Chemistry - Chemical Kinetics Question with Solution | TestHub

A (Zero order) → 1, 2

${\mathrm{t}}_{1/2}\propto [\mathrm{A}{]}_0\Rightarrow$ doubling $[\mathrm{A}{]}_0$ doubles ${\mathrm{t}}_{1/2}$

Units: $[\mathrm{k}]=\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$

B (First order) $\to 3$

$\ln [\mathrm{A}]=\ln [\mathrm{A}{]}_0-\mathrm{kt}\Rightarrow$ straight line

C (Second order, 2A) → 4

For 2A:

$\frac{1}{[\mathrm{A}]}=\frac{1}{[\mathrm{A}{]}_0}+\mathrm{kt}$

At $50\%$ completion: $[\mathrm{A}]=\mathrm{a}/2\Rightarrow {\mathrm{t}}_{50\%}=\frac{1}{\mathrm{ka}}$

At $75\%$ completion: $[\mathrm{A}]=\mathrm{a}/4\Rightarrow {\mathrm{t}}_{75\%}=\frac{3}{\mathrm{ka}}$

So, $\frac{t_{75\%}}{t_{50\%}}=3$

D (Pseudo-first order) $\to 3,5$

behaves as first order in $\Rightarrow \ln [\mathrm{A}]$ vs t linear

effective first order constant ${\mathrm{k}}^{\prime }=\mathrm{k}[\mathrm{B}{]}_0$