Question

The first-order rate constant for the decomposition of N₂O₅ to NO₂ and O₂ at 70°C is 6.82 × 10⁻³ s⁻¹. Suppose we start with 0.300 mol of N₂O₅(g) in a 0.500-L container. How many moles of N₂O₅ will remain after 1.5 min?

a. 0.067 mol
b. 0.086 mol
c. 0.112 mol
d. 0.145 mol

Answer 1

To determine the number of moles of N₂O₅ remaining after 1.5 minutes at 70°C given the first-order rate constant, we utilize the first-order rate law. After converting time to seconds and using the rate law formula, we find that approximately 0.145 mol of N₂O₅ remains, corresponding to answer choice d.

Step-by-step explanation:

The given problem involves a first-order decompositon reaction of N₂O₅ to NO₂ and O₂. The rate law for a first-order reaction can be expressed as ln([A]t/[A]0) = -kt, where [A]0 is the initial concentration, [A]t is the concentration at time t, k is the rate constant, and t is the time. Given that the initial concentration of N₂O₅ is 0.300 mol in a 0.500 L container, which is 0.600 M, and the rate constant at 70°C is 6.82 × 10⁻³ s⁻¹, we need to find the concentration after 1.5 minutes (90 seconds).

First, convert the rate constant to the appropriate time unit: Rate constant k = 6.82 × 10⁻³ s⁻¹. Converting 1.5 minutes to seconds gives us t = 90 seconds. Using the integrated rate law for a first-order reaction, we can solve for [A]t:

ln([A]t/0.600 M) = -(6.82 × 10⁻³ s⁻¹)(90 s)

After solving, we find the concentration of N₂O₅ remaining to be approximately 0.145 mol. Therefore, the correct answer is d. 0.145 mol.