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# Explain half-life for zero-order and second-order reactions

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## The half-life of Zero-order:The zero-order reaction refers to a condition in which the rate of the reaction does not vary with the change in concentration of the reactant.Half-life refers to the amount of time required to reduce the amount of a substance to half of its initial value.Let. $\mathrm{A}\to \mathrm{B}+\mathrm{C}$; The rate law is given as: Rate = $\mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{n}}$; where n is the order of the reaction. for zero-order n=0The differential rate law is given as: $\mathrm{k}=-\frac{\mathrm{d}{\left[\mathrm{A}\right]}^{0}}{\mathrm{dt}}$rearranging: $\mathrm{d}{\left[\mathrm{A}\right]}^{0}=-\mathrm{kdt}$Integrating both side: ${\int }_{{\left[A\right]}_{0}}^{\left[A\right]}d\left[A\right]=-{\int }_{0}^{t}kdt\phantom{\rule{0ex}{0ex}}⇒{\left[A\right]}_{{\left[A\right]}_{0}}^{\left[A\right]}=-kt\phantom{\rule{0ex}{0ex}}⇒\left[A\right]-{\left[A\right]}_{0}=-kt\phantom{\rule{0ex}{0ex}}⇒\left[A\right]={\left[A\right]}_{0}-kt----->\left(i\right)$For half-life: at $\mathrm{t}={\mathrm{t}}_{\frac{1}{2}}$; $\left[A\right]=\frac{{\left[A\right]}_{0}}{2}$Substituting in eq(i)$⇒\frac{{\left[A\right]}_{0}}{2}={\left[A\right]}_{0}-k{t}_{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}⇒{t}_{\frac{1}{2}}=\frac{{\left[A\right]}_{0}}{2k}$.The half-life for second-order:Let, $2\mathrm{A}\to \mathrm{B}$the differential rate law is given as: $-\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}=\mathrm{k}{\left[\mathrm{A}\right]}^{2}$rearranging and integrating.$\frac{d\left[A\right]}{{\left[A\right]}^{2}}=-kdt\phantom{\rule{0ex}{0ex}}⇒{\int }_{{\left[A\right]}_{0}}^{\left[A\right]}\frac{d\left[A\right]}{{\left[A\right]}^{2}}=-{\int }_{0}^{t}kdt\phantom{\rule{0ex}{0ex}}⇒{\left[-\frac{1}{\left[A\right]}\right]}_{{\left[A\right]}_{0}}^{\left[A\right]}=-kt\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\left[A\right]}-\frac{1}{{\left[A\right]}_{0}}=kt\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\left[A\right]}=kt+\frac{1}{{\left[A\right]}_{0}}$For half-life: at $\mathrm{t}={\mathrm{t}}_{\frac{1}{2}}$; $\left[\mathrm{A}\right]=\frac{{\left[\mathrm{A}\right]}_{0}}{2}$substituting in the above equation$⇒\frac{1}{\frac{{\left[A\right]}_{0}}{2}}=k{t}_{\frac{1}{2}}+\frac{1}{{\left[A\right]}_{0}}\phantom{\rule{0ex}{0ex}}⇒{t}_{\frac{1}{2}}=\frac{1}{k{\left[A\right]}_{0}}$

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