Enthalpy

Document Sample
Enthalpy Powered By Docstoc
					Chemistry
Session opener   Nuclear chemistry, sun & life
Session objectives

1.   Radioactivity
2.   Nuclear reactions
3.   Kinetics of radioactive decay
4.   Radioactivity equilibrium
5.   Fission reaction
6.   Fusion reaction
7.   Radiocarbon dating
8.   Medical applications
Magic numbers

RADIOACTIVITY     AND NUCLIDE STABILITY
protons     neutrons      stable nuclides
even                   even                     157
even                   odd                      52
odd                    even                     50
odd                    odd                       5


“Magic numbers” analgous to the noble gas electronic
configurations occur at: 2, 8, 20, 28, 40, 50, 82, 126, 184
The most stable isotopes have “magic numbers” of both
protons and neutrons.
n/p too large
 beta decay



   X




                  Y
                 n/p too small
       positron decay or electron capture


                                            23.2
Stability of nucleus

It has been observed that for light elements of
atomic number upto 20, n/p ratio is 1

When the n/p ratio is too high

This nucleus would be unstable and
would tend to lower the ratio by emitting
  -particles and thus moves towards the
zone of stability.


  1
0n    1 H1 
                    1 e
                         0

 neutrons proton β - particle (electron)
Stability of nucleus

When the n/p ratio is too low

This is effected either by the emission of an
alpha particle or a positron or by capturing an
orbital electron


11 Na20  10 Ne20 + +1e0
         
                       positron

37 Rb82 + -1e0  36 Kr 82
                


92 U235  90 Th231 + 2He4
         
                        particle
Nuclear reactions

Loss of –particle



 z Am z2 Bm4 2 He4   (  particle)



Loss of   –particle



z Am z1 Bm 1 e0
Group displacement laws

The emission of an -particle results in the
formation of an element which lies two places
to the left and the emission of a  -particle
results in the formation of an element which
lies one place to the right in the periodic table
Illustrative Example

92 U238 is radioactive and emits a and b particles to
form
       82 Pb206 . What is the respective values of a and b?

Solution


Let x number of  particles and y number of
 particles get emitted.



 92U238  x
                  2 He
                           4
                                 y   1e
                                              0
                                                     82Pb
                                                              206



 238 = 4x + 206 or x = 8
  92 = 2x – y + 82 or y = 6
Solution Cont.



          212Podecays by alpha emission. Write the balanced
          nuclear equation for the decay of 212Po.


      212Po        4He   + AX
       84          2       Z


  212 = 4 + A           A = 208

  84 = 2 + Z            Z = 82

     212Po        4He   + 208Pb
      84          2        82




                                                              23.1
Disintegration series

         Name of the Series   Initial element   Stable element   Value of   Value of n
                                                                   n for      for the
                                                                  initial     stable
                                                                 element     element
                                                                              Series
 4n       Thorium series      Thorium–232         Lead–208         58          52



4n + 1   Neptunium series     Neptunium–        Bismuth–209        59          52
                                 237


4n + 2    Uranium series      Uranium–238         Lead–206         59          51



4n + 3    Actinium series     Uranium–235         Lead–207         58          51
4n series




            In the 4n series,all
            nuclides have mass
            number that are
            multiples of 4.
4n +1 series




               In the 4n + 1
               series,all nuclides
               have mass numbers
               that are one number
               greater than
               multiples of 4.
4n+2 series




In the 4n + 2 series, all nuclides have mass number that are
two numbers greater than multiples of 4.
4n+3 series




In the 4n + 3 series, all nuclides have mass numbers that
are three numbers greater than multiples of 4.
Nuclear Binding Energy
Nuclear binding energy
The energy required to break up a nucleus into
its component protons and neutrons.

BE+19 F  91p+101 n
   9       1    0


E = mc2

BE = 9 x (p mass) + 10 x (n mass) –     19F   mass

BE (amu) = 9 x 1.007825 + 10 x 1.008665 – 18.9984


BE = 0.1587 amu               1 amu = 1.49 x 10-10 J

 BE = 2.37 x 10-11J
                                Binding energy
Binding energy per nucleon=
                              Number of nucleons
            2.37 x 10-11 J
          =                =1.25 x 10-12 J
            19 nucleons
Average Binding Energy as a Function of
Atomic Number
Kinetics of radioactive decay


        M        D

at t = 0 N0
at t = t N



 N0=Initial no. of radioactive particles

 N=No. of radioactive particles after time t
Kinetics of radioactive decay

The rate of disintegration of
                  dN
M into D is equal to
                   dt
      dN        dN
i.e.,      N or       N      (I)
       dt         dt
        N
             dN
 or        N
                   dt
        No

          No
 or     n     t      (II)
          N                                N  Noet      (iv)
       2.303       N 
           log10  0       (iii)
         t          N
Half life period
           .693
t1 / 2 
             
Average or Mean life Period 

   1  t1 / 2
           1.44 t1 / 2
    0.693

Units

1 curie (C) = 3.7 × 1010 dps
S.I unit is Beequerel (Bq)
        (Bq) = 1 disintegration per second.
Illustrative example
Calculate the activity in terms of dpm of 0.001 g sample of
Pu239 (half life = 24300 years).

