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Question: Find the value of half life ,For a first order reaction : (A) \[\dfrac{0.53}{k}\] (B) \[0.693\ti...

Find the value of half life ,For a first order reaction :
(A) 0.53k\dfrac{0.53}{k}
(B) 0.693×K0.693\times K
(C) 0.693K\dfrac{0.693}{K}
(D) None above this

Explanation

Solution

In order to answer this question we should first know about the 1st1^{st} order reaction and half life of a reaction. The relation between half life and first order reaction is t12=0.693k{{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, [A]=[A]0ekt\left[ A \right] = \left[ A \right]0{{e}^{-kt}}
Now when the time t=t12t={{t}_{\dfrac{1}{2}}}, [ Half life =t12= {{t}_{\dfrac{1}{2}}}]
Here, [A]\left[ A \right] is concentration of reaction and [A]0\left[ A \right]0 is the initial concentration
Now by substituting [A]\left[ A \right] with [A]02\dfrac{\left[ A \right]0}{2} and tt with t12{{t}_{\dfrac{1}{2}}}, in the integration form of first order reaction we get,
[A]02=[A]0ekt12\Rightarrow \dfrac{\left[ A \right]0}{2}=\left[ A \right]0{{e}^{-k{{t}_{\dfrac{1}{2}}}}}
12=ekt12\Rightarrow \dfrac{1}{2}={{e}^{-k{{t}_{\dfrac{1}{2}}}}}
By eliminate “e” from the above equation we get-
ln(12)=kt12\Rightarrow \ln \left( \dfrac{1}{2} \right)=-kt\dfrac{1}{2}
t12=0.693k\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 t12=[R]02k{{t}_{\dfrac{1}{2}}}=\dfrac{\left[ R \right]0}{2k} . Unit of [R]0\left[ R \right]0 is [mol.L1][mol.{{L}^{-1}}] that is the initial reactant concentration .