Question

The oxidation of ammonia produces nitrogen and water via the following reaction: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) Suppose the rate of formation of H2O(l) is 3.0 mol/(L ∙ s). Which of the following statements is true? a. The rate of consumption of NH3 is 0.50 mol/(L ∙ s). b. The rate of consumption of NH3 is 2.0 mol/(L ∙ s). c. The rate of formation of N2 is 2.0 mol/(L ∙ s). d. The rate of formation of N2 is 1.3 mol/(L ∙ s). e. The rate of consumption of O2 is 2.0 mol/(L ∙ s).

Answer 1

Answer:

The rate of consumption of $NH_{3}$ is 2.0 mol/L.s

Explanation:

Applying law of mass action to this reaction-

$-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

where $-\frac{\Delta [NH_{3}]}{\Delta t}$ represents rate of consumption of $NH_{3}$, $-\frac{\Delta [O_{2}]}{\Delta t}$ represents rate of consumption of $O_{2}$, $\frac{\Delta [N_{2}]}{\Delta t}$ represents rate of formation of $N_{2}$ and $\frac{\Delta [H_{2}O]}{\Delta t}$ represents rate of formation of $H_{2}O$.

Here rate of formation of $H_{2}O$ is 3.0 mol/(L.s)

From the above equation we can write-

$-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

Here $\frac{\Delta [H_{2}O]}{\Delta t}=3.0 mol/(L.s))$

So, $-\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

Hence, $-\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\times 3.0 mol/(L.s)=2.0 mol/(L.s)$