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The rate law equation for the reaction A to B is found to be -(d[A])/(dt)=k[A]^(1//2) If [A]_(0) were the initial concentration of A, derive expressions for (i) rate constant in the integrated form(ii) half-life period of the reaction |
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Answer» Solution :(i) `-(d[A])/(DT)=k[A]^(1//2)" or "-(d[A])/([A]^(1//2))=k" dt"` Integrating both sides of the equation, `-int[A]^(1//2)d[A]=int k" dt or "-([A]^(1//2))/(1//2)=kt+I` or `-2[A]^(1//2)=kt+I""...(i)` When `t=0,[A]=[A_(0)]:.-2[A_(0)]^(1//2)=0+I` Substittuting this value of I in EQN. (i), we get `-2[A]^(1//2)=kt-2[A_(0)]^(1//2)" or "kt=2"{"[A_(0)]^(1//2)-[A]^(1//2)"}"" or "k=(2)/(t){[A_(0)]^(1//2)-[A]^(1//2)}` (ii) `t=t_(1//2)" when "[A]=[A_(0)]//2` `:.t_(1//2)=(2)/(k){[A_(0)]^(1//2)-([A_(0)]^(1//2))/(2^(1//2))}=(2)/(k)[A_(0)]^(1//2){1-(1)/(sqrt(2))}=(2)/(k)[A_(0)]^(1//2)(sqrt(2)-1)/(sqrt(2))" or "t_(1//2)=(sqrt(2)"("sqrt(2)-1")")/(k)[A_(0)]^(1//2)` |
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