Question

Ammonia is manufactured in large amounts by the reaction N21g2 + 3 H21g2 ¡ 2 NH31g2 (a) How is the rate of consumption of H2 related to the rate of consumption of N2? (b) How is the rate of formation of NH3 related to the rate of consumption of N2?

Answer 1

The rate of consumption of H2 is related to the rate of consumption of N2 in a 3:1 ratio. The rate of formation of NH3 is also related to the rate of consumption of N2 in a 3:1 ratio.

Step-by-step explanation:

The rate of consumption of H2 is related to the rate of consumption of N2 in a 3:1 ratio. This means that for every 3 moles of H2 consumed, 1 mole of N2 is consumed. This ratio is determined by the balanced equation of the reaction, which shows that 3 moles of H2 are required to react with 1 mole of N2 to produce 2 moles of NH3.

The rate of formation of NH3 is also related to the rate of consumption of N2 in a 3:1 ratio. For every 1 mole of N2 consumed, 2 moles of NH3 are formed. This ratio is determined by the stoichiometry of the balanced equation, which shows that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

Answer 2

`\large \boxed{\text{(a) three times; (b) twice}}`

Step-by-step explanation:

1N₂ + 3H₂ ⟶ 2NH₃

`\text{rate} = -(1)/(1)\cfrac{\text{d[N$_(2)$]}}{\text{d}t}= -(1)/(3)\frac{\text{d[H$_(2)$]}}{\text{d}t} = +(1)/(2)\frac{\text{d[NH$_(3)$]}}{\text{d}t}`

(a) H₂ vs N₂

`-(1)/(1)\frac{\text{d[N$_(2)$]}}{\text{d}t}= -(1)/(3)\frac{\text{d[H$_(2)$]}}{\text{d}t}\\\\\frac{\text{d[H$_(2)$]}}{\text{d}t} = 3\frac{\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.}`

(b) NH₃ vs N₂

`-(1)/(1)\frac{\text{d[N$_(2)$]}}{\text{d}t} = +(1)/(2)\frac{\text{d[NH$_(3)$]}}{\text{d}t}\\\\\frac{\text{d[NH$_(3)$]}}{\text{d}t} = -2\frac{\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.}`