Chemistry - Chemical Kinetics Question with Solution | TestHub

From the given graph Slope $=\frac{(\text{ rate }{)}^{1/2}}{[A]}=4$ $\because$ rate $(r)=k[A{]}^n$ Where, $k=$ rate constant $[A]=$ concentration at time $t$ (in min). $n=$ order. According to the graph $(n=2)$ $\therefore k=\frac{\text{ rate }}{[A{]}^n}=[4{]}^2=16$ Hence, $n=2$ and $$k=16.0{\mathrm{dm}}^3{\mathrm{mol}}^{-1}\stackrel{-1}{\min }$$ and option (d) is the correct answer.