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).

[tex]-\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}[/tex]

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

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

From the above equation we can write-

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

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

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

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