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For the decomposition of azoisopropane to hexane and nitrogene at 543 K, the following data are obtained.

t(sec)

P (mm of Hg)

0

35.0

360

54.0

720

63.0

Calculate the rate constant.


As we have given the data 
Thus,

K = 2.303tlogp1p2    = 2.303360log3554    = 2.20 × 10-3 51
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During nuclear explosion, one of the products is 90Sr with half-life of 28.1 Years. If 1 μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically. 

All radioactive decay are first order process.
 decay constant.

     λ=  0.693t1/2 = 0.69328.1 = 0.0247 year-1

(i) where t  = 10 years,

         λ = 2.303tlogN0Nt = 10 yearsN0 = 1 microgram      = 1×10-6g, N = ?10 = 2.3030.0247log1×10-6N

log1×10-6N=10×0.02472.303=0.1072        1 × 10-6N =Antilog 0.1072 = 1.1280      N = 1×10-61.1280 = 0.7842 log


(ii) 
       t =60 years         60 = 2.3030.0247log1×10-6Nlog1×10-6N = 60×0.02472.303 = 0.6435      1×10-6N = 60×0.02472.303 = 0.6435      1×10-6N = Antilog 0.6453 = 4.400         N = 1×10-64.400 = 0.227 μg


Thus, after 10 years and 60 years, 0.7842 log and 0.227 log of 
90Sr will be left as undecayed.
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For first order reaction show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

For the first order reaction, time required for 99% completion.

t1/2 = 2.303tlogaa-x

Ist case    
                a=100%x = 90%, (a-x) = 100 -90 = 10%
     t90% = 2.303klog10010 = 2.303klog 10t90% = 2.303k                                 ...(i)

IInd case

                
a=100%x = 99%, (a-x) = (100-99) = 1%t12 = 2.303klog1001      = 2.303k×2

       t99% = 2.303k×2             ...(ii)

Dividing eqn. (ii) by eqn. (i)


t99%t90% = 2.303k×2×k2.303 = 2

Hence, time required for 99% completion is twice for the time required for the completion of 90% reaction.
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The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the C14 activity found in living tree. Estimate the age of the sample or calculate the age of the artefact.

We have given
t
1/2 = 5730 years

NN0 = 80% = 0.80

Decay constant,

              λ = 0.693t1/2 = 0.6935730   = 1.21 × 10-4 year-1

All radioactive nuclear decay are first order process.

Therefore, decay constant,

                  λ = 2.03tlogN0Nλ = 2.30tlog108

or       

             time, t = 2.303λlog 1.25            = 2.3031.21 × 10-4 log 1.25            = 2.3031.21×10-4×0.0969            = 1845 years.Age of sample = 1845 years.            
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A first order decomposition reaction takes 40 minutes for 30% of decomposition. Calculate t1/2 value for it.

For the first order reaction 

K = 2.303tlogaa-x    = 2.30340log10070    = 2.30340log 100 - log 70    = 2.30340(2-1.8808)   = 2.30340×0.1192 = 0.0068 min-1

 (ii) 0.0068 = 2.303tlog100500.0068 = 2.303tlog 20.0068 = 2.303t×0.3010  t = 2.303×0.30100.0068 = 101.94 min.
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