Answer:
a. 113 min
Explanation:
Considering the equilibrium:-
2N₂O₅ ⇔ 4NO₂ + O₂
At t = 0 125 kPa
At t = teq 125 - 2x 4x x
Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x
125 - 3x = 176 kPa
x = 17 kPa
Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given that:
The rate constant, k = [tex]2.8\times 10^{-3}[/tex] min⁻¹
Initial concentration [tex][A_0][/tex] = 125 kPa
Final concentration [tex][A_t][/tex] = 91 kPa
Time = ?
Applying in the above equation, we get that:-
[tex]91=125e^{-2.8\times 10^{-3}\times t}[/tex]
[tex]125e^{-2.8\times \:10^{-3}t}=91[/tex]
[tex]-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)[/tex]
[tex]t=113\ min[/tex]
