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20 August, 09:25

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?

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  1. 20 August, 09:43
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    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
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