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Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of ·0.0038Ms−1: 2NH3(g)→N2(g)+3H2(g) Suppose a 450.mL flask is charged under these conditions with 150.mmol of ammonia. How much is left 20.s later? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.

Respuesta :

Answer:

[tex]1.2\times 10^{2}[/tex] mmol of [tex]NH_{3}[/tex] is left after 20 s.

Explanation:

Initial concentration of [tex]NH_{3}[/tex] = [tex]\frac{150\times 10^{-3}}{450}\times 10^{3}[/tex] M = 0.333 M

The integrated rate law for the given zero order reaction:

                             [tex][NH_{3}]=-kt+[NH_{3}]_{0}[/tex]

where, [tex][NH_{3}][/tex] represents concentration of [tex]NH_{3}[/tex] after "t" time, k is rate constant and [tex][NH_{3}]_{0}[/tex] is initial concentration of [tex]NH_{3}[/tex].

Here, k = 0.0038 M/s, [tex][NH_{3}]_{0}[/tex] = 0.333 M and t = 20 s

So, [tex][NH_{3}]=(-0.0038M.s^{-1}\times 20s)+0.333M[/tex]

   or, [tex][NH_{3}][/tex] = 0.257 M

So, number of mol of [tex]NH_{3}[/tex] left after 20 s = [tex]\frac{0.257}{1000}\times 450[/tex] mol = 0.11565 mol

So, number of mmol of [tex]NH_{3}[/tex] left after 20 s = 115.65 mmol = [tex]1.2\times 10^{2}[/tex] mmol (2 sig. digits)