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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

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 .