Chemistry - Chemical Kinetics Question with Solution | TestHub

$t_{1/2}\propto a_0^{1-n}\Rightarrow t_{1/2}=c\cdot {\left(a_0\right)}^{1-n}[c=$ constant $]$

$$\log t_{1/2}=\log c+(1-n)\log a_0\therefore [1-n]=\tan 45^0\Rightarrow 1-n=1$$

$$\Rightarrow n=0$$

Now, for K, As we know for zero order reaction,

$${\mathrm{t}}_{1/2}=\frac{{\mathrm{a}}_0}{2\mathrm{K}}\Rightarrow \log {\mathrm{t}}_{1/2}=\log \left(\frac{{\mathrm{a}}_0}{2\mathrm{K}}\right)=\log {\mathrm{a}}_0+\log \frac{1}{2\mathrm{K}}$$

From graph, $\log \frac{1}{2\mathrm{K}}=2\mathrm{K}=\frac{1}{200}{mol1}^{-1}{\sec }^{-1}$

$$\therefore 250\mathrm{K}=250\times \frac{1}{200}=1.25\mathrm{mol}/1/\sec$$