Answer:
5.9 s
Explanation:
Let's consider the following reaction.
2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)
The rate law is:
rate = 2.11 M⁻¹s⁻¹[NH₃]²
where
2.11 M⁻¹s⁻¹ is the rate constant (k)
2 is the order of reaction for NH₃
The initial concentration of NH₃ ([NH₃]₀) is 0.590 M. If it decreases by 88.0%, the final concentration ([NH₃]) will be 100.0% - 88.0% = 12.0% of the initial concentration, that is, 0.0708 M.
For a second order reaction we can use the following expression.
[tex]\frac{1}{[NH_{3}]} =\frac{1}{[NH_{3}]_{0}} + k.t\\\frac{1}{0.0708M} =\frac{1}{0.590M} +2.11M^{-1} s^{-1} .t\\t=5.89s[/tex]
