Answer:
[tex]\large \boxed{\text{(a) three times; (b) twice}}[/tex]
Explanation:
1N₂ + 3H₂ ⟶ 2NH₃
[tex]\text{rate} = -\dfrac{1}{1}\cfrac{\text{d[N$_{2}$]}}{\text{d}t}= -\dfrac{1}{3}\dfrac{\text{d[H$_{2}$]}}{\text{d}t} = +\dfrac{1}{2}\dfrac{\text{d[NH$_{3}$]}}{\text{d}t}[/tex]
(a) H₂ vs N₂
[tex]-\dfrac{1}{1}\dfrac{\text{d[N$_{2}$]}}{\text{d}t}= -\dfrac{1}{3}\dfrac{\text{d[H$_{2}$]}}{\text{d}t}\\\\\dfrac{\text{d[H$_{2}$]}}{\text{d}t} = 3\dfrac{\text{d[N$_{2}$]}}{\text{d}t}\\\\\text{The rate of consumption of hydrogen is $\large \boxed{\textbf{three times}}$}\\\text{the rate of consumption of nitrogen.}[/tex]
(b) NH₃ vs N₂
[tex]-\dfrac{1}{1}\dfrac{\text{d[N$_{2}$]}}{\text{d}t} = +\dfrac{1}{2}\dfrac{\text{d[NH$_{3}$]}}{\text{d}t}\\\\\dfrac{\text{d[NH$_{3}$]}}{\text{d}t} = -2\dfrac{\text{d[N$_{2}$]}}{\text{d}t}\\\text{The rate of formation of ammonia is $\large \boxed{\textbf{twice}}$}\\\text{the rate of consumption of nitrogen.}[/tex]
