Answer
$a) -\frac{[\Delta H]}{\Delta t}$ = $-\frac{[\Delta I_{2}]}{\Delta t}$ = $\frac{1}{2}$ $\frac{[\Delta HI]}{\Delta t}$
$b) -\frac{1}{5}$$\frac{[\Delta Br^{-}]}{\Delta t}$ = -$\frac{[\Delta BrO_{3}]}{\Delta t}$ = -$\frac{1}{6}$$\frac{[\Delta H^{+}]}{\Delta t}$ = $\frac{1}{3}$$\frac{[Br_{2}]}{\Delta t}$ = $\frac{1}{3}$$\frac{[H_{2}O}{\Delta t}$
Work Step by Step
Given an equation aA → bB, use the relation:
$\frac{1}{a}\frac{[\Delta A]}{\Delta t}=\frac{1}{b}\frac{[\Delta B]}{\Delta t}$
$a) -\frac{[\Delta H]}{\Delta t}$ = $-\frac{[\Delta I_{2}]}{\Delta t}$ = $\frac{1}{2}$ $\frac{[\Delta HI]}{\Delta t}$
$b) -\frac{1}{5}$$\frac{[\Delta Br^{-}]}{\Delta t}$ = -$\frac{[\Delta BrO_{3}]}{\Delta t}$ = -$\frac{1}{6}$$\frac{[\Delta H^{+}]}{\Delta t}$ = $\frac{1}{3}$$\frac{[Br_{2}]}{\Delta t}$ = $\frac{1}{3}$$\frac{[H_{2}O}{\Delta t}$
