Question) (i) If slope is equal to $-2\times {10}^{-6} se{c}^{-1}$, what will be the value of rate constant?
(ii) How does the half - life of zero order reaction relate to its rate constant?

Dear student,

(i)This graph is for first order reaction.

$$\begin{gathered} A=A_0{e}^{-kt} \\ kt=\ln \frac{\left|A_0\right|}{\left|A\right|} \\ \frac{1}{2.303}kt=\log \frac{\left|A_0\right|}{\left|A\right|} \\ \log \left|A_0\right|-\log \left|A\right|=\frac{kt}{2.303} \\ \log \left|A_0\right|=\log \left|A_0\right|-\frac{kt}{2.303} \left(where \frac{-k}{2.303}=slope\right) \\ so k =2.303\times 2.0\times {10}^{-6}=4.606\times {10}^{-6}{s}^{-1}. \end{gathered}$$

(ii) For zero order, unit of rate constant = unit of rate of reaction
Half life time is the time in which the concentration of a particular reactant is reduced to half of the initial concentration.
$$\begin{gathered} \left[A\right]=\frac{1}{2}{\left[A\right]}_0 \\ u\sin g integrated form we can develop relation between zero order reaction and half life \\ \left[A\right]={\left[A\right]}_0-kt \\ now \frac{1}{2}\left[A\right]={\left[A\right]}_0-kt\frac{1}{2} \\ or t_{1/2}=\frac{{\left[A\right]}_0}{2k} \\ half life of zero order depends on the initial concentration of reac\tan t and the rate cons\tan t. \end{gathered}$$


Regards