At 25°C the decomposition of N2O5 (g) into NO2 (g) and O2 (g) follows first-order kinetics with k = 3.4*10-5 s-1. How long will it take for a sample originally containing 2.0 atm of N2O5 to reach a partial pressure of 380 torr?

6.1 h = 6 h and 8 min

Explanation:

First, let's found the rate of disappearing of N2O5. Knowing that it's a first-order reaction, it means that the rate law is:

rate = k*pN2O5

Where k is the rate constant, and pN2O5 is the initial pressure of N2O5 (2.0 atm), so:

rate = 3.4x10⁻⁵*2.0

rate = 6.8x10⁻⁵ atm/s

Thus, at each second, the partial pressure of the reagent decays 6.8x10⁻⁵ atm. The rate is also the variation of the pressure divided by the time. Because it is decreasing, we put a minus signal in the expression.

1 atm = 760 torr, so 380torr/760 = 0.5 atm

rate = - Δp/t

6.8x10⁻⁵ = - (0.5 - 2.0) / t

t = 1.5/6.8x10⁻⁵

t = 22,058 s (:60)

t = 368 min (:60)

t = 6.1 h = 6 h and 8 min