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Kitty [74]
3 years ago
8

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).
Chemistry
1 answer:
Sonbull [250]3 years ago
6 0

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)  

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