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A→→C(Half life=1 minute), B→→C(Half life=2minutes). At t=0, concentration of A was 4N｡ whereas concentration of B was N｡ at t=0. Find- (i) the concentration of C at the instance when concentration of A and B becomes equal. Also find(ii) the concentration of C at the instance when activity of A becomes equal to activity of B.
Asked by patra04011965 | 30 Mar, 2020, 08:02: PM
answered-by-expert Expert Answer
Concentration of A at time t ,  begin mathsize 14px style N subscript A equals space 4 space N subscript o space e to the power of negative lambda subscript A t end exponent space end style........................  (1)
Concentration of B at time t , begin mathsize 14px style N subscript B equals space space N subscript o space e to the power of negative lambda subscript B space t end exponent space end style.........................( 2 )
where λA is decay constant of A and it is given by, λA = ln2 min-1 
where λB is decay constant of B and it is given by, λB = (1/2) ln2 min-1

Time t for getting equal concentration of A and B is obtained by equating above equation and solving for time t
 
If NA = NB ,  then we have , begin mathsize 14px style 4 space N subscript o space e to the power of negative lambda subscript A t end exponent space space equals space N subscript o space e to the power of negative lambda subscript B t end exponent space end style
begin mathsize 14px style ln left parenthesis 4 right parenthesis space minus space lambda subscript A t space equals space minus lambda subscript B t end style
begin mathsize 14px style t space equals space fraction numerator ln left parenthesis 4 right parenthesis over denominator lambda subscript A minus lambda subscript B end fraction space equals space fraction numerator ln left parenthesis 4 right parenthesis over denominator ln left parenthesis 2 right parenthesis space minus space begin display style 1 half end style ln left parenthesis 2 right parenthesis end fraction space equals space 4 space m i n end style

Hence concentration of C at t = 4 min  given by,  
begin mathsize 14px style C subscript t equals 4 end subscript space equals space 4 N subscript o space e to the power of negative space 4 space ln 2 end exponent space plus space N subscript o space e to the power of negative 2 space ln 2 end exponent space equals N subscript o subscript space end subscript open square brackets 4 over 16 space plus space 1 fourth close square brackets space equals space N subscript o over 2 end style
----------------------------------------------------------
 
Activity of A , begin mathsize 14px style fraction numerator d space N subscript A over denominator d t end fraction space equals space minus lambda subscript A N subscript A equals space minus lambda subscript A space 4 space N subscript o space e to the power of negative lambda subscript A t end exponent end style
Activity of B , begin mathsize 14px style fraction numerator d space N subscript B over denominator d t end fraction space equals space minus lambda subscript B N subscript B equals space minus lambda subscript B space N subscript o space e to the power of negative lambda subscript B t end exponent end style
If both activities are equal, then we have,
 
begin mathsize 14px style lambda subscript A space 4 space N subscript o space e to the power of negative lambda subscript A t end exponent space equals space lambda subscript B space N subscript o space e to the power of negative lambda subscript B t end exponent space end style
begin mathsize 14px style e to the power of open parentheses lambda subscript A space minus space lambda subscript B close parentheses t end exponent space equals space 4 space lambda subscript A over lambda subscript B space equals space 8 end style
begin mathsize 14px style open parentheses lambda subscript A space minus space lambda subscript B close parentheses space t space equals space ln left parenthesis 8 right parenthesis space space space o r space space t space equals space fraction numerator ln left parenthesis 8 right parenthesis over denominator open parentheses lambda subscript A space minus space lambda subscript B close parentheses end fraction space equals space fraction numerator 2 space ln left parenthesis 8 right parenthesis over denominator ln left parenthesis 2 right parenthesis end fraction space equals space 6 space m i n end style
 
Hence concentration of C at t = 6 min  given by,  
begin mathsize 14px style C subscript t equals 6 end subscript space equals space 4 N subscript o space e to the power of negative space 6 space ln 2 end exponent space plus space N subscript o space e to the power of negative 3 space ln 2 end exponent space equals N subscript o subscript space end subscript open square brackets 4 over 64 space plus space 1 over 8 close square brackets space equals space fraction numerator 3 N subscript o over denominator 16 end fraction end style
Answered by Thiyagarajan K | 30 Mar, 2020, 10:40: PM

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