Question

The first-order rate constant for the decomposition of N₂O₅,

2N₂O₅(g) → 4NO₂(g) + O₂(g)
at 70∘C is 6.82×10⁻³ s⁻¹. Suppose we start with 2.10×10⁻² mol of N₂O₅(g) in a volume of 2.5 L .
How many moles of N₂O₅ will remain after 7.0 min ?
Express the amount in moles to two significant digits.

Answer 1

After 7 minutes, approximately 1.60×10⁻² moles of dinitrogen pentoxide will remain. This calculation uses the first-order decay equation with the given rate constant and the initial amount of N₂O₅.

Step-by-step explanation:

The question concerns the first-order decomposition reaction of dinitrogen pentoxide (N₂O₅) into nitrogen dioxide (NO₂) and oxygen (O₂) gases. Given a first-order rate constant (k) and initial conditions, we need to find the amount of N₂O₅ remaining after a certain time has elapsed.

First-order reactions follow the equation ln([A]t/[A]0) = -kt, where [A]0 is the initial concentration, [A]t is the concentration at time t, and k is the rate constant. Considering we start with 2.10×10⁻² moles of N₂O₅ in a 2.5 L container, the initial concentration ([A]0) is 8.40×10⁻³ M. After 7.0 minutes, or 420 seconds, we apply the given rate constant (6.82×10⁻³ s⁻¹) to find the remaining concentration. We can then calculate the remaining moles of N₂O₅ present in the container.

By solving the equation, we find that the remaining concentration of N₂O₅ ([A]t) is about 3.37×10⁻³ M, which equates to approximately 1.60×10⁻² moles remaining when rounded to two significant digits.