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  • What will be the expression of half -life when reaction follows rate law, rat

What will be the expression of \mathrm{t_{1 / 2}} half -life when reaction follows rate law, rate \mathrm{=k [concentration ]^2}
Here, k is the rate constant a is the initial concentration of the reactant.

Option: 1

\mathrm{1 / k_a}


Option: 2

\mathrm{k} \mathrm{a}^2


Option: 3

3 / 2 \mathrm{ka}^2


Option: 4

1 / \mathrm{ka}^2


Answers (1)

best_answer

The expression for integrated rate law of \mathrm{n^{\text {th }}} order reaction is
\mathrm{ {\left[\frac{1}{(a-x)^{n-1}}\right]-\left[\frac{1}{a^{n-1}}\right]=(n-1) k t \ldots . . e q n(1)} \\ }

\mathrm{ \text { time }=0 \quad A \rightarrow \text { Product(s) } \\ }

\mathrm{ \text { time }=(t) a-x \\ }

\mathrm{\text { At } t=t_{1 / 2}, \\ }

\mathrm{x=\frac{a}{2} \text { (reaction is half completed) } }

Substituting the value of X in equation (1) gives,
\mathrm{ {\left[\frac{1}{\left(a-\frac{a}{2}\right)^{n-1}}\right]-\left[\frac{1}{a^{n-1}}\right]=(n-1) k t_{1 / 2}} \\ }

\mathrm{ {\left[\frac{2^{n-1}}{a^{n-1}}-\frac{1}{a^{n-1}}\right]=(n-1) k t_{1 / 2}} }

On rearranging we get,
\mathrm{ t_{1 / 2}=\frac{1}{k(n-1)}\left[\frac{2^{n-1}-1}{a^{n-1}}\right] \ldots \ldots \operatorname{eqn}(2) }

In general,
\mathrm{ \operatorname{Rate}(R)=k[A]^n }
Given,
\mathrm{ Rate =k[\mathrm{~A}]^2 }
So, order =2

Substituting n=2 in equation (2) gives,
\mathrm{ t_{1 / 2}=\frac{1}{k(2-1)}\left[\frac{2^{2-1}-1}{a^{2-1}}\right] \\ }

\mathrm{ t_{1 / 2}=\frac{1}{k}\left[\frac{2-1}{a}\right] \\ }

\mathrm{ t_{1 / 2}=\frac{1}{k a} }

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Nehul

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