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?

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

tallinn tallinn · Answer 2 · 2022-01-28T09:35:09+01:00

The correct answer is 6.1 h = 6 h and 8 min

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

Then rate is = k*pN2O5

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

Then rate = 3.4x10⁻⁵*2.0

After that rate = 6.8x10⁻⁵ atm/s

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

Then 1 atm = 760 torr, so 380torr/760 = 0.5 atm
After that rate = -Δp/t
Then 6.8x10⁻⁵ = -(0.5 - 2.0)/t
Then t = 1.5/6.8x10⁻⁵
Now t = 22,058 s (÷60)
Then t = 368 min (÷60)

Now answer is t = 6.1 h = 6 h and 8 min

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