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 In given graph between \ln t_{\frac{1}{2}} v / s \ln A_0  Which line represents the first order reaction?

Option: 1

AB
 


Option: 2

AC


Option: 3

AD

 


Option: 4

None of these


Answers (1)

General formula for half life of n^{t h} order reaction: t_{\frac{1}{2}}=\frac{2^{n-1}-1}{\left(A_0\right)^{n-1}(n-1) \kappa}
Thus we can writet_{\frac{1}{2}}=\frac{K}{\left(A_0\right)^{n-1}}, where, K= constant
On integrating above equation: \ln t_{\frac{1}{2}}=\ln K+\ln \left(\frac{1}{\left(A_0\right)^{n-1}}\right)
Or: \ln t_{\frac{1}{2}}=\ln \left(\frac{1}{\left(A_0\right)^{n-1}}\right)+C,where, C= constant

For first order reaction: \ln t_{\frac{1}{2}}=\ln \left(A_0\right)^0+C, thus t_{\frac{1}{2}} is independent of A_0
On comparing with line equation: Y=m X+C, \mathrm{slope (\mathrm{m})} of graph between\ln t_{\frac{1}{2}} \mathrm{v} / \mathrm{s} \ln A_0 is zero.
Thus the \mathrm{AC} line represents the first order reaction.
Also, line \mathrm{\mathrm{AB} \: shows\: t_{\frac{1}{2}} } is directly proportional to A_0, which represent order of reaction is less than one.
And, line \mathrm{\mathrm{AD}\: shows \: t_{\frac{1}{2}}} is inversely proportional to A_0, which represents the order of reaction is more than one.

Posted by

Sumit Saini

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