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------------------------------------------------ The decomposition of [tex]$N_2O_5$[/tex] follows first-order kinetics. The reaction is:

[tex]$2N_2O_5 \rightarrow 4NO_2 + O_2$[/tex]

The rate constant is [tex]$3.38 \times 10^{-5} \text{ s}^{-1}$[/tex].

Calculate the following:

a) The half-life of [tex]$N_2O_5$[/tex].

b) If the initial pressure is 500 torr, what will the pressure be after 10 seconds?

c) What will the pressure be after 10 minutes?

d) What will the pressure be immediately after the initiation of the reaction?

a) The half-life of a first-order reaction can be calculated using the formula t1/2 = (0.693 / k), where k is the rate constant. In this case, k = 3.38 * 10⁻⁵ 1/sec. Therefore, t1/2 = (0.693 / 3.38 * 10⁻⁵) ≈ 20,522 seconds.

b) To calculate the pressure of N₂O₅ after 10 seconds, you can use the first-order rate equation P(t) = P₀ * e^(-kt), where P(t) is the pressure at time t, P₀ is the initial pressure, k is the rate constant, and t is the time. Plugging in the values, P(10) = 500 torr * e^(-3.38 * 10⁻⁵ * 10) ≈ 250.19 torr.

c) After 10 minutes, which is equivalent to 600 seconds, you can use the same equation P(600) = 500 torr * e^(-3.38 * 10⁻⁵ * 600) ≈ 2.47 torr.

d) After the initiation of the reaction, the pressure of N₂O₅ will continuously decrease as it follows first-order kinetics. The pressure will decrease exponentially with time until it reaches zero, but the exact pressure at any given time can be calculated using the rate equation mentioned in part (b). First-order kinetics and reaction rate equations.

Learn more about pressure of N₂O₅

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