Chemistry - Chemical Kinetics Question with Solution | TestHub

$2.5+17.5+120+360=500$

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

Since $t_{7/8}=3\times t_{1/2}$, then $y=1$.

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

$t_{7/8}=7\times t_{1/2}$, then $x=2$.

Now for Exp. (2) & Exp. (3):

$t_{1/2}=\frac{1}{k[B_0]}=\frac{1}{k^{\prime }[B_0][A_0]}$

$a=\frac{10}{2\times 2}\Rightarrow 2.5$

And $b=7\times 2.5\Rightarrow 17.5$

For Exp. (1) and (4):

$t_{1/2}=\frac{\ln 2}{k^{\prime }}=\frac{\ln 2}{k[B_0{]}^2}$