumb question</sub>

Q:what is criteria for elementry process? Ans: rate law with order to molecularity (necessarily integer) are elementry process but

for given rxn.order of rxn=1+3/2=5/2 since molecularity can never be fraction so, given rxn is not elementry process

Q7:2N0+O₂2NO₂

RATE= K[NO]²[O₂] hOW will rate of rxn change if

vol. of rxn vessel is reduced to 1/4th of its original value?

Ans: Rat=k[no²][0₂

let amole of no and b mole of o2 be taken to start arxn in vessel of v at any

time. if vol of vessel is reduced to1/4 then for same mole N0 +o₂





by eq(i)and (ii)





Q8: if rate depends on conc according to eq.



,what will be the order of rxn where (a) conc. is very high (b) is very low





(a) if c is very high thn 1/c is smallest & then neglible.



= const

so, order of rxn is zero



(b) if c is very low 1+k₂c =k'





so order of rxb is unity

Q9: for 2A + B + C(excess)products.

calculate (1) rate expression

(ii) effect on rate if conc of ais doubled & of

b is tripled & c is doubled of conc of c

(i) rate= k[A],²[B]¹[c]⁰

(ii) let intial conc of a, b,c be a,b, c respectively



r 1= k(a)²(b)(c)⁰-------(1)



now,[A]=2a;[B]=3b [c]=2c

r₂=k(2a)²(3b)(2c)⁰



r₂12(a)²(b)(c)⁰



by eq (1) and (ii)





if intial conc. of [A]& [oh^-] are 0.002m & 0.3m respectively & if it takes for 1y. A to react at 25c, calculate rate constant for rxn.



Ans:- A + On^- Products







Question - 1. A requires activates energy of 20 . When a 20% sol. of A mas kept at 25⁰C for 20 min, 25% decomposition took place. What will be % decomposition in same time in a 30% sol. maintained at 40⁰C. Assume E_q remains constant ?



Ans:_ Given A B

& 20% sol. of A decomposes 25% in 20 min. at 25⁰C

Let Initial amount, a = 20

amount left (a - x) = 20 X = 15

t = 20)

= 0.0144 min^-1

t = 20)

Now, Suppose amount m is left in 30% sol. in 20 min. at 40⁰C.



% discount =

= 67.21 %



If initial conc. of A is A₀, Calculate.

(a) integrated form of rate expresion.

(b) half life period.



Ans:- (a)





at t=0, A=A₀



By eqn(1) and (ii)



if t=t1/2, A=



Q3: for homogeneous gaseous phase rXb.

2A-> 3b+c , intial pressure of reactant was p while pressure at time t was p. find pressure after time 2t. rXn is ist order

Ans: 2a(9)->3B+c(9)



let total pressure at t=2t be b, then p⁰-2x+3x+x=B⁰+2x=B





by eq (ii) and (iv)

Q4: if rxn Aproducts, concentration

of reactant

A are coaco,a²co,

a³co.... after time interval 0, t, 2t, 3t... where a is constant

. show that rxn is (1) order . also calculate relation in k, a& t.

Ans:- for i order rxn k=



k comes out constant so, rxn is ist order also,

rate constant of decomposition is 4.70x10^-3min^-1

calculate rate of intial rate of diffusion to rate of diffusion after 45 hr.

of intial of decomposition

assume composition of gas pressent & gas deffusion remains same.

Ans:- CH₃oCH₉CH₄(9)+H₉(9)

pressure at t = 0          0     0     0

at t= 4.5 hr 0.4-a=A a     a     a



A=0.11 0.4-a=0.11a=0.29 now mol wt of mix

=18.78



Hard type Q1: show that time t_1/2/ts/4 for nth order rxn is function of n alow. tz/a is time required for conc. to become 1/4th of original conc.



ans: Aproduct



at t=0, c=c_o



by eq.(1) and(2)



from eq (3) if t= t1/2, c= c0/2

kt_3/4,c=c0/4







from eq (3) if t= t_3/4, c=co/4





Q2.some PH3(9) is introduced into a flask at 600c containing an insert gas. PH3 proceeds to decomposed into P4 (9) & H2(9) & rxn. gas to completion. Total pressure is given below as function of time. Find order of rxn. & calculate rate constant.

p(mmh_g 262.4 272.9 279.51 276.4

ans:- 4pH₃(9)+ inertgasp₄(9)+ 6H₂(9) + inert gas

at t=0

t=60

t=120

t=



also at t=0, p+p'=276.4----------(2)

p=18.66mm &p'=243.74mm

at t=60, pressure=272=(p-a)+p'+a/4+6a/4

272.9=18.66-A+243.74+7A/4

=A=14mm



at t=120, pressure 272.51=p-A'+p'+A'/4+6A'/4

275.51=18.66-a'+243.74+7A'/4

A'=17.48mm



rxn is ist order

in process it contains bnth to b^{(n+2) & B}(n+4) t0 b<sup>(n-1)



vol of reactant consumed is 25ml & at t=10min, vol used is 32 ml.

calculate rate constant for conversion of Bn+ of B(n+4)

assuming it to be a ist order

Ans:- bn+B(n+4)

a          0

(a-x)      x

Bn+B(n+2)++2e- nfactor

= 2 B(n+4)++5e-B(n-1)+nfactor=5

Let normality be n for reducing agent so, at t=0 ax2= Nx25(by law of equivalence)



at t=t,(a-x)x2+5.x=nx32-------(11) for Bn+

for B(n+4)x from (i)

and (ii)



2.07x10-2min-1



____________Intentanious rate

average rate

order of reaction

Rate of reaction

Rate constant

Law of mass action

Molecularity of rxn.

Machanism of rxn.

Half life

Initial rate method

Preuodo Ist order rxn.

Inversion of regur

Activation energy

Collision frequency

Arrhenins Eq.

Transition state

Q.4: [Concept of sequential reaction:]

The reactant A with ini9tial concentration A goes to B. B instantly starts forming

at a different rate. Find the time at which max conc. Of B occurs and this concentration?

Given k1 and k2

sol :-

Rate laws:

Dumb question:

why is

sol:- b forms at a rate of k,a it instanteously goes to c with arate k₂

now a=A₀ l-kt(on integration)

put in (ii)



















at t=0, b=0

so





Now applying mass balance :

A0= A + B + c

so c= A₀-A-B



the maximum conc of b occurs when



so k₁A=k₂B[from 2]</sup>