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The rate constant for a first order reaction is 60 S–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value ?

Rate constant of reaction,
k = 60 S–1
t15/16 = ?

Rate constant of first order reaction is given by.

k=2.303tlogaa-xt=2.303klogaa-x

where 15 / 16th the reaction is over the

a-x = 1-116M if a = 1 M.

log15/16 = 2.30360 S-1 log 11-15/16               = 2.30360log 16               = 2.30360×1.2041               = 0.046 S = 4.6 × 10-2sec.

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Calculate the half life of a first order reaction from their rate constants given below:
(i) 200 s–1       (ii) 2 min–1         (iii) 4 years–1

For the first order reaction.


                             t1/2 = 0.693k
(a) 
         K = 200 s-1                    t1/2 = 0.693200 = 3.465 × 10-3 s


(b)  

          K = 2 min-1            t1/2 = 0.6932=0.3465 min


(c) 

         K = 4 years -1         t1/2 = 0.6934 = 0.1732 years.
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What is the effect of temperature on the rate constant of reaction? How can this effect of temperature on rate constant be respresented quantitatively?

The rate constant of reaction increases with increase of temperature. This increase is generally two fold to five fold for 10 K rise in temperature. This is explained on the basis of collision theory. The main parts of collision theory are as follows:

(i) For a reaction to occur, there must be collision between the reacting species.

(ii) Only a certain fraction of total collisions are effective in forming the products.

(iii) For effective collisions, the molecule must possess the sufficient energy (equal or greater
than threshold energy) as well as proper orientation.
On the basis above conclusions, the rate of reaction is given by
Rate =f x 2 (where f is the effective collisions and is total number of collisions per unit volume per second).

Quantitatively, the effect of temperature on the rate of a reaction and hence on the rate constant k was proposed by Arrhenius. The equation, called Arrhenius equation is usually written in the form

K=Ae-Ea/RT            ...(i)
.
where A is a constant called frequency factor (because it gives the frequency of binary collisions of the reacting molecules per second per litre, E
0 is the energy of activation, R is a gas constant and T is the absolute temperature. The factor e–Ea/RT gives the fraction of molecules having energy equal to or greater than the activation energy, Ea.

The energy of activation (Ea) is an important quantity and it is characteristic of the reaction. Using the above equation, its value can be calculated.

Taking logarithm or both sides of equation (i), we get,
In k = In A - EaRT1

If the value of the rate constant at temper-atures T
1 and T2 are k1 and k2 respectively, then we have

 In k1 = In A - EaRT1            ...(ii)In k2 = In A - EaRT2            ...(iii)

Subtracting eqn. (i) from eqn. (ii), we get

   In k2 - In k1 = -EaRT2+EaRT1                       = EaRT1+EaRT2
     

or                          Ink2k1 = EaR+1T1-1T2                                         = EaRT2-T1T1T2

or       logk2k1 = Ea2.303RT2-T1T1 T2

Thus knowing the values of the constant k
1 and k2 at two different temperature T1 and T2, the value of Ea can be calculated.

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The experimental data for decomposition of N2O5 [2N2Os → 4NO2 + O2] in gas phase at 318 k are given below:

(a) Plot [N2O5] again t.
(b) Find the half life period for the reaction.
(c) Draw a graph between log [N2O5] and t.
(d) What is rate law?
(e) Calculate the rate constant.
(f) Calculate the half life period from K and compare it with (ii). 


(a) Plot of [N2O5] vs. time.

(a) Plot of [N2O5] vs. time.
(b) Time taken for the concentration of

(b) Time taken for the concentration of N2O5 to change from 1.63 x 10–2 mol L–1 to half the value.
t0.5 = 1420 s [from the graph] (c) Plot of log (N2O5) vs. time.

t

log10[N2O5]

0

400

800

1200

1600

2000

2400

2800

3200

3600

- 1.7918

- 1.8665

- 1.9431

- 2.0315

- 2.1079

- 2.1938

- 2.2757

- 2.3752

- 2.4559

- 2.5376

 


(a) Plot of [N2O5] vs. time.
(b) Time taken for the concentration of

(a) Plot of [N2O5] vs. time.
(b) Time taken for the concentration of
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e reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

exp.

 

[A]/
mol L–1

[B]/M

Initial rate of formation
of D/mol L–1 min–1

I

0.1

0.1


2.0 x 10–2

II

-

0.2

4.0 x 10–2

 

III

0.4

0.4

-

IV

-

0.2

2.0 x 10–2


Rate law for the reaction is given by:
Rate = k [A]1 [B]0 = k[A]
2.0 x 10–2 mol L–1 min–1 = k[0.1 mol L–1]
Rate constant,

k=2.0×10-2mol L-1 min-10.1 mol L-1 or

Rate constant = k = 0.2 min–1.
(i) In experiment II Rate = k[A]

A = ratek

 

       = 4.0×10-2M min-10.2 min-1=0.20 M       


(ii) In experiment III

Rate = k[A]
= 0.2 min–1 x 0.4 M
= 0.08 min–1
= 8.0 x 10–2 M min–1

(iii) In experiment IV

Rate = k[A]

A = ratek=2.0×10-2M min-10.2 min-1 = 0.1 M


Thus the completed table is

exp.

 

[A]/
mol L–1

[B]/M

Initial rate of formation
of D/mol L–1 min–1

I

0.1

0.1


2.0 x 10–2

II

0.2

0.2

4.0 x 10–2

 

III

0.4

0.4

8.0 x 10-2

IV

0.1

0.2

2.0 x 10–2

 

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