The first-order reaction,2 N₂O(g)→ 2 N₂(g)+ O₂(g),has a rate constant equal to 0.76 s^-1 at 1000 K.How long will it take for the concentration of N₂O to decrease to 42% of its initial concentration?

Question

Quizplus · Accepted Answer

The integrated rate law for a first-order reaction is:

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

where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, k is the rate constant, and t is the time.

We are given the rate constant k = 0.76 s^-1 and we want to find the time it takes for [A]t/[A]0 = 0.42.

Substituting the given values into the integrated rate law, we get:

ln(0.42) = -0.76t

Solving for t, we get:

t = ln(0.42)/(-0.76)

t ≈ 1.1 s

Therefore, the answer is choice B.

The First-Order Reaction,2 N2O(g)→ 2 N2(g)+ O2(g),has a Rate Constant

The First-Order Reaction,2 N₂O(g)→ 2 N₂(g)+ O₂(g),has a Rate Constant