Solution:
                                        0.693
Calculation of decay constant  
                                         t1 / 2
           24300  3.15  107
t1 / 2                       minutes
                  60
            0.693  60
                          min1
         24300  3.15  107
                                             103  6.02  1023
Number N of the nuclei in the given sample 
                                                    239
       dN
Hence,      N
        dt
dN      0.693  60        103  6.023  1023
                                             = 1.37 × 108 dpm.
 dt   24300  3.15  107           239
Illustrative example
One of the hazards of nuclear explosion is the generation of Sr90 and
its subsequent incorporation in bones. This nuclide has a half-life of
28.1 years. Suppose 1m g was absorbed by a new born child, how
much Sr90 will remain in his bones after 20 years?

Solution

                  0.693 0.693
 We have                      0.0246 yr 1.
                   t1/ 2   28.1

          2.303      a          2.303      a 
 As t          log      20         log    
                   ax        0.0246     ax

      a
         1.64
     ax

 Here a = 10–6 g
  a – x = 6.09 × 10–7 g
Illustrative example
Calculate the weight of Na24 which will give
radioactivity of one curie (half-life of Na24 is 15 hr).
1C = 3.7 × 1010 dps

Solution
            0.693       0.693
 Now                            1.28  105
             t1/ 2   15  60  60

  3.7  1010  N .....(i)


   N = 2.89 × 1015


  6.023 × 1023 of Na24 atoms = 24 g

                                24  2.89  1015
  2.89  10 15
                  of Na atoms 
                       24
                                                  1.15  10 7 g
                                 6.023  1023
Illustrative example
The half-life for the decay of U238 to Th234 is 4.6 × 109 years.
How many a particles are produced per second in a sample
containing 3 × 1020 atoms of U238 .
Solution
                         0.693
 We know that t1/ 2 
                           
          0.693
or                1.5  1010 yr 1
         4.6  109
           dN
 Now          N  1.5  1010  3  1020
           dt
      dN
         4.52  1010 atoms yr 1
      dt

Number of  particles emit per second

      4.52  1010
                      1433
   365  24  60  60
Artificial transmutation of elements

Process of transformation of one element into the other by artificial
means i.e. by bombarding the nuclei of atom by high speed
subatomic particles is called artificial transmutation of elements:

Rutherford 1919.

 14N
  7
        + 4He →
          2
                       17O
                        8
                              + 1H
                                1


 Irene Joliot-Curie.

 24Al
 13
        + 4He →
          2
                       30P
                       15
                              + 1n
                                0
           15          14       +1
          30P
                →      30Si   + 0
                                             Shared Nobel Prize 1938
Transuranium Elements

Element coming after uranium [Z = 92] in the
periodic table i.e. with atomic number greater
than 92 are called transuranic elements. e.g.

238
 92 U
         +
             1
             0n   →   239      +        
                       92 U

         239      →
                      239
                       93 Np
                               +    0   
          92 U                     -1



249
 98 Cf
         + 15 N →
            7
                      260
                      105 U
                                   1
                               + 4 0n
Fission of   U-235
Fission of         U-235

                             56 Ba 36   Kr  31 n
                             140    93
                                               0




  92 U 0   n  92 U
  235   1       236
                             54 Xe 38   Sr  21 n
                             144    90
                                               0



                             55 Co 37   Rb  21 n
                             144    90
                                               0




The minimum amount of fissionable material required to
continue a nuclear chain reaction is called critical mass.
Nuclear Reactors
Breeder Reactors
A breeder reactor is one that produces more
fissionable nuclei than it consumes i.e. it
increases the concentration of fissionable nuclie.


  238U
   92
          + 1 0n →
              1       239U
                       92


          →               -1 
   239U       239Np   +    0
    92         93


          →               -1 
  239Np       239Pu        0
   93          94     +
Fusion




2     2
1           4
  H  1 H  2He    energy.
Difference between fission and fusion

    Nuclear fission              Nuclear fusion
 1. It involves breaking   1. It involves fusion of
 up of a heavier           two or more lighter
 nucleus into lighter      nuclei to form a heavier
 nucles.                   nucleus.
 2. Large number of        2. It is difficult to control
 radioisotopes are         this process.
 formed.
 3. It is a chain          3. It is not a chain
 process.                  process.

 4. This does not          4. This is initiated by
 require high              very high temperature.
 temperature
 5. It can be controlled 5. It cannot be
 and energy is           controlled.
 released during the
 reactions.
Rate law for reactions
involving parallel reactions
              K1
                     B (80%)   Main reaction


A


                     A (20%)   Side reaction
               K2


         d{A}
Rate         K1[A]  K 2 [A]  (K1  K 2 )[A]
          dt
Illustrative example
227Ac  has a half-life of 22 years with respect to radioactive decay.
The decay follows two parallel paths, one leading to 227Th and
the other to 223Fr. The percentage yields of these two daughter
nuclides are 2 and 98 respectively. What are the decay constants
(l) for each of the separate path?

