On the basis of the unit of the reaction we can see that the reaction is a zero order reaction.

$$ \begin{aligned} k & =\frac{\text { Rate }}{[A]^x[B]^v} \\ & =\frac{\text { concentration }}{\text { time }} \times \frac{1}{(\text { concentration })^{\prime \prime}} \quad(\text { where }|A|=|B|) \end{aligned} $$

Taking SI units of concentration, mol $\mathrm{L}^{-1}$ and time, s, the units of $k$ for different reaction order are listed in Table 3.3

$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text { Table 3.3: Units of rate constant } $$

Reactien	order	thits of rete conatant
Zero order reaction	0	$\frac{mol L^{- 1}}{s} \times \frac{1}{{(mol L^{- 1})}^{6}} = mol L^{- 1} s^{- 1}$
First order reaction	1	$\frac{mol L^{- 1}}{s} \times \frac{1}{{(mol L^{- 1})}^{1}} = s^{- 1}$
Second order reaction	2	$\frac{mol L}{}^{- 1} \times \frac{1}{s} \times {mol}^{- 1} L s^{- 1}$

IAT (IISER) 2024 Chemistry - Chemical Kinetics Question 3 English Explanation

Fig. 3.3: Variation in the concentration vs time plot for a zero order reaction

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