Chemistry - Chemical Kinetics Question with Solution | TestHub

Rate constant = magnitude of slope. $k_{\mathrm{I}}=\tan 45^{\circ }=1$ M/min and $k_{\mathrm{II}}=\tan 60^{\circ }=\sqrt{3}$ M/min.

$(t_{100\%}{)}_{\mathrm{I}}=\frac{2}{1}=2$ min and $(t_{100\%}{)}_{\mathrm{II}}=\frac{1}{\sqrt{3}}$ min. Ratio $=2\sqrt{3}$: (A) true.

$[\mathrm{A}]=2-t$ and $[\mathrm{C}]=1-\sqrt{3}t$ (until it reaches 0). The difference is $1+(\sqrt{3}-1)t$, which is never zero, so the lines never meet. At 0.5 min, $[\mathrm{A}]=1.5$ M and $[\mathrm{C}]=0.134$ M: (B) false.

$[\mathrm{A}]=1$ M at $t=1$ min. C ran out at 0.577 min: (C) true.

With both at 1 M: $t_{50\%}=\frac{0.5}{k}$, so the ratio $\frac{(t_{50\%}{)}_{\mathrm{I}}}{(t_{50\%}{)}_{\mathrm{II}}}=\frac{k_{\mathrm{II}}}{k_{\mathrm{I}}}=\sqrt{3}$: (D) true.

Answer: (A), (C), (D).