 Solution
                    
                    1     2
                          27
                            h%
                             (
                            T 2)
2 c
 7
2A
                    
                    2     2
                          23
                            r 8
                             ( %
                            F9 )

            0.693                                 On solving,
 total           0.0315 per year
              22                                1  6.3  10 4 per year
1   2  0.0315
                                                 2  3.08  10 2 per year
         2 98
while          48
        1   2
Radioactive equilibrium
 A B C D

If the rate of decay of B is the same
as its rate of formation from A, then
the amount of B will remain constant.

 dNA   dNB
     
  dt    dt
 NA   B
    
 NB   A
Age of minerals and rocks

The end product in the natural disintegration series is
an isotope of lead. Each disintegration step has a
definite decay control. By finding out the amounts of
percent radioactive element and the isotope of lead
(e.g. 92U238 and 82Pb206) in a sample of rock and
knowing the decay constant of the series, the age of
rock can be calculated.
Illustrative Example
An uranium containing ore pitch blende contains
0.055 g of 82Pb206 for 1 g of 92U238. Calculate the
age of the ore. (Half-life = 4.6 × 109 years)

Solution
     2.303    N
t         log o ,
             N
                 0.693
We have                  0.15  109
                4.6  109

Here N = 1 g
                                   238
Now 0.055 g of Pb206  0.055          gm of U238  0.0635 g
                                   206

 No  1  0.0635  1.0635 g

         2.303         1.0635
t                log         4.10  108 yrs.
       0.15  109        1
Radio carbon dating

The principle is based upon the formation of C14
by neutron capture in the upper atmosphere.



 7 N14 + 0n1  6C14 + 1H1
               


This carbon-14 is radioactive with a half life period of 5700 years.
This is changed into nitrogen by the emission of   particles.


                                          
   14  N14 + e0           t = 5700 yrs 
 6C     7     -1            1            
                             2            
Radio carbon dating

The age of the object can be estimated.



          2.303     initial activity (amount of C14 in a living object)
tiem(t)        log
                    Final activity (amount of C14 in a dead object)


This method cannot be applied to estimate the age of an
object which is about 20,000 to 50,000 yrs old.
Illustrative Example
A piece of charcoal from the ruins of a settlement in Japan was
found to have 14C/12C ratio that was 0.617 times that formed in
living organism. How old is this piece of charcoal. Given the half life
of 14C is 5,770 years. [log 1.62 = 0.210]

Solution

  We know

       0.693 2.303    N
                log o
        t1/ 2   t     Nt

  0.693 2.303       1
             log
  5770    t       0.617

   t = 4027 years
Application: Tracers
Radioisotopes are used as tracers to find
the reaction mechanism.

 For example


         O                              O
C6H5 C             *
              + CH3OH          C6H5 C       +   H2O
         OH                             *
                                        OCH3
Radioisotopes in Medicine
•   1 out of every 3 hospital patients will undergo a nuclear
    medicine procedure
•   24Na,    t½ = 14.8 hr,  emitter, blood-flow tracer
•   131I,   t½ = 14.8 hr,  emitter, thyroid gland activity
•   123I,   t½ = 13.3 hr, -ray emitter, brain imaging




Brain images with      123I-labeled

compound




                                                                23.7
Illustrative Example
The half life of W-238, which decays to Pb-206 is
4.5 x 109 years. A rock containing equal numbers of
atoms of two isotopes would be how much old?
Solution

W238  Pb206
                            0.693
t1 / 2  4.5  109 yrs. 
                              k
       2.303    N
k          log o
         t      Nt
No  Initial no. of atoms of W238
Nt  No. of atoms of W238 after certain time
t  Age of the rock

 0.693      2.303    2N
                 log t
4.5  109     t      Nt
     2.303
t          4.5  109 log2  4.5  109 yrs.
     0.693
Illustrative Example
A small amount of solution containing Na24 radioisotope with
activity 2 × 103 dps was injected into blood of a patient. After 5 hr,
a sample of blood drawn out from the patient showed an activity of
15 dpm ml–1 half-life of Na24 is 16 hr. Find the volume of blood in
the patient.
Solution

Let the volume of blood be V ml.

Activity of Na24 in V ml blood = 2 ×103 dps = 2 × 60 × 103 dpm

Activity of Na24 in V ml blood after 5 hr = 15 × V dpm
     0.693 0.693
               0.0462 hr 1
      t1/ 2   16


  Now

          0.693    N
     t         log 0
                  N

            0.693      120  103
     or 5         log
            0.0462      15  V

     V  3.713  103
Thank you

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:11
posted:3/1/2012
language:English
pages:49