Question

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?

Answer 1

To determine the time it takes for a sample of N2O5 to reach a certain partial pressure, we can use the first-order rate equation and solve for time.

Step-by-step explanation:

The decomposition of N2O5(g) into NO2(g) and O2(g) at 25°C follows first-order kinetics. The rate constant, k, is given as 3.4×10−5 s−1.

To determine how long it will take for a sample initially containing 2.0 atm of N2O5 to reach a partial pressure of 380 torr, we can use the first-order rate equation:

ln([N2O5]t/[N2O5]0) = -kt

Rearranging the equation and substituting the given values, we get:

ln(380/760) = -(3.4×10−5 t)

Solving for t, we find that it will take approximately 2,370 seconds or about 39.5 minutes.

Answer 2

6.1 h = 6 h and 8 min

Step-by-step 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