Answer
$a) -\frac{1}{2}\frac{[H_{2}]}{\Delta t}= -\frac{[O_{2}]}{\Delta t}=\frac{1}{2}\frac{[\Delta H_{2}O]}{\Delta t}$
$b)-\frac{1}{4}\frac{[NH_{3}]}{\Delta t}=-\frac{1}{5}\frac{[\Delta O_{2}]}{\Delta t}=\frac{1}{4}\frac{[\Delta NO]}{\Delta t}=\frac{1}{6}\frac{[\Delta H_{2}O]}{\Delta t}$
Work Step by Step
Given an equation aA → bB, use the relation:
$-\frac{1}{a}\frac{[\Delta A]}{\Delta t}=\frac{1}{b}\frac{[\Delta B]}{\Delta t}$
$a) -\frac{1}{2}\frac{[H_{2}]}{\Delta t}= -\frac{[O_{2}]}{\Delta t}=\frac{1}{2}\frac{[\Delta H_{2}O]}{\Delta t}$
$b)-\frac{1}{4}\frac{[NH_{3}]}{\Delta t}=-\frac{1}{5}\frac{[\Delta O_{2}]}{\Delta t}=\frac{1}{4}\frac{[\Delta NO]}{\Delta t}=\frac{1}{6}\frac{[\Delta H_{2}O]}{\Delta t}$