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In a zero-order reaction, if the initial concentration of the reactant is doubled, the time required for half the reactant to be consumed
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increases two-fold
$\begin{array}{l}
\mathrm{K}=\frac{\mathrm{A}_{0}-\mathrm{A}_{\mathrm{t}}}{\mathrm{t}} \\
\mathrm{K}=\frac{\mathrm{A}_{2}}{2 \mathrm{t}_{1 / 2}} \\
\mathrm{t}_{1 / 2}=\frac{\mathrm{A}_{0}}{2 \mathrm{~K}} \quad \text { Zero order }
\end{array}$
$\mathrm{t}_{1 / 2} \propto$ initial concentration so double times
\mathrm{K}=\frac{\mathrm{A}_{0}-\mathrm{A}_{\mathrm{t}}}{\mathrm{t}} \\
\mathrm{K}=\frac{\mathrm{A}_{2}}{2 \mathrm{t}_{1 / 2}} \\
\mathrm{t}_{1 / 2}=\frac{\mathrm{A}_{0}}{2 \mathrm{~K}} \quad \text { Zero order }
\end{array}$
$\mathrm{t}_{1 / 2} \propto$ initial concentration so double times
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