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Find the value of half life ,For a first order reaction :
(A) \[\dfrac{0.53}{k}\]
(B) \[0.693\times K\]
(C) \[\dfrac{0.693}{K}\]
(D) None above this

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Last updated date: 24th Feb 2024
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Hint:In order to answer this question we should first know about the \[1^{st}\] order reaction and half life of a reaction. The relation between half life and first order reaction is \[{{t}_{\dfrac{1}{2}}} = \dfrac{0.693}{k}\].

Complete step-by-step solution:
When rate of reaction is directly dependent on the concentration of a single reactant the phenomena is as First order reaction. On the other hand, the amount of time taken for the concentration of a reaction to reach half of its initial concentration is called the Half life reaction. From the integrated form of first order reaction we get, \[\left[ A \right] = \left[ A \right]0{{e}^{-kt}}\]
Now when the time \[t={{t}_{\dfrac{1}{2}}}\], [ Half life \[= {{t}_{\dfrac{1}{2}}}\]]
Here, \[\left[ A \right]\] is concentration of reaction and \[\left[ A \right]0\] is the initial concentration
Now by substituting \[\left[ A \right]\] with \[\dfrac{\left[ A \right]0}{2}\] and \[t\] with \[{{t}_{\dfrac{1}{2}}}\], in the integration form of first order reaction we get,
\[\Rightarrow \dfrac{\left[ A \right]0}{2}=\left[ A \right]0{{e}^{-k{{t}_{\dfrac{1}{2}}}}}\]
\[\Rightarrow \dfrac{1}{2}={{e}^{-k{{t}_{\dfrac{1}{2}}}}}\]
By eliminate “e” from the above equation we get-
\[\Rightarrow \ln \left( \dfrac{1}{2} \right)=-kt\dfrac{1}{2}\]
\[\therefore t\dfrac{1}{2}=\dfrac{0.693}{k}\]
Therefore option (C) is the right answer.

 Note: For zero order reaction half life will be \[{{t}_{\dfrac{1}{2}}}=\dfrac{\left[ R \right]0}{2k}\] . Unit of \[\left[ R \right]0\] is \[[mol.{{L}^{-1}}]\] that is the initial reactant concentration .
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