Chemistry - Chemical Kinetics Question with Solution | TestHub

$\mathrm{Exp}(1):r=K^{\prime }[B{]}^y$ as $[A_0]\gg [B_0]$

$$\because t_{7/8}=3\times t_{1/2}\Rightarrow y=1$$

$\mathrm{Exp}(2):r=K^{\prime \prime }[A{]}^x$ as $[A_0]\ll [B_0]$

$$\because t_{7/8}=7\times t_{1/2}\Rightarrow x=2$$

Now, for $\exp (2)$ and (3), $t_{1/2}=\frac{1}{K^{\prime \prime }[A_0]}=\frac{1}{K[B_0][A_0]}$

$\therefore a=\frac{10}{2\times 2}=2.5$ and $b=7\times 2.5=17.5$

And for $\exp (1)$ and (4), $t_{1/2}=\frac{\ln 2}{K^{\prime }}=\frac{\ln 2}{K[A_0]}$

$\therefore c=30\times 2=60$ and $d=3\times 60=180$