Answer:
Explanation:
[tex]\mathbf{From \ the \ information \ given:} \\ \\ \mathbf{The \ rates \ of \ the \ f ollowing \ reactions \ can \ be \ expressed \ as \ follows:}[/tex]
(a)
[tex]\mathbf{3O_2 \to 2O_3} \\ \\ \\ \mathbf{-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{2}{3} \dfrac{d[O_3]}{dt}}[/tex]
(b)
[tex]\mathbf{C_2H_6 \to C_2H_4 + H_2} \\ \\ \\ \mathbf{ -\dfrac{d[C_2H_6]}{dt}= \dfrac{d[C_2H_4]}{dt}=\dfrac{d[H_2]}{dt}}[/tex]
(c)
[tex]\mathbf{ClO^-+Br^- \to BrO^-+Cl^-} \\ \\ \\ \mathbf{ -\dfrac{d[ClO^-]}{dt}= -\dfrac{d[Br^-]}{dt} = \dfrac{d[BrO^-]}{dt} = \dfrac{d[Cl^-]}{dt} }[/tex]
(d)
[tex]\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\ \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}[/tex]
(e)
[tex]\mathbf{2AsH_3 \to 2As + 3H_2} \\ \\ \\ \mathbf{-\dfrac{1}{2}\dfrac{d[AsH_3]}{dt}=\dfrac{1}{2}\dfrac{d[As]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}}[/tex]
