Chemistry - Chemical Kinetics Question with Solution | TestHub

Order of Reactions: Linear concentration-time plots confirm both reactions are zero-order, where rate constants are $k_I=\tan (60^{\circ })=\sqrt{3}$ and $k_{II}=\tan (30^{\circ })=\frac{1}{\sqrt{3}}$.

Option A (True): $t_{100\%}=\frac{[R{]}_0}{k}⟹(t_{100\%}{)}_I=\frac{1}{\sqrt{3}}$ and $(t_{100\%}{)}_{II}=\frac{0.5}{1/\sqrt{3}}=\frac{\sqrt{3}}{2}$, satisfying $(t_{100\%}{)}_I=\frac{2}{3}(t_{100\%}{)}_{II}$.

Option B (True): Equating $[A]=[C]⟹1-\sqrt{3}t=0.5-\frac{t}{\sqrt{3}}$ solves directly to $t=\frac{\sqrt{3}}{4}min$.

Option C (False): When $[B]=[A]=0.5M$ at $t=\frac{1}{2\sqrt{3}}min$, $[C]=\frac{1}{3}M$ and $[D]=\frac{1}{6}M$, meaning $[C]\ne [D]$.

Option D (True): Since $t_{50\%}=\frac{[R{]}_0}{2k}$ with equal initial concentrations ($1M$), the half-life ratio is $\frac{(t_{50\%}{)}_I}{(t_{50\%}{)}_{II}}=\frac{k_{II}}{k_I}=\frac{1/\sqrt{3}}{\sqrt{3}}=\frac{1}{3}$.